Changing Data inside XML element

Alan Kilborn

I think I misspoke before with my “end Exclusion Region” phraseology.
I really meant (I think) to not use the word “exclusion”.
Sure, that’s what the functionality in the overall regex is, but from a user instruction perspective, it confuses. User is better to think in terms of start and end rather than start and exclusion.

How about:

BSR (Begin Search-region Regex)
ESR (End Search-region Regex)
FR (Find Regex)
RR (Replace Regex)

Of course, in the end, totally up to you.

astrosofista

@Alan-Kilborn said in Changing Data inside XML element:

I think the regexes shown above are victims of this site not being able to slow backslash followed by [ correctly!

Not in my post, I used the \Q \E escape sequence to avoid that.

Cheers

Alan Kilborn

@guy038

Also, can we encapsulate the (?s) and (?-s) difference somehow, so you don’t have to publish TWO regexes that differ only in that leading element?

Actually, shouldn’t it be up to the user to use (?s) or (?-s) as appropriate in their SR, ER, etc expressions?

You have one use of . in the generic expression, perhaps it should have its own (?s) or (?-s) specifier that only relates to it?

I suppose this gets even more complicated, but I’ll leave it to a “big regex thinker” like you. :-)

@astrosofista said :

Not in my post, I used the \Q \E escape sequence to avoid that.

Right, I was of course meaning the others.
But, it is probably bad to modify the regex you want to write, just so that it meets some criterion of the site it is posted on!

guy038

Hello, @matthew-allshorn, @alan-kilborn and All,

Back to my @matthew-allshorn’s search regex :

SEARCH (?-i)(<!\\[CDATA\\[|(?!\A)\G)[^]\r\n]*?\K/

I could have slightly modified it, as Regex_A, below which, in turn, could also be expressed, according to the generic syntax, as Regex_B

Regex_A :   (?x-i)  (?:  <!\\[CDATA\\[  |  (?!\A)\G)      [^]]      *?  \K  /

Regex_B :   (?x-i)  (?:  <!\\[CDATA\\[  |  (?!\A)\G)  (?: (?!]).)   *?  \K  /

Now, Alan, I fully endorse your new abbreviations ! So :

Let FR (Find Regex ) be the regex which defines the char, string or expression to be searched
Let RR (Replacement Regex ) be the regex which defines the char, string or expression which must replace the FR expression
Let BSR ( Begin Search-region Regex ) be the regex which defines the beginning of the area where the search for FR, must start
Let ESR ( End Search-region Regex) be the regex which defines, implicitly, the area where the search for FR, must end

Then, the generic regex can be expressed :

SEARCH (?-i:BSR|(?!\A)\G)(?s:(?!ESR).)*?\K(?-i:FR)

REPLACE RR

Important :

You must use, at least, the v7.9.1 N++ release, so that the \A assertion is correctly handled
You must, move the caret at the very beginning of current file ( Ctrl + Home ), if you plane a Replace All operation
If all instances of FR, to match, are located in the same line as BSR, use (?-s) ( instead of (?s) ), in the appropriate non-capturing group
If the BSR expression to match is independent from case, use (?i) ( instead of (?-i) ), in the appropriate non-capturing group
If the FR expression to match is independent from case, use (?i) ( instead of (?-i) ), in the appropriate non-capturing group.
Depending of the FR contents, when case is not concerned, you may, either :
- Omit its non-capturing group and, if necessary, use the $0 syntax, in the replacement regex RR
- Define a group for some part of FR, which will be re-used in the replacement regex RR

Alan, do you like it that way ?

Best Regards,

guy038

Alan Kilborn

@guy038 said in Changing Data inside XML element:

Alan, do you like it that way ?

I do have some suggestions; will post back after I have some more time to consider it.

Alan Kilborn

@guy038 said in Changing Data inside XML element:

(?-i:BSR|(?!\A)\G)(?s:(?!ESR).)*?\K(?-i:FR)

So I experimented a bit, and I found that it could be judged a bit “fragile”, from its higher-level intent. Example:

overall regex:

(?x)    (?-i:    (?s:B.*?S.*?R)    |(?!\A)\G)    (?s:(?!E.*?S.*?R).)    *?\K    (?s-i:F.*?R)

data set 1:

B
S
R

FR

F
R

E
S
R

FR

result 1 (intended):

data set 2:

B
S
R

FR

F


E
S
R

FR

result 2 (unintended):

I certainly understand the reason for failure of result 2.

guy038

@Alan-kilborn and All,

Alan, you’re cheating a bit ! Let’s me explain :

I’ll use this sample text :

B
S
R
FR

F
R

F

F


R


R

F

FR
E
S
R
FR

F
R

F

F


R


R

F

FR

FR

Your FR regex is (?s-i:F.*?R) which means : “Search for the shortest range, possibly empty, of any char, EOL included, between a F letter and a R letter, with that case”

So, using only the regex (?s-i:F.*?R), it matches, in the BSR •••••••• ESR area :

The string FR, line 9
The string FCRLFR, lines 11 and 12
The string FCRLFCRLFFCRLFCRLFCRLFR, between the lines 14 and 19
The string FCRLFCRLFFR, between the lines 24 and 26

Now, let’s consider this other FR regex (?s-i:F\R*R). This regex searches for a F, with that case, followed with a range of consecutive EOL chars, possibly empty, till a R letter, with that case. I think this definition is close to what we expect to :

Using this new version (?s-i:F\R*R), it matches, in the BSR •••••••• ESR area :

The string FR, line 9
The string FCRLFR, lines 11 and 12
The string FCRLFCRLFCRLFR, between the lines 16 and 19
The string FR, line 26

So, if we use this overall regex ( only the FR part have been changed ) :

(?x)    (?-i:    (?s:B.*?S.*?R)    |  (?!\A)\G)    (?s:(?!E.*?S.*?R).)    *?\K    (?s-i:F\R*R)

it correctly matches and marks :

The string FR, line 9, right after the BSR
The string FCRLFR, lines 11 and 12
The string FCRLFCRLFCRLFR, between the lines 16 and 19
The string FR, line 26, right before the ESR
And every other match of the simple regex (?s-i:F\R*R), located after the ESR, are discarded, as expected ;-))

Refer the picture, below, where I’m using the regex in line 3 :

In summary, this generic regex seems quite robust !!

Best Regards,

guy038

Alan Kilborn

So the part that bothers me (slightly) is that as a user, I want to think of specifying the search region and the find regexes as independent entities.
Maybe another way of saying it is that I’d like the search region to have precedence over the find.
In my example 2 this doesn’t happen, and maybe there is no way to have it happen.

With this, I run the risk of writing a bad replacement, that could be data dependent.
Meaning that I could craft something that works well for a few cases at the top of a file where I’m testing it, but perhaps not all cases farther down in the file (or worse, multiple files).
So I do my “few cases” analysis, deem it working, and then tell it to blast away at the rest of the file(s), and find out some time later that I’ve corrupted my data with errant replacements.

Normally, in such a risky situation – if I know/suspect it to be so – I could tell myself “Don’t use this technique in a Replace All situation, just do a Replace (with implied Find Next) so that you can verify each replacement.” But… this technique uses \K and individual-replace doesn’t work (certainly not your fault @guy038 – are the devs ever going to make it work?).

Perhaps there are just always going to be such risks with certain techniques.

guy038

Hi, @alan-kilborn and All,

I tried to study the \K behavior, in case of a simple replacement with the Replace button and, unfortunately, I cannot deduce a general rule for that assertion :-((

Open a new tab and paste the simple expression abgcd ef ghi j
Select one line of the list , below, containing a regex
Open the Replace dialog( Ctrl + H )
For all the examples, the Replace with field contains only the @ character
Click once on the Find Next button
Then, click several times on the Replace button … till no replacement occurs

=>

Results, after :

    - A "CTRL + HOME" operation

    - SELECTION of ONE regex or LINE, below

	- A "CTRL + H" operation

    - ONE click on the "FIND NEXT" button

    - SEVERAL clicks on the "REPLACE" button  ( NOT the "REPLACE ALL" button ! )

against the string  "abgcd ef ghi j", PASTED in a NEW tab :


(?x-s).+?\K\x20           # KO  ( 3 SPACE chars NOT changed )
(?x-s).*?\K\x20           # OK  ( 3 SPACE chars CHANGED into @ )

(?x-is)(ab.*?\Kg|.*?\Kp)  # KO  ( TWO "g" NOT changed )
(?x-is)(ab.*?\Kg|.*?\Kh)  # KO  ( TWO "g" NOT changed, "h" CHANGED into @ only  )
(?x-is)(ab.*?\Kg|.*?\Kg)  # OK  ( TWO "g" CHANGED into @ )

(?x-is)(.*?\Kg|.*?\Kp)    # OK  ( TWO "g" CHANGED  into @ )
(?x-is)(.*?\Kg|.*?\Kh)    # OK  ( TWO "g" and "h" CHANGED into @  )
(?x-is)(.*?\Kg|.*?\Kg)    # OK  ( Two "g" CHANGED into @ )

(?x-is)a.*?\Kg            # KO  ( TWO "g" NOT changed )
(?x-is)\l.*?\Kg           # KO  ( TWO "g" NOT changed )

(?x-is).*?\Kg             # OK  ( TWO "g" CHANGED into @ )
(?x-is)(ab|)*?\Kg         # OK  ( TWO "g" CHANGED into @ )
(?x-is)(ab|\G)*?\Kg       # OK  ( TWO "g" CHANGED into @ )

Although the \K behavior is rather difficult to interpret, it seems, however, from the last example (?x-is)(ab|\G)*?\Kg, that the generic regex is compatible with several “step by step” replacements, using the Replace button only. So :

Paste the text, below, in a new tab :

F

R

FR
B
S
R
FR

F
R

F

F


R


R

F

FR
E
S
R
FR

F
R

F

F


R


R

F

FR

FR

Move back to the very beginning ( Ctrl + Home )
Open the Replace dialog ( Ctrl + H )

SEARCH    (?x)  (?: (?s-i:B.*?S.*?R)  |  (?!\A)\G )  (?s-i:(?!E.*?S.*?R).)*?  \K  (?s-i:F\R*R)   #  With my FR version

REPLACE   @

Click once on the Find Next button ( it should select the first FR, after “BSR” )
Then, click four times on the Replace button

=> One at a time, the selected matched zone is replaced with the @ sign ;-))

and you should get that picture :

Now, Alan, it quite possible to avoid the \K assertion, with that syntax :

SEARCH    (?x)  (  (?: (?s-i:B.*?S.*?R)  |  (?!\A)\G )  (?s-i:(?!E.*?S.*?R).)*?  )  (?s-i:F\R*R)  #  With my FR version

REPLACE   \1@

Remark that the replacement is \1@ ( instead of @ ). After running this S/R, in the same conditions as above, we get similar results ;-)

However, note that, if we just search for the FR or try to mark the FR zones, that second syntax, without \K, is quite annoying and not really exact, as it does select FR, but it also selects :

The BSR part and anything till FR
Anything since last match till FR

Thirdly, Alan, regarding the precedence of the region to search for, over the effective FR match :

As your FR regex (?s-i:F.*?R) contains a range of any char, EOL included, you need to restrict this area to an area not containing the ESR, too ! Thus, this fact changes your regex to :

SEARCH    (?x)  (?: (?s-i:B.*?S.*?R)  |  (?!\A)\G )  (?s-i:(?!E.*?S.*?R).)*?  \K  (?s-i:F((?!E.*?S.*?R).)*?R)   #  With your FR version
                                                                                  <--- Find Regex---- FR --->
REPLACE   @

Of course, your regex does not match exactly like my FR regex ! But the matches are correct and correctly limited in the BSR •••• ESR area. Moreover, the step by step replacement is still effective and give the picture below :

Best Regards,

guy038

P.S. :

An other example :

Let’s consider this text :

/

/

The licenses/for most {software are/designed to/take away your} freedom/to share/and change it.

{By contrast, the/GNU General Public/License is } intended to/guarantee/your {freedom/to share and/change} free/software

This General/Public License/applies to {most of/the Free Software/Foundation's software}

{/}

/
/

Paste it in a new tab
Move back to the beginning ( Ctrl + Home )

Now, let’s suppose that we want to change any / character into a @ character, but ONLY in areas delimited by curly braces { •••• }. Note that, in our sample, the second line contains two { ••••• } zones. So :

FR = /
RR = @
BSR = \{
ESR = \}

And gives this overall S/R :

SEARCH    (?x-si) (?:  \{  |  (?!\A)\G  )  (?: (?!\}). )*?  \K  /

REPLACE   @

Remark that the part (?:(?!ESR).)*? , implicitly, defines an area which does not contain any } character till the next / char to search for. In other words, it tries to find a / before a possible further } character !

=> After 1 click on the Find Next button and 9 clicks on the Replace button, you’ll see that all occurrences of /, which are included in areas between curly braces ONLY, have been replaced, as expected, with the @ symbol ;-))

Alan Kilborn

@guy038

So I was intrigued by your \K results, but let me confine my response to just this small part of your posting:

(?x-s).+?\K\x20           # KO  ( 3 SPACE chars NOT changed )
(?x-s).*?\K\x20           # OK  ( 3 SPACE chars CHANGED into @ )

abgcd ef ghi j
          1111  <- tens, and
01234567890123  <- ones position in doc

But rather than looking at it as a user and regex guru as you did, I cheated – because I’m not the same guru – and looked at the N++ source code.

So, what Notepad++ does, when you click on the Replace button is, it runs its Find Next code search (transparently) and sets a variable called “nextFind”. This search starts at the minimum position of the current selection (or just the caret position if no selection is active) and proceeds toward higher positions in the doc.

If what comes out of that search is a match of the same selected text position range as you began with, then the replacement is made.

If what results from the Find Next search is some different text selected (or you had nothing selected originally), the replace is not done, but the new text selection is now the nearest match higher in the doc (presumably convenient for your next press of Replace!).

On actual replacement, it then moves the selection to the following match it finds (meaning that internally it runs another Find Next). For the case where the replacement was made, the following search starts at the very righthand side of the newly inserted text.
That’s kind of a wordy explanation for this part, but hopefully it makes some sense.

Okay, so what does this all this mean for regexes using \K, specifically yours above where the first one doesn’t work and the second one does ?

analysis of “first” regex:

(?x-s).+?\K\x20

move caret to start-of-file, data as stated above: abgcd ef ghi j

first press of Replace button:
active selection range at time of button press: (0,0)
code runs at Replace press:
calculates “nextFind” selection range = (5,6)
no (selection range) equivalency to (0,0) so NOTHING IS REPLACED
selection moves to (5,6) which is the space between d and e

second press of Replace button:
active selection range at time of button press: (5,6)
code runs at Replace press:
calculates “nextFind” selection range = (8,9)
no (selection range) equivalency to (5,6) so NOTHING IS REPLACED
selection moves to (8,9) which is the space between f and g

third press of Replace button:
active selection range at time of button press: (8,9)
code runs at Replace press:
calculates “nextFind” selection range = (12,13)
no (selection range) equivalency to (8,9) so NOTHING IS REPLACED
selection moves to (12,13) which is the space between i and j

etc.

analysis of “second” regex:

(?x-s).*?\K\x20

move caret to start-of-file, data as stated above: abgcd ef ghi j

first press of Replace button:
active selection range at time of button press: (0,0)
code runs at Replace press:
calculates “nextFind” selection range = (5,6)
no (selection range) equivalency to (0,0) so NOTHING IS REPLACED
selection moves to (5,6) (the space between d and e)

second press of Replace button:
active selection range at time of button press: (5,6)
code runs at Replace press:
calculates “nextFind” selection range = (5,6)
have (selection range) equivalency to (5,6) so REPLACEMENT IS MADE
selection moves to (8,9) which is the next match between the f and g

third press of Replace button:
active selection range at time of button press: (8,9)
code runs at Replace press:
calculates “nextFind” selection range = (8,9)
have (selection range) equivalency to (8,9) so REPLACEMENT IS MADE
selection moves to (12,13), which is the next match between the i and j

etc.

Here’s the conclusion I draw from this:

If you use \K in a regex, if the part of the regex to the left of the \K matches as zero-length, the Replace button press WILL work to replace data. If, however, the part of the regex to the left of \K matches one or more characters, a press of Replace will NOT perform a textual substitution, but will rather just move to the next higher match.

Following this rule: Because the “first” regex demands that a minimum of one character be matched to the left of \K, no Replace ments are made. Because the “second” regex has a minimum-of-zero requirement, when it does match zero, that match comes to the left of \K, so a Replace is allowed to actually replace.

Probably a zero-length match to the left of \K in a regex isn’t very useful often; perhaps that is why I don’t recall hearing of \K and single-step replace “working only sometimes” in the past? Typically, it is, “doesn’t work”.

guy038

Hello, @alan-kilborn and All,

You said :

If you use \K in a regex, if the part of the regex to the left of the \K matches as zero-length, the Replace button press WILL work to replace data. If, however, the part of the regex to the left of \K matches one or more characters, a press of Replace will NOT perform a textual substitution, but will rather just move to the next higher match.

Not totally exact, Alan !

For instance, let’s imagine this text, in a new tab, beginning with two blank lines :



78464 13232178913 894654465464 12231

52632abc9526271 026238121 945135 s1658

6479123789 456134 978941 13454

46464646l 4567861341 128978 10313

111386460abc9564 6240 17868913345100544 4867864

Note that the lines 5 and 11, only, contains the string abc

And let’s use a simple regex, derived from our generic regex, presently studied in the previous posts :

(?-s)(abc|(?!\A)\G).*?\K\x20

Now, in this new tab, containing the sample text :

Move back to the beginning ( Ctrl + Home )
Open the Replace dialog Ctrl + H )
- SEARCH (?-s)(abc|(?!\A)\G).*?\K\x20
- REPLACE @
Click once on the Replace button ( identical to a click on the Find Next button ! )

=> The first space character, of line 5, is matched. As the caret was initially on a blank line, it cannot match any standard character. So the part (?-s)(?!\A)\G.*?\K\x20, with the second alternative, has not been used by the regex engine

Thus, the regex engine uses, necessarily, the first alternative ( the regex (?-s)abc.*?\K\x20 ), which indeed, contains a part, before \K, which is not a zero-length area, as equal to (?-s)abc.*?

However, if you click a second time on the Replace button, the first space char of line 5 is, as expected, changed into the @ symbol !

Of course, any further space char, located in lines containing the abc string, are replaced, in the same way, after successive clicks on the Replace button

You also said :

Probably a zero-length match to the left of \K in a regex isn’t very useful often;

But it just so happens that our generic regex S/R :

SEARCH (?s-i:BSR|(?!\A)\G)(?s-i:(?!ESR).)*?\K(?s-i:FR)

REPLACE RR

seems to work correctly with “step-by-step” replacement ;-))

Note that the search syntax above corresponds to the particular case where :

The search is sensible to the case of letters
The search may extend to a multi-lines area
No subset of FR, is needed in the replacement regex => All groups are non-capturing ones

Best Regards,

guy038

Alan Kilborn

@guy038 said in Changing Data inside XML element:

Not totally exact, Alan !

But it just so happens that our generic regex S/R…

Ok, then, back to the drawing board.
More analysis work on this for me!
I’ll return with the answers.
Well…if I find them.

Alan Kilborn

@guy038

So this is all very confusing until one breaks it down! :-)
And even then it is still confusing. :-(

Following your steps:

Now, in this new tab, containing the sample text :
Move back to the beginning ( Ctrl + Home )
Open the Replace dialog Ctrl + H )
SEARCH (?-s)(abc|(?!\A)\G).*?\K\x20
REPLACE @
Click once on the Replace button ( identical to a click on the Find Next button ! )

Then, without doing anything else (without pressing any more buttons), you say:

Thus, the regex engine uses, necessarily, the first alternative ( the regex (?-s)abc.?\K\x20 ), which indeed, contains a part, before \K, which is not a zero-length area, as equal to (?-s)abc.?

Which I wholly agree with.

But… the point where I talked (in my previous post) about a zero-length string to the left of \K has not come into play (yet)!

We are left sitting and looking at:

Note the single space selected in line 5.

At this point Replace is pressed a second time.
The internal find is again run and the match that occurs is going to effectively be this part of the regex:

(?!\A)\G.*?\K\x20

and that IS going to be zero-length to the left of the \K !

The key point might be that \G will match because it by definition “matches at the start of string-to-search at the first attempt”. And because this search is indeed a new search, we are AT the first attempt. The start of the string-to-search is going to be the left side of the selected space character.

So in summary, to the left of the \K for the match we have:

(?!\A) -> match of zero-length
\G -> match of zero-length (from “key point” just above)
.*? -> match of zero-length (because \x20 will match next and since we are a minimal match, we match zero)

Add up all those zeroes to obtain: 0+0+0=0

And thus my postulate from the previous posting seems to hold: If the match to the left of \K is a zero-length match, the replacement WILL be made. So in this case the @ replaces the space.

Any time the Replace button is pressed, it only replaces if the current selection matches the find-expression – do we agree on this? Regex or not…right?

Because \K “cancels out” what comes before it in a regex (and it must cancel it very deeply in the regex engine – because (?<...) doesn’t have the same difficulty \K does), the only way a current selection is going to match an expression using \K is if what is to the left of the \K has no length to it.

I’d be happy to have you poke holes in my logic here! :-)

guy038

Hello, @alan-kilborn and All,

Alan, I did numerous tests and, I’m afraid, that all that story does not depend on the \G assertion, anyway but only on the \K one !! Yes, really not easy :-((

So here is, below, how I imagine the process, after clicking, either, on the Find Next or the Replace of the Replace dialog, whatever the search mode. Note that we will not speak about the Replace All behavior, at all !

After any click on the Find Next button, the regex engine starts searching for a match of the search regex :

From the end of a present normal selection
From the present position of the caret, if NO selection exists

Then :

IF NO match exists, the overall process stops
IF a match has been found, it is selected

After any click on the Replace button, the regex engine starts searching for a match of the search regex :

From the start of a present normal selection
From the present position of the caret, if NO selection exists

Then :

IF NO match exists, the overall process stops
IF a match has been found :
- IF the match is strictly identical to the previous selection :
  - The selection/match is replaced with the replacement regex
  - The current position is reset to the location, right after the end of the replacement regex
  - And the regex engine re-starts searching, further on, for a next match of the search regex :
    - IF such a match can be found, it is selected
    - IF NO match exists, the overall process stops
- IF the match is different from the previous selection OR IF no previous selection exists :
  - This match is just selected, without any replacement process

So, the complete solution, in order that the step by step replacement works correctly, is that the current search, in any mode, matches the previous selection, whatever the way that selection was obtained :

By a previous click on the Replace button ( standard case of subsequent replacements )
By a previous click on the Find Next button ( standard case of a first replacement )
By a manual selection or move of the caret, generated intentionally, by the user !

Thus :

In case of a search, in the Normal or Extended mode OR if no \K assertion is used, in Regex mode :

=> The previous selection, generated by a click on the Find Next or Replace button AND the present match, met at the start of the selection, are always identical, meaning that the Step by Step replacement feature is always functional !

In the case of the particular use of the \K assertion, in Regex mode, the present selection, before clicking on the Replace button, is the part of the regex, located after \K structure. So the rule is :

=> A simple replacement can occur ONLY IF the overall regex can match the part of the regex located after the \K syntax !

In other words, if we consider the general regex form ........\K••••••••, this means that :

The overall regex ........\K•••••••• must match, exactly, the same expression than the •••••••• sub-regex

For instance :

(A) The regex (?-s).*?\Kabc does match the sub-regex abc, after \K => The step-by-step replacement is possible
(B) Similarly, the regex (?-s).*?\Kabc\d does match the sub-regex abc\d, after \K => The step-by-step replacement is OK
(C) And the regex (?-s).*?\K\dabc does match the sub-regex \dabc, after \K => The step-by-step replacement will work

But, on the contrary :

(D) The regex (?-s)\d.*?\Kabc does not match the sub-regex abc, after \K => The step-by-step replacement is not possible
(E) Similarly, the regex (?-s)\d.*?\K\dabc does not match the sub-regex \dabc, after \K => The step-by-step replacement is KO
(F) Now, the regex (?-s)\d.*?\K\d?abc never matches the sub-regex \d?abc, after \K, too => The step-by-step replacement will not work. However, by clicking successively on the Replace button, it produces :
- Firstly, a selection of the \dabc string, if that string is preceded, itself, with a digit
- Secondly, a selection of the abc string
(G) Note that the use of the regex (?-s)\d?.*?\K\d?abc, is more difficult to understand, although logical : by clicking successively on the Replace button, it produces, either :
- A selection of abc, if the previous selection was \dabc
- A replacement by the @ symbol, if the previous selection was abc
(H) To end with, remark that, making the first \d? expression lazy, the resulting regex (?-s)\d??.*?\K\d?abc does match the sub-regex \d?abc, after \K. So :
- The step-by-step replacement works nice
- Any click on the Replace button, from the second, does change the matched string \dabc into a @ symbol

Test all these regexes, above, against the text 123abc456abc0abc789abc0, pasted in a new tab
The Replace regex is just @, in all cases
Select the Regular expression search mode
Click only on the Replace or Find Next button ( Do not use the Replace All button )

Alan, the crucial point is to move the caret or create a normal selection, by yourself, before one or some click(s) on the Replace button, and observe the results with the hope of deducing a logical behavior, as I did ! This will allow you to confirm or deny my assertions ;-))

Best Regards,

guy038

guy038

Hi, All,

See my results of testing the @sasumner’s build, about the \K assertion and the step by step replacement :

https://github.com/notepad-plus-plus/notepad-plus-plus/issues/8434#issuecomment-784583153

See, also, an interesting helping post, of @ArkadiuszMichalski, regarding how to test new N++ builds !

https://github.com/notepad-plus-plus/notepad-plus-plus/issues/9493#issuecomment-781344297

Best Regards,

guy038

Michael Vincent

@guy038 said in Changing Data inside XML element:

See, also, an interesting helping post, of @ArkadiuszMichalski, regarding how to test new N++ builds !
https://github.com/notepad-plus-plus/notepad-plus-plus/issues/9493#issuecomment-781344297

Seems to be official now:

https://github.com/notepad-plus-plus/notepad-plus-plus/wiki/Testing

Good job @ArkadiuszMichalski !

Cheers.