Find & Replace issues

guy038

For newby people, about regular expressions concept and syntax, begin with that article, in N++ Wiki :

http://docs.notepad-plus-plus.org/index.php/Regular_Expressions

In addition, you’ll find good documentation, about the new Boost C++ Regex library, v1.55.0 ( similar to the PERL Regular Common Expressions, v1.48.0 ), used by Notepad++, since its 6.0 version, at the TWO addresses below :

http://www.boost.org/doc/libs/1_48_0/libs/regex/doc/html/boost_regex/syntax/perl_syntax.html

http://www.boost.org/doc/libs/1_48_0/libs/regex/doc/html/boost_regex/format/boost_format_syntax.html

The FIRST link explains the syntax, of regular expressions, in the SEARCH part
The SECOND link explains the syntax, of regular expressions, in the REPLACEMENT part

You may, also, look for valuable informations, on the sites, below :

http://www.regular-expressions.info

http://www.rexegg.com

http://perldoc.perl.org/perlre.html

Be aware that, as any documentation, it may contain some errors ! Anyway, if you detected one, that’s good news : you’re improving ;-))

Cheers,

guy038

Kirk Weir

Hi @guy038,

Excellent, thanks a lot for the info. You are most kind!

Regards,
Kirk

AdrianHHH

Another way of inserting (missing) leading zeros for numbers. This takes two search-and-replace operations. The first step is to insert the wanted number of leading zeros at the front of every number. The second step is to remove any unneeded zeros.

For the example, where numbers matching C1| are to be changed to P001|, i.e. adding two zeros.
First step: Replace (C)(\d{1,2}\|) with \100\2. (As Notepad++ only allows nine groups there is no ambiguity with the \100 part, it means \1 then 00.)
Second step: Replace (C)0+(\d{3}\|) with \1\2.

Scott Sumner

@AdrianHHH

As Notepad++ only allows nine groups…

I didn’t check your proposed solution, but rather I just wanted to point out that Notepad++ can handle more than 9 captured groups. For example, there’s a regex replacement in this thread that uses 52 capture groups!

guy038

Hello, @adrianhhh, @kirk-weir, @scott-sumner and All,

As soon as I saw your post, and understood your “philosophy” to add missing leading 0’s, my previous regex S/R, below, looks excessively complicated !!

SEARCH ((\d)?(\d)?(\d)?(\d)?(\d)?(\d)?(\d)?\d)

REPLACE (?2:0)(?3:0)(?4:0)(?5:0)(?6:0)(?7:0)(?8:0)\1

Indeed, your method looks better and more simple. In addition, I succeeded to simplify your two regex S/R :-))

So, let’s start with the original text, below, with some numbers, preceded by the letter C :

I omitted the last | character, which is useless, for our discussion. Now, as we’re searching for a formatted list of eight digits numbers, we need to insert a seven 0’s string, right after the letter C. To do so, I use the following S/R :

SEARCH (?<=C)

REPLACE 0000000

And I get the text :

C000000012345
C0000000123456
C0000000123
C00000001234567
C00000001234
C000000012
C000000012345678
C00000001

Notes :

As the search is only a look-behind construction, it matches the zero-length position, right after the letter C
And, at that position, it simply inserts, in replacement, the 0000000 string !!

Now, to get the aligned table of numbers, padded out with some 0’s, I chose the following S/R, which suppresses the unnecessary 0, rather than rewriting the letter C and the different numbers to keep !

SEARCH (?-s)(?<=C).*(?=\d{8})

REPLACE Leave EMPTY

We obtain, at once, the correct list below :

Magic, isn’t it !

Notes :

As usual, the modifier (?-s), ensures that the special dot character will match standard characters, only !
The search regex looks for any amount, even empty, of standard characters ( .* ), if two conditions are true :
- This range of characters must be preceded by the C letter ( (?<=C) )
- This range of characters must be followed by an eight digits number ( (?=\d{8}) )
As the replacement part is EMPTY, this range is just deleted

To sump up, in order to obtain an aligned list of n digits numbers, padded out with a particular character :

Choose the fix string, located right before the padded characters to insert. Note that the first replacement zone could have contained n spaces or n dots or any other padded character !
Repeat the look-behind in the second search zone and use the (?=\d{n}) look-ahead

As you see, Adrian, it’s a good lesson ! Very often, two simple consecutive S/R are better that a single complicated one :-D

Best Regards,

guy038

AdrianHHH

Thanks for the clarification @Scott-Sumner. My wording may have been poor; Notepad++ allows more than nine groups but the backslash only allows nine.

I have just re-checked the Boost page on replacements (see http://www.boost.org/doc/libs/1_48_0/libs/regex/doc/html/boost_regex/format/boost_format_syntax.html ). The table of escape sequences shows \D as “If D is a decimal digit in the range 1-9, then outputs the text that matched sub-expression D.” Thus this backslash form only allows 9 captures. Additional captures can be accessed as shown in the Placeholder Sequences table by using ${n}which "Outputs what matched the n’th sub-expression".

AdrianHHH

@guy038 I have seldom used look-behinds or look-aheads and I do not remember their syntax. For the times I have needed complicated search and replaces the performance difference between the non-look-behind(or ahead) form and the form with look-behinds(or aheads) is much less than the mental effort it would take me to change my approach. Having said that, I am very happy that you have found another, possibly neater, way of using my idea.

James Phoden

Hello sorry to jump onto this thread but it thought it better then starting a new one as my issue is related.

I need to find and replace this
Original String
“Prod_Data:Logos:Race Logos:illing Logo&Maps:Maps:AWT:6f AWT.eps”
Required String
“\\grp-pserv-01wl\Prod_Data\Logos\Race Logos\illing Logo&Maps\Maps\AWT\6f AWT.eps”

So basically replace the "Prod_Data: with "\\grp-pserv-01wl\Prod_Data\
and : with \ but only replace them if the original string is within within the quote marks “”.

Scott Sumner

@James-Phoden

So the part that makes your situation ugly is that I presume there can be a variable amount of :xxxx in your real data–you didn’t say… For example, maybe all of the following are valid things you want to match for replacement:

"Prod_Data:Logos:Race Logos:illing Logo&Maps:Maps:AWT:6f AWT.eps"
"Prod_Data:Logos:Race Logos:illing Logo&Maps:Maps:AWT AWT.eps"
"Prod_Data:Logos:Race Logos:illing Logo&Maps:Maps AWT.eps"
"Prod_Data:Logos:Race Logos:illing Logo&Maps AWT.eps"
"Prod_Data:Logos:Race Logos AWT.eps"
"Prod_Data:Logos AWT.eps"

If this is NOT the case and you always have SIX sets of :xxxx then the situation is a lot less ugly. But, moving forward with a variable (but bounded) count (from 1 to 10 occurrences, for example), try this [and PLEASE use copy-n-paste… :-) ]:

Find what zone: (?-s)"Prod_Data(?::(.+?))(?::(.+?))?(?::(.+?))?(?::(.+?))?(?::(.+?))?(?::(.+?))?(?::(.+?))?(?::(.+?))?(?::(.+?))?(?::(.+?))?"
Replace with zone: "\\\\grp-pserv-01wl\\Prod_Data(?1\\${1})(?2\\${2})(?3\\${3})(?4\\${4})(?5\\${5})(?6\\${6})(?7\\${7})(?8\\${8})(?9\\${9})(?10\\${10})"
Search mode radio-button: Regular expression

What this is doing is matching whatever follows the individual colons (after your "Prod_Data leading text is matched) into the capture groups 1-10. At replacement time, the colons (converted to backslashes) plus the captured groups are conditionally inserted into the output stream. The conditional syntax is necessary because of the variable number of occurrences of \xxxx that might be needed.

Thus, each (?::(.+?))? in the FW string captures a :xxxx – the second occurrence of : in this is the real/literal colon…the first colon is part of (?: which is just syntax saying “group the stuff in the parentheses but don’t capture it for later use”.

And each (?y\\${y}) (where y = 1…10) in the RW string represents a backslash and the original xxxx. When group 8 (for example) doesn’t exist the syntax (?8 will evaluate to false and whatever occurs between the (?8 and the next )` will NOT be part of the replacement data.

Thus the data above will convert to the following:

"\\grp-pserv-01wl\Prod_Data\Logos\Race Logos\illing Logo&Maps\Maps\AWT\6f AWT.eps"
"\\grp-pserv-01wl\Prod_Data\Logos\Race Logos\illing Logo&Maps\Maps\AWT AWT.eps"
"\\grp-pserv-01wl\Prod_Data\Logos\Race Logos\illing Logo&Maps\Maps AWT.eps"
"\\grp-pserv-01wl\Prod_Data\Logos\Race Logos\illing Logo&Maps AWT.eps"
"\\grp-pserv-01wl\Prod_Data\Logos\Race Logos AWT.eps"
"\\grp-pserv-01wl\Prod_Data\Logos AWT.eps"

If this (or ANY posting on the Notepad++ Community site) is useful, don’t reply with a “thanks”, simply up-vote ( click the ^ in the ^ 0 v area on the right ).

guy038

Hi, @Scott-sumner and @james-phoden,

Scott, why don’t you use the simple S/R, below :

SEARCH :|(?-i)(Prod_Data)

REPLACE \\(?1\\grp-pserv-01wl\\\1)

Of course, it works, strictly, with your original text. May be, you want to avoid colons, placed outside a "...." block, don’t you ?

Cheers,

guy038

Scott Sumner

@guy038

Yea, it’s the difference between wanting to help and making too many/few assumptions about a questioner’s data. In this case I wouldn’t go so far to assume that colons only appear in these places, but who knows?

guy038

Hello, @scott-sumner and All,

I’ve studied the general case of searching a specific character, ONLY IF, located inside a range of characters with delimiters.

Now, two cases are possible :

Case A : an area with a same starting and ending character, as, for instance, '.....' or "....."
Case B : an area with a different starting and ending character, as, for instance, (.....), [.....], {.....} or <.....>

Notes :

For our discussion, we are supposed to look for the colon character
For case A, I chose the double quotes delimiter ", as common boundary
For case B, I chose the start delimiter < and the end delimiter >

Let’s begin with the easier form !

Case B : A possible regex would be :

SEARCH :(?=[^<\r\n]*>)

REPLACE Any string or character, even EMPTY

Note that this regex looks for a colon character, ONLY IF followed by a range, possibly empty, of characters, different from the first delimiter < and from the EOL characters \r and \n , till the ending delimiter >

On the test example, below, the regex finds all colon characters, located inside the <.....> areas exclusively ( the ones which are underlined ) Fine !

1:23:<This:is a: tiny>text:to :see<if :my :logic:>is: correct:<I: hope:that>: all: will:be<::fine,: indeed>: ! :<:>78:9
          ¯    ¯                      ¯   ¯     ¯               ¯     ¯                    ¯¯     ¯              ¯

Now, in case A, the annoying thing is that it’s impossible to distinguish the two delimiters ! So, we’re going to cheat a bit ! First, we’ll replace, for instance, any area "....." by an oriented area as, for instance, #"....."@. Of course, these new boundaries must be absent from the present contents of the file !

So, assuming the original text :

1:23:"This:is a: small"text:to :see"if :my :logic:"is: correct:"I: hope:that": all: will:be"::fine,: indeed": ! :":"78:9

The simple S/R :

SEARCH ".*?"

REPLACE #$0@

would get the following text :

1:23:#"This:is a: small"@text:to :see#"if :my :logic:"@is: correct:#"I: hope:that"@: all: will:be#"::fine,: indeed"@: ! :#":"@78:9

Accordingly, the correct regex becomes :

SEARCH :(?=[^#\r\n]*@)

REPLACE Any string or character, even EMPTY

Again, only the colons, inside the areas, which are underlined, are matched by the regex !

1:23:#"This:is a: small"@text:to :see#"if :my :logic:"@is: correct:#"I: hope:that"@: all: will:be#"::fine,: indeed"@: ! :#":"@78:9
           ¯    ¯                         ¯   ¯     ¯                 ¯     ¯                      ¯¯     ¯                ¯

To end with, use the simple regex :

SEARCH #|@

REPLACE Empty

in order to get the original areas "....."

Et voilà !

Cheers,

guy038