Regarding regex operators precedence, taken from the link,

https://www.boost.org/doc/libs/1_55_0/libs/regex/doc/html/boost_regex/syntax/perl_syntax.html#boost_regex.syntax.perl_syntax.operator_precedence

The table, below, gives the hierarchy of these operators, listed from the highest priority to the lowest priority :

1. POSIX based Bracket Character set : [:Class character:], [=Equivalent Class=], and [.Collating element.]
2. Escaped characters : \...
3. Bracket Character set, ( negative or not ) : [^.....] and [.....]
4. Grouping, ( capturing or not ) : (.....) and (?:.....)
5. Quantifiers : *, +, ?, {n}, {m,n} and {m,}
6. Concatenation ( Implicit )
7. Anchoring : ^ and $
8. Alternation : |

Here are some examples to verify this hierarchy :

• Between level 1 and level 2 :

The regex [[=\=]] matches the reversed slash \, only and NOT the regex [[==]], which is, besides, invalid !

• Between level 2 and level 3 :

The regex \[1] means the regex \[ , so the string [, followed with the string 1] and NOT the regex \1, as [1] represents the 1 digit., which,finally, matches the 1 digit

• Between level 3 and level 4 :

The regex [(123)45] matches 1, 2, 3, 4 and 5 digits, as well as the parentheses ( and ), and NOT the number 123, as a group, or the digits 4 or 5, which can be found with the regex (123)|[45]

• Between level 4 and level 5 :

The regex (123)+ represents the number 123, possibly repeated, and NOT the 12 number, followed with any range of consecutive digit(s) 3, which can be found with the regex 123+

• Between level 5 and level 6 :

The regex 123+45+ matches the 12 number, followed with any range of consecutive digit(s) 3, followed with 4 number, followed with any range of consecutive digit(s) 5 and NOT any range of the 123 number, followed with any range of the 45 number, which can be obtained with the regex (123)+(45)+

• Between level 6 and level 7 :

I have not been able to detail differences between implicit concatenation of regexes ( for instance, regex a, followed with regex b resulting in the regex ab ) and anchoring which defines zero-length regexes, matching specific locations in file contents !

Indeed, if we consider the simple regex ^123, to my mind, the regex ^1, immediately followed with the regex 23 or the regex ^12, immediately followed with the regex 3 and the regex ^123, or even the zero-lengh regex ^ followed with the regex 123, seem all identical !?

A bit off topic : just notice that string concatenation does NOT represent the same concept as regex concatenation ! For instance, the regex [12], followed with the regex [34] matches all elements of the set { 13, 14, 23, 24 }, whereas the string 12, followed with string 34, represents the single-element set { 1234 }

• Between level 7 and level 8 :

The regex ^12|34$ matches the 12 number, beginning a line OR the 34 number, ending a line ( and NOT a line with number 12 OR number 34, only ( which can be found with the regex ^(12|34)$ ) NEITHER a line beginning with the 1 digit, ending with the 4 digit and between, either, digit 2 OR 3 ( which can be found with the regex ^1(2|3)4$ )

Best regards,

Merry Christmas and Happy Holidays to all ;-))

guy038

P.S. :

I’ve, also, found out a great article on operators precedence, regarding the main progamming or script languages ;-)) Just click below :

https://rosettacode.org/wiki/Operator_precedence

I think I see what you are saying. You are saying that to some people with misunderstanding would think that the (?i) only affects friday in the example above.

In general this brings up a good point, or a question. What is the precedence of regexes?

For example if you have the regex friday|saturday|sunday we know that it truly means (friday)|(saturday)|(sunday) – without the capturing of course. But it could mean frida(y|s)aturda(y|s)unday I suppose – but I know it doesn’t. But I don’t know the real rules of when one needs non-capturing parens…and when one doesn’t.

Maybe this is a hard question to ask, and it isn’t really Notepad++ related…

Ah…, yes, Robin, you’re quite right ! It’s the usual drawback of finding out a regex, without having the real text to test the regex against !

I’m not an HTML coder and, may be, I’m going to tell a nonsense but let’s suppose the following text, without the ending tag, at line 4. So :

<p class=“amigo”>1. Blah blah blah <br>
      2. Blah blah blah <br>
      3.  Blah blah blah <br>
      4.  Blah blah blah
      ....

How can I decide between this case A :

<p class=“amigo”>1. Blah blah blah </p>
      2. Blah blah blah <br>
      3. Blah blah blah <br>
      4. Blah blah blah <br>
      ....

And this case B, below ?

<p class=“amigo”>1. Blah blah blah <br>
      2. Blah blah blah <br>
      3. Blah blah blah <br>
      4. Blah blah blah    </p>
      ....

Of course, feel free to send me an e-mail, with your true text, if you don’t mind, just telling me where you would like to replace the <br> with </p>

To @alan-kilborn,

Of course, I didn’t say that it was an hidden rule or a work-around ! I just wanted to point out, for beginners to “regex world”, the fact that, if the search pattern is an alternative, with several branches, either :

• In a non-capturing group, as (?i:friday|saturday|sunday) or (?:(?i)friday|saturday|sunday)

• In a capturing group, as ((?i)friday|saturday|sunday)

The modifier, (?i) in this example, affects any branch of the alternative

Best Regards,

guy038

Because alternative branches are tried from left to right, and options are not reset until the end of the subpattern is reached, an option setting in one branch does affect subsequent branches

Is there something new or surprising here?

In the example you gave, this part:

(?i:friday|saturday|sunday)

It corresponds to my “revised Wiki” entry of:

(?flags:searchpattern)

where in this case:

searchpattern = friday|saturday|sunday

but it is still just a re pattern…

I feel like I am missing something here because my first thought is of course this is the way it works and nothing is new here.

guyo38 made also a beautiful regex. But, there is a case where the formula does not fit. Suppose:

Case 1

<p class=“amigo”>1. Blah blah blah <br>
      2. Blah blah blah <br>
      3.  Blah blah blah <br>
      4.  Blah blah blah       </p>

Case 2

<p class=“amigo”>1. Blah blah blah <br>
      2. Blah blah blah <br>
      3.  Blah blah blah <br>
      4.  Blah blah blah        <br>
 new sentence here </p>

In this 2 cases your regex (?s)<p class[^<]+\K(?!</p>)<(?-s).+> replace by </p> will replace the first instance of <br>, and that will ruin the html code.

So, in this 2 cases, I would only like to search and replace only those tags that contains only one instance of <br>. Or, if there are more <br>, the replacement with </p> would not take place anymore.

May be a bit late but here is a regex S/R which is able to detect and replace any ending tag, different of </p>, with this one, in any mono or multi-lines range <p class.....>.............<...>, whatever its location on current line

SEARCH (?s)<p class[^<]+\K(?!</p>)<(?-s).+>

REPLACE </p>

Remark : I’s important to point out that, due to the \K syntax, this regex S/R works if you click on the Replace All button , exclusively ! ( Any step-by-step replacement , with the Replace button, will not work )

So, assuming this test text :

bla bla   <p class=“amigo”>My mother is at home.<br>   bla bla

bla blah

bla bla   <p class=“amigo”>My mother
 is at 

home.<br>   bla bla

bla blah

bla bla   <p class=“amigo”>My mother
 is at home.<h>   bla bla

bla blah

bla bla   <p class=“amigo”>My mother is at home.</p>   ====== NOT CHANGED ======

bla blah

bla bla   <p class=“amigo”>My mother is at home.</h>   bla bla

bla blah

bla bla   <p class=“amigo”>My
 mother
 is at
 home.</p>   ====== NOT CHANGED ======

bla blah

bla bla   <p class=“amigo”>My mother is at home.</a>   bla bla

bla blah

bla bla   <p class=“amigo”>My mother is at home.<p>   bla bla

bla blah

bla bla   <p class=“amigo”>My mother is at home.<br>   bla bla

bla blah

You would obtain :

bla bla   <p class=“amigo”>My mother is at home.</p>   bla bla

bla blah

bla bla   <p class=“amigo”>My mother
 is at 

home.</p>   bla bla

bla blah

bla bla   <p class=“amigo”>My mother
 is at home.</p>   bla bla

bla blah

bla bla   <p class=“amigo”>My mother is at home.</p>   ====== NOT CHANGED ======

bla blah

bla bla   <p class=“amigo”>My mother is at home.</p>   bla bla

bla blah

bla bla   <p class=“amigo”>My
 mother
 is at
 home.</p>   ====== NOT CHANGED ======

bla blah

bla bla   <p class=“amigo”>My mother is at home.</p>   bla bla

bla blah

bla bla   <p class=“amigo”>My mother is at home.</p>   bla bla

bla blah

bla bla   <p class=“amigo”>My mother is at home.</p>   bla bla

bla blah

Best regards,

guy038

P.S. : To continue the @alan-kilborn discussion on flags, from the link below :

https://gammon.com.au/pcre/pcrepattern.html#SEC11

It is said :

Because alternative branches are tried from left to right, and options are not reset until the end of the subpattern is reached, an option setting in one branch does affect subsequent branches

So, for instance, the regex (?-i)WEDNESDAY|(?i:friday|saturday|sunday)|Monday would match :

• WEDNESDAY and Monday, in that exact case

• Friday, as well as Saturday and Sunday in any case

So checking the Notepad++ regex wiki: http://docs.notepad-plus-plus.org/index.php/Regular_Expressions

I see that this syntax IS there, but I guess I never understood it because it is SO f*cked up. :-)

Here’s what the Wiki says:

(?:flags-not-flags ...), (?:flags-not-flags:...)
Applies flags and not-flags to search inside the parentheses. Such a construct may have flags and may have not-flags - if it has neither, it is just a non-marking group, which is just a readability enhancer.

Here’s how I would write it so that it is (hopefully) understandable:

(?flags-notflags:searchpattern)
Applies flags and notflags to the searchpattern inside the parentheses and forms a non-capturing group. Such a construct may optionally have flags and may optionally have -notflags ; if it has neither, it is just a simple non-capturing group.  Note that the effect of flags/nonflags only applies to the searchpattern inside the enclosing parentheses.

(?-i)b(?i:b)b to match bbb or bBb but not Bbb, BBB, bBB, etc.

Note that the ?i only applies to what is inside the enclosing round brackets. After the closing one, the outer leading (?-i) goes back into being in effect.

Before this new knowledge I would have achieved the same thing this way: (?-i)b((?i)b)b or even messier (?-i)b(?:(?i)b)b or (?-i)b(?i)b(?-i)b

Note also that (?i:pattern) is a non-capturing group. And i could be s or whatever is legal (see prior post).

Before some wiseguy points out that b[bB]b works just as well…these are just made up examples to show a technique, not true real-life searches.

Of these:

(?adluimnsx-imnsx)

Boost does not support adlun, reducing what it does support to:

(?imsx-imsx)

If used one of the invalid ones, Notepad++ will say “Find: Invalid regular expression”

@Terry-R used the word “bracketed” to mean…

Yes I meant the “round brackets” ( and ). As Peter stated there are SO many different varieties it’s very easy to get confused. Your regex101 link above would have shown you what each capture group referred to and there were the “round brackets” mentioned.

I would strongly suggest you study the various characters used that have special meaning, following the links above that Peter and myself mentioned. Unless you get these basics sorted you will have ALL sorts of problems trying to create regexes successfully.

Terry

PCRE (Perl Compatible Regular Expresions) have their origin in Perl, and I learned my regex through Perl, so when I’m confused, I go to Perl’s perlre manpage. That shows a fully-expanded non-capturing group with enabled modifiers, disabled modifiers, and a non-capturing pattern: (?adluimnsx-imnsx:pattern)

• adluimnsx are the possible pattern modifiers to enable (where you can have 0 or more of the modifiers),
• -imnsx are the possible pattern modifiers to disable (where you can have 0 or more of the modifiers after the -),
• pattern is the part of the regex pattern that you want to group and match, but not capture.
• If you don’t have modifiers, it shortens to (?:pattern);
• if you don’t have a pattern and just want modifiers, it shortens to (?adluimnsx-imnsx).

Oh, I think I just realized your confusion. @Terry-R used the word “bracketed” to mean “surrounded by parentheses (...)”, but you interpreted it to mean “surrounded by angle brackets <...>”.

Given different people’s terminology in reference to parentheses/parenthesis, braces, brackets, curly brackets, square brackets, angle brackets, I try to be as explicit as possible, and often give examples of which I mean (though I sometimes fail at this).

But please, so to understand better this replace on your regex

I think you misunderstand the brackets around the ?-s. This is NOT a capture group, it is a modifier (I think that’s the correct wording). It modifies the parameters that the regex uses when searching for text. The first bracket for capturing is the one for the opening tag sequence (.*<p class="amigo">.*).

Might I suggest you need to read up a bit more on what metacharacter sequences are used in regex to modify the parameters. This site I use often and has good examples.
http://rexegg.com/regex-quickstart.html
Bear in mind that most regex engines will be slightly different in how they use these metacharacters so not everything on this site will fit exactly to how Notepad++ works, but nonetheless the information is still helpful.

Terry

The regex (?-s)(.*<p class="amigo">.*)<br>(.*)$ only has two capturing parenthesis-pairs. The (?-s) is a non-capturing command to the regex engine, and doesn’t capture anything. The next parenthesized group is a capture group, and captures the ...<p...>... into \1. The <br> is not captured (it’s not inside parentheses). The next parenthesized group is a capture group, and grabs everything beyond the <br> to the end of the line.

(?-s)(.*<p class="amigo">.*)<br>(.*)$
^^^^^ = not a capture group (it's a command to regex)

(?-s)(.*<p class="amigo">.*)<br>(.*)$
     ^^^^^^^^^^^^^^^^^^^^^^^ = first capture group, into \1

(?-s)(.*<p class="amigo">.*)<br>(.*)$
                            ^^^^ = not in parentheses, so not captured

(?-s)(.*<p class="amigo">.*)<br>(.*)$
                                ^^^^ = second capture group, into \2

-----
This FAQ gives lots of good pointers to regex documentation. You can study more about regular expressions and capture groups vs. non-capture groups in the links provided there.

so,

\1 is the first bracket
\2 second bracket
\3 third bracket

but, If I replace your regex just with \1it will delete <br>
So, seems to me that those \1 \2 \3 .. selects the words/code which is not in brackets. Correct ?

Note that the following portion (after the <br>) is now bracketed. Although none of the examples actually show any text following the <br>, if any did occur they would be captured and returned.

Terry

Ok, but if I want to make this search and replace just the <br> from all those lines, how can I do this?

So, to find all lines that contains those 2 tags, and replace only tha last one, <br> with </p> Something like this, but not too good:

Search: (?-s).*<p class="amigo">(.*)(<br>).*$

Replace by: 1\2\3</p>

Can you help me a little bit?

<p class="amigo">My mother is at home.<p>

asdas

<p class="amigo">My mother is at home.<br>

asfa

asffa

will match from the first p-amigo line to the end of the ...<br> line.
The examples I showed will correct that, too.

edit: this is true only if dot-matches-newline checkbox is turned on in the dialog, of course… if it’s turned off, then it doesn’t matter whether .* is greedy or not. (edits #2-n: clarified wording a couple times)

A couple of nitpicks with your solution:

1. (?s) and (?-s) are opposites: by having one followed by the other, this turns on “dot matches newline” then immediately turns off “dot matches newline”. You just need (?-s) to turn it off.

2. <br>*$: this matches the literals <, b, r, followed by 0 or more occurrences of >, followed by end of line. Which means it would match <br, or <br>>>>>>>>>, not just <br>.

3. >*$: Also, because of the $ anchor immediately after the 0-or-more >, then nothing – no whitespace, no nothing – will be allowed after the <br> at the end of the line. I hope that’s what you intended. If you really meant to match it, even when more stuff comes after the <br> on the same line, then you should use .*<p class="amigo">(?-s)(.*)<br>.*$

4. .*<p: just a note: if you happen to have the ☑ . matches newline option checked in your Find window, then it will match from the beginning of the first line to the end of the last line with amigo-paragraph and <br>. This can be seen with the example data here:

    <p class="amigo">My mother is at home.<br>
   
    asfa
   
    asffa
   
    <p class="amigo">My mother is at home.<br>
   
    asfa
   
    asffa
   
    <p class="amigo">My mother is at home.<p>
   
   
    afasf
   
    <p class="amigo">My mother is at home.<br>
   
    asfa
   
    asffa
   
    <p class="amigo">My mother is at home.<p>
   
    asdas
   
    <p class="amigo">My mother is at home.<br>
   
    asfa
   
    asffa
   

   … where it would find 1 occurrence, when there are really four. To fix this, move the (?-s) to the beginning of the pattern, instead of after the amigo-paragraph.

My recommended regular expression, which will allow text after the <br>, will not match <br, and will keep the instances separate: (?-s).*<p class="amigo">(.*)<br>.*$

The sequence (?s)(?-s) makes little sense in a regular exp. I think what you really wanted was to drop both of those and change .* to .*?

.*<p class="amigo">(?s)(?-s)(.*)<br>*$

see here:

https://regex101.com/r/We7Afi/1

no, it’s not about escape. My regex is almost good, but something else is missing :)