Regex question

[justnewbie]

Hi guys,

I have this text:

Code: Pascal  [Select][+][-]

1. something retvalue -1 anything anyt=55; nothing var1=5;

By using the RegExpr, I want to get the text in this form (ie. the anything and nothing should be in a new line):

Code: Pascal  [Select][+][-]

1. something retvalue -1
2. anything anyt=55;
3. nothing var1=5;

The https://regex101.com/ site gives me the good result if I write the \n\0 in the substitution field (see the attached picture).
But the Lazarus cannot handle it.

My code:

Code: Pascal  [Select][+][-]

1. reBefore := '(anything|nothing)';
2. reAfter := '\n\0';
3. Memo1.Text := ReplaceRegExpr(reBefore, Memo1.Text, reAfter, true);

This code doesn't do what I need. How can I solve this?

[engkin]

replace '\n\0' with LineEnding

[justnewbie]

replace '\n\0' with LineEnding

Do you mean

Code: Pascal  [Select][+][-]

1. reAfter := LineEnding;

?
This doesn't work.

[engkin]

Code: Pascal  [Select][+][-]

1. program Project1;
2.  
3. {$mode objfpc}{$H+}
4.  
5. uses
6.   RegExpr;
7.  
8. var
9.   s1:string='something retvalue -1 anything anyt=55; nothing var1=5;';
10.   s2, reBefore, reAfter: String;
11. begin
12.   reBefore := '(anything|nothing)';
13.   reAfter := LineEnding+'$0';
14.   s2 := ReplaceRegExpr(reBefore, s1, reAfter, true);
15.   WriteLn(s2);
16.   ReadLn;
17. end.

[justnewbie]

Great, thank you, works!
If I may ask, why did not work the original version? >> '\n\0'
What does '$0' mean?

Also, I tried this way

Code: Pascal  [Select][+][-]

1. reAfter := '\n$0';

and it worked as well.

[engkin]

Great, thank you, works!
why did not work the original version? >> '\n\0'

It did work, but did not give what you want.

What does '$0' mean?

It refers to the text inside the brackets, in this expression it may be anything or nothing. To see it, try this change:

Code: Pascal  [Select][+][-]

1.   reAfter := '-->$0<--';

Also, I tried this way

Code: Pascal  [Select][+][-]

1. reAfter := '\n$0';

and it worked as well.

I just checked RegExpr code, it supports:

Code: Pascal  [Select][+][-]

1.      't': Result := #$9;  // \t => tab (HT/TAB)
2.      'n': Result := #$a;  // \n => newline (NL)
3.      'r': Result := #$d;  // \r => carriage return (CR)
4.      'f': Result := #$c;  // \f => form feed (FF)
5.      'a': Result := #$7;  // \a => alarm (bell) (BEL)
6.      'e': Result := #$1b; // \e => escape (ESC)
7.      'x': begin // \x: hex char
8.  

[justnewbie]

Thank you for your help, engkin! 

[justnewbie]

Now I have a bit more complicated task.
In my example there is a loop "header" that is in 3 lines, and I need it only in 1 line.
The original text:

Code: Pascal  [Select][+][-]

1.    string something="";
2.    bool anything=true;
3.    for(int i=0;
4.    i<var1;
5.    i++)
6.      {
7.       string nothing=text[i];
8.      }

I want the for(int i=0; i<var1; i++) in 1 line.
Important: I need it in a general formula, so eg. it can be either for(int var2=x; var2<var3; var2++) or for(int var4=100; var4>=var1; var4--) and similar ...

By using this regex for\s*\([^\n]+\n[^\n]+\n[^\n]+ I can get the loop "header", but I don't know how could I get the result without those 2 \n (highlighted with orange).
Any help?

[engkin]

To remove all line endings use:

Code: Pascal  [Select][+][-]

1.   reBefore := '([^\r\n]+)[\r\n]+';
2.   reAfter := '$1';
3.   s2 := ReplaceRegExpr(reBefore, s1, reAfter, true);

To remove the two line endings in a for statement:

Code: Pascal  [Select][+][-]

1.   reBefore := '(for\s*\([^\r\n]+)[\r\n]+([^\r\n]+)[\r\n]+([^\r\n]+\))';
2.   reAfter := '$1$2$3';
3.   s2 := ReplaceRegExpr(reBefore, s1, reAfter, true);

[justnewbie]

Thank you engkin, this is exactly what I need for!
(Soon I will have new question ...) 

[justnewbie]

New issue:
I need those lines that do not start with the word 'something'.
I know what ^something does, but I need for its opposite.

Edit: maybe I figured it out:
^[^\bsomething\b].*

[justnewbie]

Next:

I have this text:

mytext mytext hjsdgfjsh 57574 //gkgaksg
//  ztwojkgk
fjfjaf//tjtrjs
nodelete fghfgkhag //zrwjsj
ghfhfhd ://uzrutziut

I want to delete the texts in red (text after the //, together the // itself), but I don't want to delete the texts in teal.
Ie., no delete if
1./ a line starts with nodelete
or
2./ there is a colon (:) right before the //

So I want to get this:

mytext mytext hjsdgfjsh 57574

fjfjaf
nodelete fghfgkhag //zrwjsj
ghfhfhd ://uzrutziut

Update:
I found the solution for the 2. case:
(?<!:)\/\/[^\n]+[\n]+
I need to expand it with the 1. rule also.

[justnewbie]

I found a solution:
(^(?!nodelete).*)(?<!:)\/\/[^\n]+[\n]+

It works well in the online regex tester https://regex101.com/, but Lazarus gives an error (unrecognized modifier).
Any help?

[engkin]

Probably time to consider using a different engine. For instance look for FLRE.

[justnewbie]

Probably time to consider using a different engine. For instance look for FLRE.

What is it? I only found 'flare regular expr' in Google.
Can you provide me a link?