Pinned topic Regexp with new lines

I don't quite get your regular expression but in DXL you can match newlines in different ways, however the "." placeholder does not match on newlines:

string sText = "\n"; 

Regexp re1 = regexp "\\\n"  // match
Regexp re2 = regexp "\\n"   // match
Regexp re3 = regexp "\n"    // match
Regexp re4 = regexp "."     // no match

if (re1 sText) print "Match re1\n"; 
if (re2 sText) print "Match re2\n"; 
if (re3 sText) print "Match re3\n"; 
if (re4 sText) print "Match re4\n";

Although I don't know what the (\\\\~| *) part of your regexp is supposed to match to, but the following test at least matches in group 1 and 3 (note that you have a space between 'Version 2' and the ':' ...

string Text_Version = "Version 2 :
Add some links
with \"Satisfy\" link module"

Regexp MyText = regexp2 "(Version [0-9]*[ ]*:)[ \n]*(\\\\~| *)*(.*)"

print "Match 0: " Text_Version [match 0] "\n" 
print "Match 1: " Text_Version [match 1] "\n"
print "Match 2: " Text_Version [match 2] "\n"
print "Match 3: " Text_Version [match 3] "\n"

Regards, Mathias

Not really following; but I will say:

• (.*)   will match any number of characters no including a new-line (EOL)
• Other RegExp references to "end of string" generally mean "end of string or next EOL".

Thus, parsing text with EOLs causes problems.  I resolve that with this

• const string cl_re_strAnyChar = "[" charOf(1) "-" charOf(255)"]"    // any character except null
• Regexp re = regexp2(whatever (cl_re_strAnyChar)* whatever)

That seesm to handle EOLs in the text

-Louie

• llandale 270001QM9N
• ‏2014-04-22T17:04:16Z

Not really following; but I will say:

• (.*)   will match any number of characters no including a new-line (EOL)
• Other RegExp references to "end of string" generally mean "end of string or next EOL".

Thus, parsing text with EOLs causes problems.  I resolve that with this

• const string cl_re_strAnyChar = "[" charOf(1) "-" charOf(255)"]"    // any character except null
• Regexp re = regexp2(whatever (cl_re_strAnyChar)* whatever)

That seesm to handle EOLs in the text

-Louie

More...

Well, I tried your const string cl_re_strAnyChar but it doesn't work...

I've simplfied my object text.

// object text : "Version 2 : Add some links with link module" (without any EOL nor special characters)
Object curro = current Object
string Text_Version = curro."Object Text"
const string str_anychar = "["charOf(1)"-"charOf(255)"]"

Regexp MyText = regexp2 "([A-Z a-z 0-9]*:)(str_anychar)*"

if (MyText Text_Version)
{
    print Text_Version [match(0)] "\n"
    print Text_Version [match(1)] "\n"
    print Text_Version [match(2)] "\n"
}

// match [0] = "Version 2 :"
// match [1] = "Version 2 :"
// match [2] = ""

Why match (0) and then match(2) are wrong ??

I expected match(0) = "Version 2 : Add some links with link module"  and match(2) = "Add some links with link module" Even without EOL and with the const string, the regexp does not match !!!

• Mathias Mamsch 2700025PVX
• ‏2014-04-23T21:14:02Z

Your line 7 is wrong. It should read:

<pre class="javascript dw" data-editor-lang="js" data-pbcklang="javascript" dir="ltr">Regexp MyText = regexp2 "([A-Za-z0-9 ]*:)(" str_anychar ")*" </pre>

Regards, Mathias

 

More...

Thank's so much !

It works fine !!!

• Mathias Mamsch 2700025PVX
• ‏2014-04-23T21:14:02Z

Your line 7 is wrong. It should read:

<pre class="javascript dw" data-editor-lang="js" data-pbcklang="javascript" dir="ltr">Regexp MyText = regexp2 "([A-Za-z0-9 ]*:)(" str_anychar ")*" </pre>

Regards, Mathias

 

More...

Beat my head against the wall yesterday and missed that.  Doh!

However, I think you should move that last asterisk * inside the parens; this gives "match 2" the entire rest of the string.  The way you have it, match 2 is just the last character, in this case "e".

const string str_anychar = "["charOf(1)"-"charOf(255)"]"

Regexp MyText = regexp2 "^([A-Za-z0-9 ]*:)(" str_anychar "*)"

void Test(string in_String)
{
 print "[" in_String "]\n"
 if (MyText in_String)
 {
     print "\t0  [" in_String [match(0)] "]\n"
     print "\t1  [" in_String [match(1)] "]\n"
     print "\t2  [" in_String [match(2)] "]\n"
     print "\t3  [" in_String [match(3)] "]\n"
 }
 else print "\tNo Match\n"
}
Test("Version 2 : Add some links with link module")
Test("Version 2 : Add some: links with link module")
Test("Version 2 : Add some: links \nwith link module")

-Louie