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Hi!
Could anyone pls help med with a regex that extracts row 3 (Summer AB) only?
805
1506678441522
Summer AB
4302334
4302334
Structure
DBT.4302334"
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I've a weird situation. I created a regex expression for one of the service. I've tested this expression at regexr.com and regex101.com and there are no problems at all. when I used this regex expression in fail2ban, it missed all the lines.
here is the log output;
[05-Oct-2021 17:09:39 +0300]: IMAP Error: Login failed for xyz@xyz.com against localhost from 95.65.143.88. AUTHENTICATE PLAIN: Authentication failed. in /usr/share/roundcube/program/lib/Roundcube/rcube_imap.php on line 204 (POST /webmail/?_task=login&_action=login)
here is the regex;
(IMAP Error: Login failed for)\s([a-zA-Z0-9_.-]+\@[a-zA-Z0-9_.-]+)\s(against localhost from)\s
is fail2ban using a different regex structure?
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fail2ban is checking application logs and prevent intruders.
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fail2ban regexes are a bit "different", particularly with the new(ish) prefixes and interpolations.
Your best bet is to use "fail2ban-regex".
Somewhere around the fail2ban site there is a tutorial about iincremental development of regexes.
It's a bit outdated, but the process works.
Software rusts. Simon Stephenson, ca 1994. So does this signature. me, 2012
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yep I'm using fail2ban-regex but before that I'm preparing my expressions at regexr.com.
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My experience with fail2ban regexes is that they are sufficiently different from the "standard" ones that the regular development/test tools don't help.
I think you're stuck with fail2ban-regex, a dummy log file with examples of good and bad entries, and a whole lot of panel-beating, depending on how complex your requirements are.
It's worth looking at the various filter files provided with fail2ban. I found some helpful constructs in ones I don't actually use.
Software rusts. Simon Stephenson, ca 1994. So does this signature. me, 2012
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f2b should have provided standard regex expressions, so people wouldn't need to learn something else.
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Hi,
im relatively new to regex and im trying to match OU from the Distinguished Name in ActiveDirectory.
This is my existing regex:
(((OU=.*?))?)OU=(?<ouname>.*?(?=,))
CN=Muster\, Max,[OU=AAD-Sync,OU=IT-Abteilung],[OU=FD] 10,DC=landkreis-Musterstadt,DC=local
So far this would match the OU I marked with [] as two separate Matches, does anyone know ow to edit my existing regex that it matches every OU as one Match instead of two separates?
Would be lovely if I could get some Help 
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I have this regex to replace accents but it fails if the text contains "|".
string Text = "word|word2";
System.Text.RegularExpressions.Regex replace_a_Accents = new System.Text.RegularExpressions.Regex("[á|à|ä|â]", System.Text.RegularExpressions.RegexOptions.Compiled);
System.Text.RegularExpressions.Regex replace_e_Accents = new System.Text.RegularExpressions.Regex("[é|è|ë|ê]", System.Text.RegularExpressions.RegexOptions.Compiled);
System.Text.RegularExpressions.Regex replace_i_Accents = new System.Text.RegularExpressions.Regex("[í|ì|ï|î]", System.Text.RegularExpressions.RegexOptions.Compiled);
System.Text.RegularExpressions.Regex replace_o_Accents = new System.Text.RegularExpressions.Regex("[ó|ò|ö|ô]", System.Text.RegularExpressions.RegexOptions.Compiled);
System.Text.RegularExpressions.Regex replace_u_Accents = new System.Text.RegularExpressions.Regex("[ú|ù|ü|û]", System.Text.RegularExpressions.RegexOptions.Compiled);
Texto_Retorno = replace_a_Accents.Replace(Texto_Retorno, "a");
Texto_Retorno = replace_e_Accents.Replace(Texto_Retorno, "e");
Texto_Retorno = replace_i_Accents.Replace(Texto_Retorno, "i");
Texto_Retorno = replace_o_Accents.Replace(Texto_Retorno, "o");
Texto_Retorno = replace_u_Accents.Replace(Texto_Retorno, "u");
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Why would you use a Regex for something so simple?
bool replacedAny = false;
char[] characters = Texto_Retorno.ToCharArray();
for (int index = 0; index < characters.Length; index++)
{
switch (characters[index])
{
case 'á':
case 'à':
case 'ä':
case 'â':
{
characters[index] = 'a';
replacedAny = true;
break;
}
case 'é':
case 'è':
case 'ë':
case 'ê':
{
characters[index] = 'e';
replacedAny = true;
break;
}
case 'í':
case 'ì':
case 'ï':
case 'î':
{
characters[index] = 'i';
replacedAny = true;
break;
}
case 'ó':
case 'ò':
case 'ö':
case 'ô':
{
characters[index] = 'o';
replacedAny = true;
break;
}
case 'ú':
case 'ù':
case 'ü':
case 'û':
{
characters[index] = 'u';
replacedAny = true;
break;
}
}
}
if (replacedAny)
{
Texto_Retorno = new string(characters);
}
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
modified 16-Sep-21 10:54am.
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Nothing to see here. :innocent-whistle-smily:
In my defence, "3" is directly above "e" on the keyboard.
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
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Given some of the nonsense that appears when I type too fast I think you do very well.
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I was dreading it, Regex doesn't do the job and doesn't replace either?
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Regex can do the job. But running five+ separate regex operations on a string just to replace a few letters with their unaccented alternatives is overkill.
The other option, which is even nastier and less obvious, is to use Unicode normalization:
static string RemoveDiacritics(string stIn)
{
string stFormD = stIn.Normalize(NormalizationForm.FormD);
StringBuilder sb = new StringBuilder();
for(int ich = 0; ich < stFormD.Length; ich++)
{
UnicodeCategory uc = CharUnicodeInfo.GetUnicodeCategory(stFormD[ich]);
if (uc != UnicodeCategory.NonSpacingMark)
{
sb.Append(stFormD[ich]);
}
}
return sb.ToString().Normalize(NormalizationForm.FormC);
}
string input = "Príliš žlutoucký kun úpel dábelské ódy.";
string result = RemoveDiacritics(input); Source[^]
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
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case 'Á' case 'À' case 'Ó' ... to be continue
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In [...] groups (alternatives) you don't want the |s.
[aeiou] as a regex will match any vowel, for example.
Software rusts. Simon Stephenson, ca 1994. So does this signature. me, 2012
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I use Regex match to find multiple words in Like mode (start and end point). I do not want the return to include the space as the first position and I do not want the final space to count as valid.
Regular Expression: "\b(?:.Road.|.ous.|.Street.)\b" Must Return: Match:9 length:5 End:13
Text: "have one house free."
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Hello there, I have a string that looks like this
\abc\rvr\sad\dqwdwq Some text \rfdqwdvr\rtgevr\rgwvrr\23rvrv Some text \rffewvr\rfewvr\rvfewrr\rvrwefewv
I want to match the parts of the string that are connected by the backslashes. I partially achieved this with
[\\][A-z0-9]+ but this only matches the first one. How can I match every part with backslashes in this string?
Thanks
modified 31-Aug-21 8:08am.
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Assuming you want to extract three strings:
- \abc\rvr\sad\dqwdwq
- \rfdqwdvr\rtgevr\rgwvrr\23rvrv
- \rffewvr\rfewvr\rvfewrr\rvrwefewv
Try:
([\\][A-Za-z0-9]+)+ Regexper[^]
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
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hi would any one have a regex that would match (aug -) or(july -) or (june -)without the brackets
but would not match just (aug)or (july) or (june)
So far i have
\W*((?i)aug -(?-i))\W*
but this also matches aug which i dont want thanks
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I tried your regex in Expresso, and it matches "hello aug -goodbye" but not "hello aug goodbye".
So ... what are you talking about?
"I have no idea what I did, but I'm taking full credit for it." - ThisOldTony
"Common sense is so rare these days, it should be classified as a super power" - Random T-shirt
AntiTwitter: @DalekDave is now a follower!
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im using regex in placemint for placement of window and brousers
if i title a folder aug placemint macthes it , but i only want match on aug -
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