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• How can I extract a value from comma separated values in Perl?

  - by Octopus

  I have a log file containing statistics from different servers. I am separating the statistics from this log file using regex only. I am trying to capture the CPU usage from the running process. For SunOS, I have below output: process,10050,user1,218,59,0,1271M,1260M,sleep,58.9H,0.02%,java Here the CPU % is at 11th field if we separate by commas (,). To get this value I am using below regex: regex => q/^process,(?:.*?),((?:\d+)\.(?:\d+))%,java$/, For the linux system I have below output: process,26190,user1,20,0,1236m,43m,6436,S,0.0,1.1,0:00.00,java, Here the CPU usage is at 10th column. What regex pattern should I use to get this value?

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• Itertools group by functionality

  - by Nil

  I want to group by on dict key >>> x [{'a': 10, 'b': 90}, {'a': 20}, {'a': 30}, {'a': 10}] >>> [(name, list(group)) for name, group in groupby(x, lambda p:p['a'])] [(10, [{'a': 10, 'b': 90}]), (20, [{'a': 20}]), (30, [{'a': 30}]), (10, [{'a': 10}])] This must group on key 10 :(

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• Linux group permissions getting overwritten by owner

  - by Andy

  I am not a user of Linux, but I am encountering some permissions problems with it that I hope someone can shed some light on. Bit of background: A colleague of mine has a Linux box (running Debian I believe) with an SVN repository on it. The repository directory and files 'owner' is my colleauge. We are both members of a group called 'users'. He manages several projects both Linux and Windows apps, while I have one Windows app. For the Windows apps, we both use TortoiseSVN via an SSH link to commit/update. Performing the command 'ls -l' shows the repository files and folders on the Linux box to have the following permissions: -rwxrwx--- john users However, when my colleauge commits to the repository, the permissions change to: -rwxrwx--- john john This then means I get 'Permission denied' when trying to access the repository myself as it appears that the group permissions have been overwritten with only 'owner' permissions. To fix this, a 'chown -R' command is applied to the files/folders to set the permissions back to owner/group, but each time he writes to the repository, the issue repeats. I'm not sure if this is solely an SVN problem, or a more fundamental owner/group issue. Anyone any clue on how to stop this happening, or where to go and look? I'm trying to help out my colleague who is having some trouble resolving this issue. Apologies for the vague info, I hope I have conveyed the problem clear enough. Like I say, I am not a Linux user, I have only put down what I have managed to pick up from looking over his shoulder. Thanks for any pointers I can pass on!

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• Listing group members using ldapsearch

  - by colemanm

  Our corporate LDAP directory is housed on a Snow Leopard Server Open Directory setup. I'm trying to use the ldapsearch tool to export an .ldif file to import into another external LDAP server to authenticate with externally; basically trying to be able to use the same credentials internally and externally. I've got ldapsearch working and giving me the contents and attributes of everything in the "Users" OU, and even filtering down to only the attributes I need: ldapsearch -xLLL -H ldap://server.domain.net / -b "cn=users,dc=server,dc=domain,dc=net" objectClass / uid uidNumber cn userPassword > directorycontents.ldif That gives me a list of users and properties that I can import to my remote OpenLDAP server. dn: uid=username1,cn=users,dc=server,dc=domain,dc=net objectClass: inetOrgPerson objectClass: posixAccount objectClass: organizationalPerson uidNumber: 1000 uid: username1 userPassword:: (hashedpassword) cn: username1 However, when I try the same query on an OD "group" instead of a "container," the results are something like this: dn: cn=groupname,cn=groups,dc=server,dc=domain,dc=net objectClass: posixGroup objectClass: apple-group objectClass: extensibleObject objectClass: top gidNumber: 1032 cn: groupname memberUid: username1 memberUid: username2 memberUid: username3 What I really want is a list of users from the top example filtered based on their group memberships, but it looks like membership is set from the Group side, rather than the user account side. There must be a way to filter this down and only export what I need, right?

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• Listing group members using ldapsearch

  - by colemanm

  Our corporate LDAP directory is housed on a Snow Leopard Server Open Directory setup. I'm trying to use the ldapsearch tool to export an .ldif file to import into another external LDAP server to authenticate with externally; basically trying to be able to use the same credentials internally and externally. I've got ldapsearch working and giving me the contents and attributes of everything in the "Users" OU, and even filtering down to only the attributes I need: ldapsearch -xLLL -H ldap://server.domain.net / -b "cn=users,dc=server,dc=domain,dc=net" objectClass / uid uidNumber cn userPassword > directorycontents.ldif That gives me a list of users and properties that I can import to my remote OpenLDAP server. dn: uid=username1,cn=users,dc=server,dc=domain,dc=net objectClass: inetOrgPerson objectClass: posixAccount objectClass: organizationalPerson uidNumber: 1000 uid: username1 userPassword:: (hashedpassword) cn: username1 However, when I try the same query on an OD "group" instead of a "container," the results are something like this: dn: cn=groupname,cn=groups,dc=server,dc=domain,dc=net objectClass: posixGroup objectClass: apple-group objectClass: extensibleObject objectClass: top gidNumber: 1032 cn: groupname memberUid: username1 memberUid: username2 memberUid: username3 What I really want is a list of users from the top example filtered based on their group memberships, but it looks like membership is set from the Group side, rather than the user account side. There must be a way to filter this down and only export what I need, right?

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• .NET equivalent to Perl regular expressions

  - by r_honey

  I need to convert a Perl script to VB.NET. I have managed almost the entire conversion, but some Perl (seemingly simple) regex are causing an headache. Can someone suggest me .NET equivalent of the following perl regex: 1) $letter =~ s/Users //,; $letter =~ s/Mailboxes //,; if($letter =~ m/$first_char/i){ 2) unless($storegroup =~ /Recovery/ || $storegroup =~ /Users U V W X Y Z/ || $storegroup =~ /Users S T/ || $storegroup =~ /Users Q R/){ The regex look simple to me. I tried to wade through perl.org but understanding a langugae regex takes sometime and I need to complete the conversion quickly.

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• django: displaying group users count in admin

  - by gruszczy

  I would like to change admin for a group, so it would display how many users are there in a certain group. I'd like to display this in the view showing all groups, the one before you enter admin for certain group. Is it possible? I am talking both about how to change admin for a group and how to add function to list_display.

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• Racket regular-expression matching

  - by Inaimathi

  I'm trying to create a regex that matches the inverse of a certain string type (so, strings not ending in ".js", for example). According to the documentation, that should be the expression #rx"(?!\\.js$)", but it doesn't seem to work. To test it out, I have this function: (define (match-test regex) (map (lambda (text) (regexp-match? regex text)) '("foo.js" "bar.css" "baz.html" "mumble.gif" "foobar"))) (match-test #rx"\\.js$") returns (#t #f #f #f #f) as expected, but (match-test #rx"(?!\\.js$)") returns (#t #t #t #t #t), where I would expect (#f #t #t #t #t). What am I doing wrong, and how do I actually get a regex in Racket to express the idea "match anything which does not contain [x]"?

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• preg_match for <?php, <?, and/or ?>

  - by SoLoGHoST

  Hello guys, I'm not very familiar with regEx's and I'm trying to find a preg_match regex for searching for any of the following strings within a file and if found it will halt it. I already have the fopen and fgets setup I just need to use a regex inside of a preg_match for the following php tags: <?php <? ?> Thanks for your help with this :)

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• Grant HTTP access based on unix user group

  - by Sander Marechal

  Is it possible to grant network access or HTTP access based on a user's group? At my company we want to set up an internal composer server using Satis to manage packages for the projects we write (e.g. on repository.mycompany.com), with the packages themselves in our SVN server (svn.mycompany.com). We have several webservers with many different users on them. Some users should be able to reach the composer and SVN server. Some should not. Users that should be able to reach these servers all belong to the same group. How can I set up Apache on the Composer and SVN server to only grant access to those users in that group? Alternatively, can I set up the webservers in such a way that only users from that group are able to make a connection to our Composer and SVN servers? The best thing we have come up with so far is using SSL client certificates. We simply place a client certificate on all servers which can be used to access Composer and SVN. Only the right usergroup will have read access to the certificate. A bit clunky but it may work. But I'm looking for something better.

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• OrderBy After Group?

  - by Soo

  Ok, so I have a table Table: Id Value If I query my table and group my result by "Value" how can I make it so each of the groups are alphabetized (a group grouped by a "Value"="a" will come before a group grouped by a "Value" = "z"). My current query looks something like this: var Result = from a in DB.Table orderby a.Value group by a.Value into b select new {Groupz = b};

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• Regular - Take all numeric characters following a text character

  - by Simon

  Given a string in the format: XXX999999v99 (where X is any alpha character and v is any numeric character and v is a literal v character) how can I get a regex to match the numeric chatacters following the v? So far I've got 'v\d\d' which includes the v but ideally I'd like just the numeric part. As an aside does anyone know of a tool in which you can specify a string to match and have the regex generated? Modifying an existing regex is one thing but I find starting from scratch painful!

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• Perl : get substring which matches refex error

  - by Michael Mao

  Hi all: I am very new to Perl, so please bear with my simple question: Here is the sample output: Most successful agents in the Emarket climate are (in order of success): 1. agent10896761 ($-8008) 2. flightsandroomsonly ($-10102) 3. agent10479475hv ($-10663) Most successful agents in the Emarket climate are (in order of success): 1. agent10896761 ($-7142) 2. agent10479475hv ($-8982) 3. flightsandroomsonly ($-9124) I am interested only in agent names as well as their corresponding balances, so I am hoping to get the following output: agent10896761 -8008 flightsandroomsonly -10102 agent10479475hv -10663 agent10896761 -7142 agent10479475hv -8982 flightsandroomsonly -9124 For later processes. This is the code I've got so far: #!/usr/bin/perl -w open(MYINPUTFILE, $ARGV[0]); while(<MYINPUTFILE>) { my($line) = $_; chomp($line); # regex match test if($line =~ m/agent10479475/) { if($line =~ m/($-[0-9]+)/) { print "$1\n"; } } if($line =~ m/flightsandroomsonly/) { print "$line\n"; } } The second regex match has nothing wrong, 'cause that is printing out the whole line. However, for the first regex match, I've got some other output such like: $ ./compareResults.pl 3.txt 2. flightsandroomsonly ($-10102) 0479475 0479475 3. flightsandroomsonly ($-9124) 1. flightsandroomsonly ($-8053) 0479475 1. flightsandroomsonly ($-6126) 0479475 If I "escape" the braces like this if($line =~ m/\($-[0-9]+\)/) { print "$1\n"; } Then there is never a match for the first regex... So I stuck with a problem of making that particular regex work. Any hints for this? Many thanks in advance.

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• Why are references compacted inside Perl lists?

  - by parkan

  Putting a precompiled regex inside two different hashes referenced in a list: my @list = (); my $regex = qr/ABC/; push @list, { 'one' => $regex }; push @list, { 'two' => $regex }; use Data::Dumper; print Dumper(\@list); I'd expect: $VAR1 = [ { 'one' => qr/(?-xism:ABC)/ }, { 'two' => qr/(?-xism:ABC)/ } ]; But instead we get a circular reference: $VAR1 = [ { 'one' => qr/(?-xism:ABC)/ }, { 'two' => $VAR1->[0]{'one'} } ]; This will happen with indefinitely nested hash references and shallowly copied $regex. I'm assuming the basic reason is that precompiled regexes are actually references, and references inside the same list structure are compacted as an optimization (\$scalar behaves the same way). I don't entirely see the utility of doing this (presumably a reference to a reference has the same memory footprint), but maybe there's a reason based on the internal representation Is this the correct behavior? Can I stop it from happening? Aside from probably making GC more difficult, these circular structures create pretty serious headaches. For example, iterating over a list of queries that may sometimes contain the same regular expression will crash the MongoDB driver with a nasty segfault (see https://rt.cpan.org/Public/Bug/Display.html?id=58500)

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• Using \b in C# regular expressions doesn't work?

  - by Nikhil

  I am wondering why the following regex does not match. string query = "\"1 2\" 3"; string pattern = string.Format(@"\b{0}\b", Regex.Escape("\"1 2\"")); string repl = Regex.Replace(query, pattern, "", RegexOptions.CultureInvariant); Note that if I remove the word boundary characters (\b) from pattern, it matches fine. Is there something about '\b' that might be tripping this up?

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• In C#, is there any way to try multiple Regexes on string to see which one matches first?

  - by Matt

  Let's say I have an arbitrary list of regexes (IList<Regex> lst; for example). Is there any way to find out which one matches first? Of course there is the straightforward solution of trying each one on the string and seeing which match has the lowest index, but this could be inefficient on long strings. Of course I can go back and pull the strings back out of each regex (Regex.ToString()) and concatenate them all together ("(regex1)|(regex2)|(regex3)"), but I find this to be an ugly solution, especially since it does not even indicate which regex was matched.

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• PHP preg_replace() pattern, string sanitization.

  - by Otar

  I have a regex email pattern and would like to strip all but pattern-matched characters from the string, in a short I want to sanitize string... I'm not a regex guru, so what I'm missing in regex? <?php $pattern = "/^([\w\!\#$\%\&\'\*\+\-\/\=\?\^\`{\|\}\~]+\.)*[\w\!\#$\%\&\'\*\+\-\/\=\?\^\`{\|\}\~]+@((((([a-z0-9]{1}[a-z0-9\-]{0,62}[a-z0-9]{1})|[a-z])\.)+[a-z]{2,6})|(\d{1,3}\.){3}\d{1,3}(\:\d{1,5})?)$/i"; $email = 'contact<>@domain.com'; // wrong email $sanitized_email = preg_replace($pattern, NULL, $email); echo $sanitized_email; // Should be [email protected] ?> Pattern taken from: http://fightingforalostcause.net/misc/2006/compare-email-regex.php (the very first one...)

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• getting names subgroups

  - by Abruzzo Forte e Gentile

  Hi All I am working with the new version of boost 1.42 and I want to use regex with named sub groups. Below an example. std::string line("match this here FIELD=VALUE in the middle"); boost::regex rgx("FIELD=(?\\w+)", boost::regex::perl ); boost::smatch thisMatch; boost::regex_searh( line, thisMatch, rgx ); Do you know how to get the content of the match ? The traditional way is std::string result( mtch["VAL"].first, mtch["VAL"].second ); but i don't want to use this way. I want to use the name of the subgroups as usual in Perl and in regex in general. I tried this, but it didn't work. std::string result( mtch["VAL"].first, mtch["VAL"].second ); Do you know how to get the value using the name of the subgroup? Thanks AFG

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• Regular Expression to match IP address + wildcard

  - by Ed Woodcock

  Hey guys I'm trying to use a RegularexpressionValidator to match an IP address (with possible wildcards) for an IP filtering system. I'm using the following Regex: "([0-9]{1,3}\\.|\\*\\.){3}([0-9]{1,3}|\\*){1}" Which works fine when running it in LINQPad with Regex.Matches, but doesn't seem to work when I'm using the validator. Does anyone have a suggestion as to either a better Regex or why it would work in test but not in situ? Cheers, Ed

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• Sed non greedy curly braces match

  - by Cesar

  I have a string in a file a.txt {moslate}alho{/moslate}otra{moslate}a{/moslate} a need to get the string otra using sed. With this regex sed 's|{moslate}.*{/moslate}||g' a.txt a get no output at all but when i add a ? to the regex s|{moslate}.*?{/moslate}||g a.txt (I've read somewhere that it makes the regex non-greedy) i get no match at all, i mean a get the following output {moslate}alho{/moslate}otra{moslate}a{/moslate} How can i get the required output using sed?

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• group object with equal collections

  - by Jeroen

  Hi, Suppose 2 classes, Person and Pet. Each person has a collection of 1 or more pets. How do i group the Person in to a collection where they share the same pets. Example: Person 1: Cat, Dog, Spider Person 2: Cat, Spider, Snake Person 3: Dog Person 4: Spider, Cat, Dog Person 5: Dog What i want as a result is this: Group 1: Person 1, Person 4 Group 2: Person 3, Person 5 Group 3: Person 2 How do i achieve this using LINQ?

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• preg_match() Unknown modifier '[' help

  - by Jonathan

  Hi, I have this regex for getting the YouTube video ID: (?<=v=)[a-zA-Z0-9-]+(?=&)|(?<=[0-9]/)[^&\n]+|(?<=v=)[^&\n]+ I get it from there: http://stackoverflow.com/questions/2597080/regex-to-parse-youtube-yid The problem is I get preg_match() Unknown modifier '[' warning. I know I have to enclose the regex delimiters but I have no idea how to do this. Any help?

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• .NET Regular Expressions - Shorter match

  - by Xavier

  Hi Guys, I have a question regarding .NET regular expressions and how it defines matches. I am writing: var regex = new Regex("<tr><td>1</td><td>(.+)</td><td>(.+)</td>"); if (regex.IsMatch(str)) { var groups = regex.Match(str).Groups; var matches = new List<string>(); for (int i = 1; i < groups.Count; i++) matches.Add(groups[i].Value); return matches; } What I want is get the content of the two following tags. Instead it returns: [0]: Cell 1</td><td>Cell 2</td>... [1]: Last row of the table Why is the first match taking </td> and the rest of the string instead of stopping at </td>?

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• How to find &laquo; (&#171;) in a string

  - by Arnej65

  I need to find « in a string with a regex. How would you add that into the following: String RegExPattern = @"^[0-9a-df-su-z]+\.\s&#171"; Regex PatternRegex = new Regex(RegExPattern); return (PatternRegex.Match(Source).Value);

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• Is there a more concise regular expression to accomplish this task?

  - by mpminnich

  First off, sorry for the lame title, but I couldn't think of a better one. I need to test a password to ensure the following: Passwords must contain at least 3 of the following: upper case letters lower case letters numbers special characters Here's what I've come up with (it works, but I'm wondering if there is a better way to do this): Dim lowerCase As New Regex("[a-z]") Dim upperCase As New Regex("[A-Z]") Dim numbers As New Regex("\d") Dim special As New Regex("[\\\.\+\*\?\^\$\[\]\(\)\|\{\}\/\'\#]") Dim count As Int16 = 0 If Not lowerCase.IsMatch(txtUpdatepass.Text) Then count += 1 End If If Not upperCase.IsMatch(txtUpdatepass.Text) Then count += 1 End If If Not numbers.IsMatch(txtUpdatepass.Text) Then count += 1 End If If Not special.IsMatch(txtUpdatepass.Text) Then count += 1 End If If at least 3 of the criteria have not been met, I handle it. I'm not well versed in regular expressions and have been reading numerous tutorials on the web. Is there a way to combine all 4 regexes into one? But I guess doing that would not allow me to check if at least 3 of the criteria are met. On a side note, is there a site that has an exhaustive list of all characters that would need to be escaped in the regex (those that have special meaning - eg. $, ^, etc.)? As always, TIA. I can't express enough how awesome I think this site is.

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