From the following string:
google.com/local/reviews?placeid\\u003dChIJ070npYRaeEgRZNoxwuYYrew\\u0026q\\u003d
To extract u003dChIJ070npYRaeEgRZNoxwuYYrew although this value will change every time.
I have tried
Regex r = new Regex(#"("(?<=\placeid\\\s+)\p{L}+");
Which does not work.
I am guilty of neglecting my knowledge is regex so I apologise if this is painfully easy.
There are no whitespace chars in the string that you want to match with \s+ and there are 2 backslashes.
Using \p{L}+ only matches any letter and the string that you want also contains numbers.
(?<=placeid\\\\\s*)[\p{L}\p{N}]+
Regex demo
For example
string pattern = #"(?<=placeid\\\\\s*)[\p{L}\p{N}]+";
string input = #"google.com/local/reviews?placeid\\u003dChIJ070npYRaeEgRZNoxwuYYrew\\u0026q\\u003d";
Match m = Regex.Match(input, pattern);
Console.WriteLine(m.Value);
Output
u003dChIJ070npYRaeEgRZNoxwuYYrew
I would like to know how to extract complete words using a Regex in C#
For example, my String input:
This$#23 is-kt jkdls
I want to get Regex Match as
This$#23
is-kt
jkdls
I need to extract non space words [which can have numbers or special characters]
by specifying Regex Match pattern
Regex myrex = new Regex("pattern")
MatchCollection matches = Regex.Matches("This$#23 is-kt jkdls", #"\S+");
foreach(Match match in matches)
Console.WriteLine(match.Value);
Use \S+ to match words.
var words = string.Split(' ');
Using DirectoryServices.AccountManagement I'm getting users DistinguishedName which looks like so:
CN=Adam West,OU=STORE,OU=COMPANY,DC=mycompany,DC=group,DC=eu
I need to get first OU value from this.
I found similar solution: C# Extracting a name from a string
And using some tweaks I created this code:
string input = #"CN=Adam West,OU=STORE,OU=COMPANY,DC=mycompany,DC=group,DC=eu";
Match m = Regex.Match(input, #"OU=([a-zA-Z\\]+)\,.*$");
Console.WriteLine(m.Groups[1].Value);
This code returns STORE as expected, but if I change Groups[1] to Groups[0] I get almost same result as input string:
OU=STORE,OU=COMPANY,DC=mycompany,DC=group,DC=eu
How can I change this regex so it will return only values of OU? SO that in this example I get array of 2 matches. If I would have more OU in my string then array would be longer.
EDIT:
I've converted my code (using #dasblinkenlight suggestions) into function:
private static List<string> GetOUs()
{
var input = #"CN=Adam West,OU=STORE,OU=COMPANY,DC=mycompany,DC=group,DC=eu";
var mm = Regex.Matches(input, #"OU=([a-zA-Z\\]+)");
return (from Match m in mm select m.Groups[1].Value).ToList();
}
Is that correct?
Your regular expression is fine (almost), you are just using a wrong API.
Remove the parts of the regexp that match up to the ending anchor $, and change the call of Match for a call of Matches, and get the matches in a loop, like this:
var input = #"CN=Adam West,OU=STORE,OU=COMPANY,DC=mycompany,DC=group,DC=eu";
var mm = Regex.Matches(input, #"OU=([a-zA-Z\\]+)");
foreach (Match m in mm)
Console.WriteLine(m.Groups[1].Value);
}
Your existing regex:
#"OU=([a-zA-Z\\]+)\,.*$"
Matches OU=, then some letters and backslashes ([a-zA-Z\\]+), then a comma, then any characters (.*) to the end of the line ($).
Thus a single match will always match the entire line after the first OU section.
Modify your regex by removing the ,.*$ at the end, at it will match each OU group:
#"OU=([a-zA-Z\\]+)"
Also note that the parentheses are a capturing group. They are useful if you also want to capture just the value part by itself, but if you are not using that, they are not necessary, and you can just have this:
#"OU=[a-zA-Z\\]+"
It's beacuse you are mixing up matches and groups
string input = #"CN=Adam West,OU=STORE,OU=COMPANY,DC=mycompany,DC=group,DC=eu";
MatchCollection mc = Regex.Matches(input, #"OU=([a-zA-Z\\]+),");
foreach(Match m in mc)
{
Console.WriteLine(m.Result("$1"));
}
Group[0] returns the full match:
Group[1] returns the first Pattern in the match [i.e. everything in the first parenthesis '(' ')' ]
So if you wanted to get exactly those 2 occurances of OU.. you could do this:
Match m = Regex.Match(input, #"OU=([a-zA-Z\\]+)\,OU=([a-zA-Z\\]+)\,.*$");
Console.WriteLine(m.Groups[1].Value);
Console.WriteLine(m.Groups[2].Value);
Group[0] returns the full match: (which you don't want)
Group[1] returns the first Pattern in the match [i.e everything in the first parenthesis '(' ')' ]
Group[2] returns the second Pattern in the match [i.e. everything in the second parenthesis '(' ')' ]
Giving:
STORE
COMPANY
But I'm assuming you don't want to be so explicit with your Regex for each Pattern you are interested in.
If you want to get multiple matches, then you need to do Regex's Matches call that returns a Matchcollection.
MatchCollection ms = Regex.Matches(...);
This still won't work with your current Regex though, because everything from STORE so the end of the line will be in the first match. If you only want to get the pattern "1-or-more-letters" after a "OU="
You only need:
#"OU=([a-zA-Z\\]+)"
So your code would be:
string input = #"CN=Adam West,OU=STORE,OU=COMPANY,DC=mycompany,DC=group,DC=eu";
MatchCollection ms = Regex.Matches(input, #"OU=([a-zA-Z\\]+)");
foreach (Match m in ms)
{
Console.WriteLine(m.Groups[1].Value);// get the string in the first "(" ")"
}
Howzit,
I need help with the following please.
I need to find tags in a string. These tags start with {{ and end with }}, there will be multiple tags in the string I receive.
So far I have this, but it doesn't find any matches, what am I missing here?
List<string> list = new List<string>();
string pattern = "{{*}}";
Regex r = new Regex(pattern, RegexOptions.IgnoreCase);
Match m = r.Match(text);
while (m.Success)
{
list.Add(m.Groups[0].Value);
m = m.NextMatch();
}
return list;
even tried string pattern = "{{[A-Za-z0-9]}}";
thanx
PS. I know close to nothing about regex.
Not only do you want to use {{.+?}} as your regex, you also need to pass RegexOptions.SingleLine. That will treat your entire string as a single line and the . will match \n (which it normally will not do).
Try {{.+}}. The .+ means there has to be at least one character as part of the tag.
EDIT:
To capture the string containing your tags you can do {{(.+)}} and then tokenize your match with the Tokenize or Scanner class?
I would recommend trying something like the following:
List<string> list = new List<string>();
string pattern = "{{(.*?)}}";
Regex r = new Regex(pattern, RegexOptions.IgnoreCase);
Match m = r.Match(text);
while (m.Success)
{
list.Add(m.Groups[1].Value);
m = m.NextMatch();
}
return list;
the regex specifies:
{{ # match {{ literally
( # begin capturing into group #1
.*? # match any characters, from zero to infinite, but be lazy*
) # end capturing group
}} # match }} literally
"lazy" means to attempt to continue matching the pattern afterwards "}}" before backtracking to the .*? and reluctantly adding a character to the capturing group only if the character does not match }} - hope that made sense.
I changed your code by modifying the regex and to extract the first matching group from the regex match object (m.Groups[1].value) instead of the entire match.
{{.*?}} or
{{.+?}}
. - means any symbol
? - means lazy(don't capute nextpattern)
i have a problem in string manipulation
here is the code
string str = "LDAP://company.com/OU=MyOU1 Control,DC=MyCompany,DC=com";
Regex regex = new Regex("OU=\\w+");
var result = regex.Matches(str);
var strList = new List<string>();
foreach (var item in result)
{
strList.Add(item.ToString().Remove(0,3));
}
Console.WriteLine(string.Join("/",strList));
the result i am getting is "MyOU1" instead of getting "MyOU1 Control"
please help thanks
If you want the space character to be matched as well, you need to include it in your regex. \w only matches word charactes, which does not include spaces.
Regex regex = new Regex(#"OU=[\w\s]+");
This matches word characters (\w) and whitespace characters (\s).
(The # in front of the string is just for convenience: If you use it, you don't need to escape backslashes.)
Either add space to the allowed list (\w doesn't allow space) or use the knowledge that comma can be used as a separator.
Regex regex = new Regex("OU=(\\w|\\s)+");
OR
Regex regex = new Regex("OU=[^,]+");