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if else condition in Regex c# and return constant value


By : Degao Li
Date : September 29 2020, 05:00 AM
it helps some times , regex only:
code :
string str = "abc";
return Regex.IsMatch(str, "^.{3}$") ? "abc" : "xyz";
return Regex.IsMatch(str, "^abc$") ? "abc" : "xyz";


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using Regex Constant to return string between two tags


By : Nathan Niles
Date : March 29 2020, 07:55 AM
Does that help Your regex is good. You just need to extract group 1 (the thing you match between parenthesis)
In C# this becomes
code :
Regex regexObj = new Regex(@"\[BEGINTAG\](.+?)\[/ENDTAG\]");
resultString = regexObj.Match(subjectString).Groups[1].Value;
\[BEGINTAG](.+?)\[/ENDTAG]

Return only one group with OR condition in Regex


By : Matt Ryan
Date : March 29 2020, 07:55 AM
I wish this help you I have to write a Regex to fetch Email Address from a sentence. I want it to be returned with Group 1 only. , You may use regular expression:
code :
(?=\S+@)([^<\s]+@.*(?=[>\]]))

Why the code not working need to return True or False from the string if regex matches with condition


By : user3348797
Date : March 29 2020, 07:55 AM
like below fixes the issue From the doc search
code :
^(?:(?:-?\d{1,3}(?:\.\d{1,3}){3}(?:/\d+)?|\b(?:any|reject)\b),?)+$
^                   # beginning of line
  (?:               # start non capture group
    (?:             # start non capture group
      (?:           # start non capture group
        (?<=\d)     # positive lookbehind, make sure we have a digit before
        -           # an hyphen
      )?            # end group, optional
      \d{1,3}       # 1 up to 3 digits
      (?:           # start non capture group
        \.          # a dot
        \d{1,3}     # 1 up to 3 digits
      ){3}          # end group, must appear 3 times
      (?:           # start non capture group
        /           # a slash
        \d+         # 1 or more digits
      )?            # end group, optional
     |              # OR
      \b            # word boundary
      (?:           # start non capture group
        any         # literally  "any"
       |            # OR
        reject      # literally  "reject"
      )             # end goup
      \b            # word boundary
    )               # end group
    ,?              # optional comma
  )+                # end group, may appear 1 or more times
$                   # end of line
import re

test1 = '''192.168.1.1,192.168.1.2/32,192.168.1.5-192.168.1.7,reject,any,,,,
    reject,192.168.1.1/32,reject,any,
    172.168.1.4-172.168.1.4,reject'''
res = bool(re.search(r'^(?:(?:(?:(?<=\d)-)?\d{1,3}(?:\.\d{1,3}){3}(?:/\d+)?|\b(?:any|reject)\b),?)+$', test1))
print(res)
False

How to write hive regex to match condition 1 OR condition 2 and return whichever matches?


By : Sahar
Date : March 29 2020, 07:55 AM
wish of those help You can use lookarounds for this.
In your first case, you want to match three digits preceded by "foobar" (use lookbehind):

Ternary operator with constant variable in return statement. return a>b?a:b + 1; or return 1 + a>b?a:b;


By : Jose Linares
Date : March 29 2020, 07:55 AM
this one helps. In return a>b?a:b + 1, a>b?a:b is evaluated first; since a is not greater than b, it returns the value of b + 1 , which is 1.
In return 1 + a>b?a:b, the compiler treats the condition as (1 + a) > b. Since it's true, it returns the value of a, which is 0.
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