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Replace URI prefix from Base64String using Regex


By : Nimzo
Date : October 14 2020, 08:10 PM
like below fixes the issue I'm working with JSZip download and there are some photo URI data which should be downloaded by FileSaver. , You can try this
code :
^.*base64,(?=\/)


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How replace an optionally occuring prefix with regex


By : Javier
Date : March 29 2020, 07:55 AM
I wish this help you Just like the way you made the last & optional using ?, you can make the first one also optional:
code :
&?phoneNum=[^&]+&?

RegEx for SQL Server string to replace with unicode prefix


By : Martial Jefferson
Date : March 29 2020, 07:55 AM
I hope this helps . In my opinion, there is a better tool for your job - the TSql100Parser class:
code :
using Microsoft.Data.Schema.ScriptDom;
using Microsoft.Data.Schema.ScriptDom.Sql;
using System.Collections.Generic;
using System.IO;
using System.Linq;

class Program
{
    static void Main(string[] args)
    {
        IList<ParseError> errors = new List<ParseError>();
        var tsql = @"
                Insert into SomeTable (someColumns) 
                values ('someValue1', 
                        N'someValue2', 
                        'someValue3',
                        'some '' escaped')";
        var result = GetLiterals(tsql);
    }

    private static List<string> 
        GetLiterals(string strQuery)
    {
        var parser = new TSql100Parser(false);
        IList<ParseError> errors = new List<ParseError>();
        var result = 
            parser.GetTokenStream(new StringReader(strQuery), errors);
        return result
            .Where(t =>
                t.TokenType == TSqlTokenType.AsciiStringLiteral ||
                t.TokenType == TSqlTokenType.UnicodeStringLiteral)
            .Select(t => t.Text)
            .ToList();
    }
}

How do I prefix elements in an array with a $ symbol, using regex and -replace


By : Lee Murphy
Date : March 29 2020, 07:55 AM
it fixes the issue I'm looking to prefix all elements in an array with a dollar symbol. Example: , You can replace the start of the string (represented by ^) with $:
code :
PS > $array = "group1","group2"    
PS > $array -replace "^","$"
$group1
$group2    
PS > 

RegEx to replace prefix and postfix


By : user3505215
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further This regex should do the trick, but you always should specify the language/framework you're using, because not all regex engines support the same features.
The number that you want to capture would be in capture group #3 ((\d+)), which most languages reference as \3
code :
(?:KnownString)([a-zA-Z])(.*?)-(\d+)\#\.[a-zA-Z]+
 (?:                     # Opens NCG
     KnownString         # Literal KnownString
 )                       # Closes NCG
 (                       # Opens CG1
     [a-zA-Z]            # Character class (any of the characters within)
                           # Anything between a and z
                           # Anything between A and Z
 )                       # Closes CG1
 (                       # Opens CG2
     .*?                 # . denotes any single character, except for newline
                           # * repeats zero or more times
                           # ? as few times as possible
 )-                      # Closes CG2
                           # Literal -
 (                       # Opens CG3
     \d+                 # Token: \d (digit)
                           # + repeats one or more times
 )                       # Closes CG3
 \#                      # Literal #
 \.                      # Literal .
 [a-zA-Z]+               # Character class (any of the characters within)
                           # Anything between a and z
                           # Anything between A and Z
                           # + repeats one or more times

How to replace prefix and postfix with regex in python


By : Sowndarya Murugesan
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further As anubhava has already commented, you should use groups as this:
code :
replaced = re.sub(r'(<h1>)(.*?)(</h1>)', r'<h2>\2</h2>', s)
replaced = re.sub(r'<h1>(.*?)</h1>', r'<h2>\1</h2>', s)
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