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Replace before and after string in vim


Replace before and after string in vim

By : user2948894
Date : November 15 2020, 06:54 AM
help you fix your problem I have a large XML file that has essential information commented out for whatever dumb reason the author decided to do it.
code :
:%s#\v(\>|/\>)\s*\<!--\s*(.{-})\s*--\># name="\2"\1#
:%s#\v(/?\>)\s*\<!--\s*(.{-})\s*--\># name="\2"\1#


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How to replace two different expressions that match start of the string by null using string.replace in javascript

How to replace two different expressions that match start of the string by null using string.replace in javascript


By : CBoz
Date : March 29 2020, 07:55 AM
will help you I have to replace the mime at the start of the string by null.. The two mime regular expressions to match the start of the string are /^data:audio/x-ms-wma\w*;base64,/ , Use | to list alternatives in a regular expression.
code :
var b64Data = text.replace(/^data:audio\w+;base64,|^data:audio\/x-ms-wma\w*;base64,/,'');
Replace part of a string in python multiple times but each replace increments a number in the string by 1

Replace part of a string in python multiple times but each replace increments a number in the string by 1


By : user2977726
Date : March 29 2020, 07:55 AM
it should still fix some issue If you use re.sub() to do the replacing, you can pass a function as the second argument rather than a plain string. The function will receive a match object based on what the regex matched. You can then use some thinking like itertools.count() (or your own object) to produce increasing numbers.
For example:
code :
import re
from itertools import count

button_number_count = 1;
htmlString = "this is sometext with more sometext and yet another sometext"

counter = count(button_number_count)

// replace sometext with sometext1, sometext2...
new_string = re.sub(r'sometext', lambda x: x.group(0) + str(next(counter)), htmlString )
'this is sometext1 with more sometext2 and yet another sometext3'
Is there a shorter way to replace some chars in a string instaed of calling String.replace successively?

Is there a shorter way to replace some chars in a string instaed of calling String.replace successively?


By : user3064603
Date : March 29 2020, 07:55 AM
To fix the issue you can do I don't think there is one you can use from the standard library, but you can use replaceChars from the apache StringUtils library, the docs can be found here.
code :
public static String replaceChars(String str,
                              String searchChars,
                              String replaceChars)
Dotnet Core - Getting a warning on 'string.Replace(string, string?)' saying use 'string.Replace(string, string?, System.

Dotnet Core - Getting a warning on 'string.Replace(string, string?)' saying use 'string.Replace(string, string?, System.


By : Tony
Date : March 29 2020, 07:55 AM
this will help just... tell it what comparison type you want; for example, for an ordinal-ignore-case replace:
code :
    .Replace("and", "&", StringComparison.OrdinalIgnoreCase)
    .Replace("substringof", string.Empty, StringComparison.OrdinalIgnoreCase)
    .Replace("(", string.Empty, StringComparison.OrdinalIgnoreCase)
    .Replace(")", string.Empty, StringComparison.OrdinalIgnoreCase)
    .Replace("'", string.Empty, StringComparison.OrdinalIgnoreCase)
    .Replace(" ", string.Empty, StringComparison.OrdinalIgnoreCase)
    .Replace("eq", ",", StringComparison.OrdinalIgnoreCase);
using php preg_replace to replace a whole character string and not replace if string is part of a longer string

using php preg_replace to replace a whole character string and not replace if string is part of a longer string


By : user3642488
Date : March 29 2020, 07:55 AM
wish help you to fix your issue The word boundary after % requires a word char (letter, digit or _) to appear right after it, so there is no replacement taking place here.
You need to replace the word boundaries with unambiguous boundaries defined with the help of (? code :
$value='0.00%';
$str = 'Price: 0.00%';
echo preg_replace('/(?<!\w)' . preg_quote($value, '/') . '(?!\w)/i', '- ', $str);
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