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Every string between given strings


Every string between given strings

By : user2950636
Date : November 17 2020, 11:55 AM
hop of those help? From the documentation for RegExp.prototype.exec():
code :
var text = "first second1 third\nfirst second2 third\nfirst second3 third";
var middles = [], md, regex = /first (.*?) third/g;

while( md = regex.exec(text) ) { middles.push(md[1]); }

middles // ["second1", "second2", "second3"]


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PHP: Get strings between string A and string B (multiple strings need to be searched for)

PHP: Get strings between string A and string B (multiple strings need to be searched for)


By : rjh.hoffmann
Date : March 29 2020, 07:55 AM
I hope this helps you . You can use string functions. But for extraction purposes a simple regular expression is advisable for minimal structure verification:
code :
 preg_match_all('/[\s]id="nothread(\d+)"/', $html, $result);
 $numbers = $result[1];
Smart way to combine multiple Strings into a single String that can later be separated into the original Strings?

Smart way to combine multiple Strings into a single String that can later be separated into the original Strings?


By : ajinkya shewale
Date : March 29 2020, 07:55 AM
like below fixes the issue With your code, you can recover empty strings using the two-argument version of split:
code :
String[] separate(String string)
{
    return string.split(SEPARATOR, -1);
}
/** Returns a String guaranteed to have no NULL character. */
String escape(String source) {
    return source.replace("&", "&").replace("\u0000", "&null;");
}

/** Reverses the above escaping and returns the result. */
String unescape(String escaped) {
    return source.replace("&null;", "\u0000").replace("&", "&");
}
Find a string that matches another string on specific position, ignore other strings (C strings)

Find a string that matches another string on specific position, ignore other strings (C strings)


By : Nishan Gurung
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further Let's say I have an array of phone numbers: char numbers[4][9] = {"555555534", "112221222", "512221222", "355555534"}; I need to find there phone numbers, that matches a specific pattern. , The function can be like the following:
code :
int number_matches(char* number, char* pattern)
{
    int i = 0, flag = 1;
    for(i = 0; i < strlen(pattern); i++)
    {
         if(pattern[i] != '*')
         {
             if(pattern[i] != number[i])
             {
                 flag = 0;
                 break;
             }
         }
    }
    return flag;
}
Difference between string, cell array of strings, vector of strings, and matrix of strings in Matlab

Difference between string, cell array of strings, vector of strings, and matrix of strings in Matlab


By : jkings52
Date : March 29 2020, 07:55 AM
should help you out You are probably being confused by not realizing that you aren't really creating an array of strings, rather, you are creating a matrix of characters.
Your string_list is a 1x14 matrix of chars: my nameis xyz! (note you are missing a space between name and is). Then within your code fprintf(fid, '%s\t', string_list(k)) is taking each element of that string and printing it with a tab character between elements, which is what you are getting.
code :
if iscellstr(string_list)
   fprintf('%s\t',string_list{:});
else
   % just a char matrix
   fprintf('%s\t',string_list);
end
string_list = ["my", " ", "name", "is", " ", "xyz!"];
if isstring(string_list)
   fprintf('%s\t',string_list)
end
Sorting a list of string to show first strings with a particular substring and then the remaining strings in their sorte

Sorting a list of string to show first strings with a particular substring and then the remaining strings in their sorte


By : Blacklight MG
Date : March 29 2020, 07:55 AM
it should still fix some issue You're quite close: use your comparator, which performs ordering according to first criterion, then compound it with natural ordering comparator:
code :
Collections.sort(list, new Comparator<LoggerDetails>() {
    @Override
    public int compare(LoggerDetails o1, LoggerDetails o2) {
        if(o1.logger.startsWith("com.abc") && !o1.logger.startsWith("com.abc")) {
            return -1;
        }
        if (o1.logger.startsWith("com.abc") && !o1.logger.startsWith("com.abc")) {
            return -1;
        } -- I just copypasted your original code, this should definitely be vice versa
        return 0;
    }  
}.thenComparing(o -> o.logger, Comparator.naturalOrder()));
comparing(o -> o.logger.startsWith("com.abc"), Comparator.reverseOrder())
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