Asked  7 Months ago    Answers:  5   Viewed   30 times

Trying to replace all occurrences of an @mention with an anchor tag, so far I have:

$comment = preg_replace('/@([^@ ])? /', '<a href="/$1">@$1</a> ', $comment);

Take the following sample string:

"@name kdfjd fkjd as@name @ lkjlkj @name"

Everything matches okay so far, but I want to ignore that single "@" symbol. I've tried using "+" and "{2,}" after the "[^@ ]" which I thought would enforce a minimum amount of matches, but it's not working.

 Answers

38

Replace the question mark (?) quantifier ("optional") and add in a + ("one or more") after your character class:

@([^@ ]+)
Wednesday, March 31, 2021
 
Anand
answered 7 Months ago
38

You need to specify a matching group using brackets in order to extract content.

preg_match("//([^/]+)$/", "5020203010101/FIS/CASH FUND/SBG091241212", $matches);

echo $matches[1];
Wednesday, March 31, 2021
 
exxed
answered 7 Months ago
11

I would not recommend using regular expressions for a task like this.
A simple function to check a string for properly written brackets is quickly written:

/**
* @param $str input string
* @returns boolean true if all brackets are properly opened and closed, false otherwise
*/
function checkBraces($str)
{
    $strlen = strlen($str); // cache string length for performance
    $openbraces = 0;

    for ($i = 0; $i < $strlen; $i++)
    {
        $c = $str[$i];
        if ($c == '<') // count opening bracket
            $openbraces++;
        if ($c == '>') // count closing bracket
            $openbraces--;

        if ($openbraces < 0) // check for unopened closing brackets
            return false;
    }

    return $openbraces == 0; // check for unclosed open brackets
}

Using this code as a basis, it shouldn't be too hard to implement a check to verify whether or not the tag name of opening and closing brackets also matches - but I'll leave that to you :-)

Wednesday, March 31, 2021
 
Sauleil
answered 7 Months ago
27

Use !==FALSE instead of !=FALSE. The problem is that stripos returns 0 if the needle is at the start of the haystack, and 0 is falsy. By using !== you are forcing it to ensure the result is actually false, and not just 0.

This is actually listed in the docs. An "RTM" might be appropriate here.

Warning
This function may return Boolean FALSE, but may also return a non-Boolean value which evaluates to FALSE. Please read the section on Booleans for more information. Use the === operator for testing the return value of this function.

Saturday, May 29, 2021
 
HamidR
answered 5 Months ago
68

This should do it

old_string = "this is going to have a full stop. some written sstuff!"
k = old_string.rfind(".")
new_string = old_string[:k] + ". - " + old_string[k+1:]
Saturday, September 4, 2021
 
sassy_geekette
answered 2 Months ago
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