String Replace in php not working as expected

debugcn 投稿 Dev

user1989

I have a String as follows:

$var = "Hello Hi World it is an awesome day. Test Hi guys Whats up";

I am trying to replace the value in my string as follows:

function get_string_between($string, $start, $end){
        $string = " ".$string;
        $ini = strpos($string,$start);
        if ($ini == 0) return "";
        $ini += strlen($start);
        $len = strpos($string,$end,$ini) - $ini;
        return substr($string,$ini,$len);
        } 

$parsed0 = get_string_between($var, "Hello", "World");

So I get the String in `$parse0`

And I am replacing as follows:

$varnew = str_replace($parsed0,"NEWVALUE", $var);

echo $varnew;

I get output as follows:

Hello NEWVALUE World it is an awesome day. Test NEWVALUE guys Whats up

My expected result is:

Hello NEWVALUE World it is an awesome day. Test Hi guys Whats up.

This is happening because in $parse0 I get result as Hi and it contains two times in that String.

Is there any way I get start and and end positions using a function and replace it using those?

Jigar

I would suggest preg_replace_callback. Check the code below.

function customReplace($string, $start, $end, $value = "NEWVALUE"){
    return preg_replace_callback('/'.$start.'(.*?)'.$end.'/', function($m) use($start, $end){
        return "$start NEWVALUE $end";
    }, $string);
}

$var = "Hello Hi World it is an awesome day. Test Hi guys Whats up";

echo customReplace($var, "Hello", "World");

Output:

Hello NEWVALUE World it is an awesome day. Test Hi guys Whats up

Problem with your code is:

$varnew = str_replace($parsed0,"NEWVALUE", $var);

$parsed0 has Hi (which you fetched from the function), and saying str_replace to replace Hi with NEWVALUE in the string $var(whole string). So str_replace is working as expected. Hope it is clear now.

UPDATE

Also check @b263 answer below. You can use preg_replace too, it doesn't require a callback.

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