unexpected behaviour of str_replace with array

Question

I've the following function and using str_replace gives an unexpected result

function repo($text) {
    $search = array("0","1","2","3","4","5","6","7","8","9");
    $replace = array("z30","z31","z32","z33","z34","z35","z36","z37","z38","z99");
    $text = str_replace($search,$replace,$text);
    return $text;
}

echo repo('0');

The expected answer is

z30

and instead I get

zz330

What am I doing wrong?

Vahe Shadunts · Accepted Answer · 2013-03-01 19:19:27Z

5

Your function works this way.

0 changes to z30, php continues loop the arrays, then z30 contains '3', and 3 changes to z33. Because of that returns 'z' + 'z33' + '0' = zz330.

answered Mar 1, 2013 at 19:19

Vahe Shadunts

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Atze Kaputtnik · Accepted Answer · 2013-03-01 19:33:08Z

2

Just like the documentation says:

Because str_replace() replaces left to right, it might replace a previously inserted value when doing multiple replacements.

You probably want to do something like this instead:

function repo($text){
    $search = array("0","1","2","3","4","5","6","7","8","9");
    $replace = array("z30","z31","z32","z33","z34","z35","z36","z37","z38","z99");
    $replacePairs = array_combine($search, $replace);
    return strtr($text, $replacePairs);
}

answered Mar 1, 2013 at 19:33

Atze Kaputtnik

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