Why does php str_replace with multiple arrays give wrong result, but for loop gives correct result?

Question

I'm trying to replace the characters (numbers and letters) in a string. When I try the "php" way, it gives the wrong result for some of the characters. Why?

PHP-WAY:

$find = array( "0","1","2","3","4","5","6","7","8","9","a","b","c","d","e","f" );
$replace = array( "a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p" );
$haystack = "a5c9a06bfacf5f12cf01ab3f202f6c78"

//This incorrectly returns: kpmjkkglpkmppplmmpklklnpmkmpgmhi
echo str_replace( $find, $replace, $haystack );

LOOP WAY:

$find = array( "0","1","2","3","4","5","6","7","8","9","a","b","c","d","e","f" );
$replace = array( "a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p" );
$haystack = "a5c9a06bfacf5f12cf01ab3f202f6c78"

//This correctly returns: kfmjkaglpkmpfpbcmpabkldpcacpgmhi
$newStr = "";
$chars = str_split( $haystack );
for ( $i = 0, $length = count( $chars ); $i < $length; $i++ )
{
    $newStr .= $replace[ array_search( $chars[ $i ], $find ) ];
}
echo $newStr;

Why is the first one incorrect? Am I using it wrong?

Order of entries in your arrays.... str_replace() will process each array entry in the order they appear in your array, so if a '1' gets replaced with 'b', then that 'b' will subsequently get replaced with 'l'; use strtr() (uk3.php.net/manual/en/function.strtr.php) rather than str_replace() if you want to prevent that behaviour — Mark Baker
– Mark Baker, Commented May 28, 2013 at 22:45

Community · Accepted Answer · 2023-11-17 20:29:45Z

3

Order of entries in your arrays.... str_replace() will process each array entry in the order they appear in your array, so if a '1' gets replaced with 'b', then that 'b' will subsequently get replaced with 'l'; use strtr() rather than str_replace() if you want to prevent that behaviour.

$find = array( "0","1","2","3","4","5","6","7","8","9","a","b","c","d","e","f" );
$replace = array( "a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p" );
$haystack = "a5c9a06bfacf5f12cf01ab3f202f6c78" ;
echo strtr($haystack, array_combine($find, $replace));

Your own code only does a single replace because it's looping against your string, not against the from/to arrays.

edited Nov 17, 2023 at 20:29

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answered May 28, 2013 at 22:48

Mark Baker

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1 Comment

Don Rhummy Over a year ago

Can you please add in the sample code (that I added in my answer)? Then i'll mark yours as the answer. Thanks.

Baba · Accepted Answer · 2013-05-28 23:01:39Z

1

Just use strtr

$haystack = "a5c9a06bfacf5f12cf01ab3f202f6c78" ;
echo strtr($haystack, implode($find), implode($replace));

Or preg_replace_callback

$find = array_flip($find);
echo preg_replace_callback('/[a-f0-9]/', function ($v) use($replace, $find) {
    return $replace[$find[$v[0]]];
}, $haystack);

Output

kfmjkaglpkmpfpbcmpabkldpcacpgmhi

edited May 28, 2013 at 23:01

answered May 28, 2013 at 22:48

Baba

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Comments

Don Rhummy · Accepted Answer · 2013-05-28 22:48:13Z

0

As specified by @MarkBaker, the answer is that str_replace does not simply move forward in the string, but instead works like a recursive .replace(). Instead, use strtr (which is equivalent to Linux tr command:

$tr = array( "0" => "a","1" => "b","2" => "c","3" => "d","4" => "e","5" => "f","6" => "g","7" => "h","8" => "i","9" => "j","a" => "k","b" => "l","c" => "m","d" => "n","e" => "o","f" => "p" );
$haystack = "a5c9a06bfacf5f12cf01ab3f202f6c78"

echo strtr( $haystack, $tr );

answered May 28, 2013 at 22:48

Don Rhummy

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