regex preg_replace in foreach loop

Question

I have the below simple replace working

<?php
$mydata= '15-2003';
$pattern = '/[-]/';
$replacement = ' ';
echo preg_replace($pattern, $replacement, $mydata);
?>

Which outputs 15 2003

However when I put it in my foreach loop it doesn't seem to work?

I have this

<?php foreach ($tests as $test): ?>
<tr>
<?php
$mydata= htmlout($test['f']);
$pattern = '/[-]/';
$replacement = '';
echo preg_replace($pattern, $replacement, $mydata);
?>
<?php endforeach; ?>

Which outputs 15-2003 Where am I going wrong here?

htmlout is the below custom function.

<?php
function html($text)
{
return htmlspecialchars($text, ENT_QUOTES, 'UTF-8');
}
function htmlout($text)
{
echo html($text);
}

When I do var_dump($mydata); I get NULL

For such a simple replacement, str_replace() will do the job just fine. I also note that in your looping example you are replacing with an empty string instead of a space like your first example. — DaveRandom
– DaveRandom, Commented Jul 16, 2012 at 12:38
what is htmlout and what does it return? (if this is converting to unicode you may not be replacing the same hyphen you think you are...) — Brad Christie
– Brad Christie, Commented Jul 16, 2012 at 12:39
First, for such a simple replace you should use str_replace, which is faster. Can you do a simple var_dump($mydata); ? — Ugo Méda
– Ugo Méda, Commented Jul 16, 2012 at 12:40

Robin Castlin · Accepted Answer · 2012-07-16 13:00:17Z

1

This doesn't work as intended because htmlout() echoes the value instead of returning it.

Consider replacing

$mydata= htmlout($test['f']);

with

$mydata= html($test['f']);

What happends in your code is that it simply prints out the original string, returns NULL to $mydata and then you echo a NULL which doesn't show anything.

answered Jul 16, 2012 at 13:00

Robin Castlin

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1 Comment

ak85 Over a year ago

Thanks I understand now. Now I have this working I will change from preg_replace to str_replace()

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