Problem with snippet call

3 Posts

dr_dru Reply #1, 12 years, 2 months ago

Hi there,
I have a problem.
For example, I created simple snippet "Test":

Who knows why in the snippet call, at the end of "Hello World" appears "1"?

This is a problem in the server or engine?

Thanks in advance for your help.

This question has been answered by m.engel. See the first response.

[ed. note: dr_dru last edited this post 12 years, 2 months ago.]

discuss.answer

128 Posts

Michael Engel Reply #2, 12 years, 2 months ago

I think MODX 2.3.1 fix this.
But the main Problem is that snippets should have to return values, not just echo it. And if you just echo there should an return ''; in the end.

<?php
return "Hello world!";

or

<?php
echo "Hello world!";
return '';

does exactly what you want, even in 2.3.0

3 Posts

dr_dru Reply #3, 12 years, 2 months ago

I recently started to learn php

thank you

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24,524 Posts

sottwell Reply #4, 12 years, 2 months ago

You can just use return; it doesn't need to be return '';

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24,544 Posts

BobRay Reply #5, 12 years, 2 months ago

I prefer return ''; as a matter of personal preference.

BTW, you generally should not use echo or print in a MODX snippet. Instead, do this:

<?php
$output = "Hello world!";
return $output;

If you need to return more than one line, do this:

$output = '<p>Line one</p>';
$output .= '<p>Line two</p>';
$output .= '<p>Line three</p>';

return $output;

It's also good form to "initialize" the variable at the top of the snippet:

$output = '';

Here's an example showing why you should do that:

<?php
$docSize = $modx->getOption('docSize', $scriptProperties, 0);

if ($docSize > 50) {
   $output = 'Large';
} elseif ($docSize < 50) {
   $output = 'Small';
}

return $output;

If docSize is exactly 50, neither of the statements that set $output will execute and the last line will throw an "Undefined Variable $output" PHP warning if E_NOTICE errors are on because $output is never set to a value.

Similarly all array variables should be initialized with something like this:

$myArray = array();

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Answered Problem with snippet call