8000 Fix function example in expression language component by raulfraile · Pull Request #3771 · symfony/symfony-docs · GitHub
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Fix function example in expression language component #3771

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Fix function example in expression language component #3771

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bc673f3
Fix function example in expression language component
raulfraile Apr 7, 2014
db710c4
Removed return statement
raulfraile Jul 3, 2014
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Fix function example in expression language component 
The is_string check needs to be part of the runtime code, not of the compilation code, because it needs to check that the argument is a string, not the compiled code to access it (which is always a string as it is source code
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raulfraile committed Apr 7, 2014
commit bc673f3907f64d2946e75cf732ccf79866231f78
6 changes: 1 addition & 5 deletions components/expression_language/extending.rst
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Expand Up @@ -35,11 +35,7 @@ This method has 3 arguments:

$language = new ExpressionLanguage();
$language->register('lowercase', function ($str) {
if (!is_string($str)) {
return $str;
}

return sprintf('strtolower(%s)', $str);
return sprintf('$result = (is_string(%1$s)) ? strtolower(%1$s) : %1$s; return $result;', $str);
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don't put a return in the compiled expression code. It should simply be (is_string(%1$s) ? strtolower(%1$s) : %1$s)

}, function ($arguments, $str) {
if (!is_string($str)) {
return $str;
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