I am getting this error in this PHP code on line 3, what could be wrong? This code has been taken from php manual user notes by frank at interactinet dot com
public function myMethod()
{
return 'test';
}
public function myOtherMethod()
{
return null;
}
if($val = $this->myMethod())
{
// $val might be 1 instead of the expected 'test'
}
if( ($val = $this->myMethod()) )
{
// now $val should be 'test'
}
// or to check for false
if( !($val = $this->myMethod()) )
{
// this will not run since $val = 'test' and equates to true
}
// this is an easy way to assign default value only if a value is not returned:
if( !($val = $this->myOtherMethod()) )
{
$val = 'default'
}
?>
Answer
The public keyword is used only when declaring a class method.
Since you're declaring a simple function and not a class you need to remove public from your code.
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