PHP函数返回NULL

Question

I have the following function:

function MyReplaceListTag($myText,$number){ 

    if (strpos($myText,'<li>')===FALSE){

        $myResult= strip_tags($myText);
        return $myResult;

    }else{
        $number++;
        $pattern= '/<li>/';
        $replacement = "\n".$number."- ";
        $myText=preg_replace (  $pattern ,  $replacement ,  $myText , 1 );

        MyReplaceListTag($myText,$number);


    }
}

I call it with:

$result = MyReplaceListTag( $testTEXT,0);

Nothing is returned, var_dump($result) gives NULL.

I must be doing something incredibly stupid, but what?

Answer 1

Your function is returning NULL on the else block because it's lacking a return statement.

Change the else block from:

MyReplaceListTag($myText,$number);

to

return MyReplaceListTag($myText,$number);

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Ideone Demo