dsxon40042 2013-03-27 17:16
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我无法在PHP中获得返回值[关闭]

I can't get the return value! The Output appears 'ERROR'.

$c = new List();
$result = $c->create();

if( $result == "a"){
   echo("A");                   
}else if($result == "b"){
   echo("B");
}else{
   echo("ERROR");
}

class List{

    function create(){

       // $rVal = ...

       if($rVal == 1){
           return "a";
       }else if($rVal == 2){
           return "b";
       }
    }
}

when I change return to echo, the output appears 'aERROR'. I don't understand. create function works but the return value is not true. Can anyone help me please? Sorry for my bad English

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1条回答 默认 最新

  • duanken7168 2013-03-27 17:19
    关注

    If the code below is correct, then you are not setting $rVal. The assignment seems commented out:

    class List{
    
        function create(){
    
           // $rVal = ...
    
           if($rVal == 1){
               return "a";
           }else if($rVal == 2){
               return "b";
           }
        }
    }
    

    So, nothing (null) will be returned by create() call... which is interpreted as false.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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