dragonmeng2002 2014-11-24 10:15
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检查功能输出的其他方法?

A question from an obvious PHP newbie here.

I have a class that has two functions inside. I am trying to check the output of the first function using the second function (like in the code below). The problem with this is the output (1) is already being displayed in the if-statement line. Here's the result:

1Flag is: 1

So the question is, how can I do this without the output being displayed inside the if statement?

class Test{

    public $flag;

    private function func_one(){
        $this->flag = 1;
        echo $this->flag;
        return $this->flag;
    }

    public function display_func(){
        if ( !empty( $this->func_one() ) ){
            echo 'Flag is: ';
            $this->func_one();
        }

        // if ( $this->func_one() === 1 ){
        //  echo 'Flag is: ';
        //  $this->func_one();  
        // }

        //var_dump( $this->func_one() );
    }
}

$classTest = new Test();
$classTest->display_func();
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6条回答 默认 最新

  • dsa1234569 2014-11-24 10:19
    关注

    Store the return val in a variable and run comparisons on that

    if ( !empty( $this->func_one() ) ){
        echo 'Flag is: ';
        $this->func_one();
    }
    

    Should be

    $val=$this->func_one();
    if (!empty($val)){
          echo "Flag is: $val";
    }
    

    Also remove the echo from func_one(), just return the value. This allows you to call the method only once and make your decisions based on the value that was returned.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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