0

我有一个脚本来注册用户或检查用户信息是否有任何错误，例如密码不匹配或用户是否已经存在等。

当我打印用户名是否已被占用的错误消息时，$errorMessage 的值始终为 0，我不知道为什么：

<?php


 //get user name and passwords


 $errorMessage = "Error: ";

 $successMessage;

 if ($p1 == ""){

    $errorMessage = $errorMessage + "You left your password blank. ";

    echo $errorMessage;

    $fail = $fail + 1;

    }
 if ($p2 == ""){


    $errorMessage = $errorMessage + "You did not type your password in a second time. ";

    $fail = $fail + 1;

    echo $errorMessage;

    }


 //hash passwords
 $hash1 = sha1($p1);
 $hash2 = sha1($p2);


 $fail = 0;

 //check if user name is taken



if($hash1 != $hash2){
     $fail = $fail + 1;

     $errorMessage = $errorMessage + "Your passwords do not match. ";
     echo $errorMessage;

     }



if ($numRows > 0){


    $errorMessage = $errorMessage + "That User name is already taken, please try another username. ";

    $fail = $fail + 1;
    echo $errorMessage;

    }


if ($userName == ""){


    $errorMessage = $errorMessage + "You left your username blank. ";

    $fail = $fail + 1;
    echo $errorMessage;

    }

if ($email == ""){


    $errorMessage = $errorMessage + "You left your email blank. ";

    $fail = $fail + 1;
    echo $errorMessage;

    }



//do not register user if fails
if( $fail > 0){

    //print fail message JSON
    $successMessage = "fail";



    }

//register user
else{
    //add user to database


//print json

    $d = array('status' => "$successMessage" , 'error' => "$errorMessage" );


    $jsonCode = json_encode($d);

    header('Content-Type: application/json');

    echo $jsonCode;



?>
4

3 回答 3

7

PHP 的 concat 运算符是句点或点 ( .) 而不是加号 ( +)。

于 2013-08-30T22:16:13.593 回答
0

您没有正确连接。在 PHP 中，您使用 .. 进行连接。我为你修复了代码：

 <?php


     //get user name and passwords


     $errorMessage = "Error: ";

     $successMessage;

     if ($p1 == ""){

        $errorMessage = $errorMessage."You left your password blank. ";

        echo $errorMessage;

        $fail = $fail + 1;

        }
     if ($p2 == ""){


        $errorMessage = $errorMessage."You did not type your password in a second time. ";

        $fail = $fail + 1;

        echo $errorMessage;

        }


     //hash passwords
     $hash1 = sha1($p1);
     $hash2 = sha1($p2);


     $fail = 0;

     //check if user name is taken



    if($hash1 != $hash2){
         $fail = $fail + 1;

         $errorMessage = $errorMessage."Your passwords do not match. ";
         echo $errorMessage;

         }



    if ($numRows > 0){


        $errorMessage = $errorMessage."That User name is already taken, please try another username. ";

        $fail = $fail + 1;
        echo $errorMessage;

        }


    if ($userName == ""){


        $errorMessage = $errorMessage."You left your username blank. ";

        $fail = $fail + 1;
        echo $errorMessage;

        }

    if ($email == ""){


        $errorMessage = $errorMessage."You left your email blank. ";

        $fail = $fail + 1;
        echo $errorMessage;

        }



    //do not register user if fails
    if( $fail > 0){

        //print fail message JSON
        $successMessage = "fail";



        }

    //register user
    else{
        //add user to database


    //print json

        $d = array('status' => "$successMessage" , 'error' => "$errorMessage" );


        $jsonCode = json_encode($d);

        header('Content-Type: application/json');

        echo $jsonCode;



    ?>
于 2013-08-30T22:49:22.213 回答
0

您必须像这样连接字符串：

$errorMessage = $errorMessage . "You left your password blank. ";
于 2013-08-30T22:17:46.793 回答