PHP变量在if语句中丢失范围

时间:2013-09-04 05:26:24

标签: php

Heloo Guys,我有一个非常讨厌的问题,我的变量一直在失去范围,似乎没有任何作用,我不确定是什么错误。

这个php的目的是让Let和Admin编辑用户的电子邮件,用户ID在链接中发送,这里收到,我选择用户向他显示当前的电子邮件,然后管理员通过一个新的电子邮件提交形式,它更新......

但是$ user_id变量在IF里面是空的..就像它将它识别为另一个变量一样,我尝试了var_dump ..它实际上是空的,尝试过全局和$ GLOBALS ..没有任何作用

$userid = $_POST['user_id'];

$mysqli_object = fQuery('users', 'SELECT', '*', null, array('user_id'), array($userid), null, null);

$mysqli_count = mysqli_num_rows($mysqli_object);

$user_id = '';
$account = '';
$name = '';
$email = '';

while($row = mysqli_fetch_array($mysqli_object)) {
    $user_id = $row['user_id'];
    $account = $row['account'];
    $name = $row['name'];
    $email = $row['email'];
}

if(isset($_POST['new_email'])) {        
    if(!filter_var($_POST['new_email'], FILTER_VALIDATE_EMAIL)) {
        header("Location: editEmail.php?R=2");
        die();        
    }

    if(!empty($_POST['new_email'])) {
        //NOTHING
    }else {
        header("Location: editEmail.php?R=2");
        die();    
    }

    $new_email = $_POST['new_email'];

    $mysqli_object = fQuery('users', 'UPDATE', array('email'), array($new_email), array('user_id'), array($user_id), null, null);

    if($mysqli_object) {
        header("Location: editEmail.php?R=1");
        die();                
    }else {
        header("Location: editEmail.php?R=3");
        die();    
    }   
}

0 个答案:

没有答案