-2

1 - 更新我的 mysql 客户端的脚本 2 - 当我离开（其中 id= id”;）时，我将另一个 id 加倍只是让它们进行相同的更改。当我编辑客户端的配置文件时。我知道问题但不知道放什么，我尝试了很多选项，但仍然无法正常工作，3 - 这是我的脚本：

<?php
include('../conect.php');
if(isset($_POST['update']))
   // Get values from form
$id=$_POST['id'];
$username=$_POST['username'];
$utilizator=$_POST['utilizator'];
$password=$_POST['password'];
$nivel=$_POST['nivel'];
$departament=$_POST['departament'];
$location=$_POST['location'];
$country=$_POST['country'];
$email=$_POST['email'];
$ip=$_POST['ip'];


$query = "UPDATE utilizatori SET username = '$username', utilizator = '$utilizator', password = '$password', nivel = '$nivel', departament = '$departament', location = '$location', country = '$country', email = '$email', ip = '$ip' where id= id";
$res = mysql_query($query);
mysql_query($update);
    echo $update;


mysql_query($query);
echo "Record Updated";

header('location:../user.php');
// close connection
mysql_close(); 
?>
4

2 回答 2

0

您没有引用$id变量，而是引用了id中断查询语句的字符串文本。添加一个美元符号来引用变量，如果它不是整数，也用单引号括起来。

$query = "UPDATE utilizatori SET username = '$username', {...}, ip = '$ip' where id= '$id' ";

于 2013-05-29T16:33:55.077 回答
0

试试这个：

    <?php
include('../conect.php');
if(isset($_POST['update'])){
// Get values from form
$id=$_POST['id'];
$username=$_POST['username'];
$utilizator=$_POST['utilizator'];
$password=$_POST['password'];
$nivel=$_POST['nivel'];
$departament=$_POST['departament'];
$location=$_POST['location'];
$country=$_POST['country'];
$email=$_POST['email'];
$ip=$_POST['ip'];
    $query = "UPDATE utilizatori SET username = '$username', utilizator = '$utilizator',
    password = '$password', nivel = '$nivel', departament = '$departament', 
    location = '$location', country = '$country', email = '$email', ip = '$ip' 
    where id= $id";
    $res = mysql_query($query);
    if ($res) {
    echo "Record Updated";
    } else {
    echo "Record not updated.";
    }
    header('location:../user.php');
    } // end of first if statement
    // close connection
    mysql_close(); 
    ?>
于 2013-05-29T16:54:41.353 回答