脚本更新客户端不工作的mysql

1 - 更新我的mysql客户端脚本 2 - 当我离开(其中id = id“;)我加倍了其他id只是使他们相同的变化。 。客户的个人资料,我知道这个问题,但不知道要放什么东西在那里,我试了MENY选项和仍然无法正常工作, 3 - 这里是我的脚本:脚本更新客户端不工作的mysql

<?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();

?>

回答:

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();

?>

回答:

你是不是引用$id变量,而是一个字符串文本id,它打破了查询语句,添加一个美元符号来引用该变量,如果它不是一个整数,它也用单引号括起来。

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

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