通过ajax

Question

这里有什么问题

JS

function cstatus(status, user, pass){
    var id = $('.pmarked').data('id');
    console.log(id); // 101
    $.ajax({
        url: 'pro-status.php',
        type: 'post',
        data: {'status': status, 'user': user, 'pass': pass, 'id': id},
        success: function(data) {
            console.log(data);
        }
    });
}

亲status.php

$sql = "update posts set status = :astatus, user = :auser, pass = :apass, where id = :aid";

$stmt = $db->prepare($sql);
$stmt->execute(array(
    ":astatus" => $_POST['status'],
    ":auser" => $_POST['user'],
    ":apass" => $_POST['pass'],
    ":aid" => $_POST['id'] // line 12
));

错误

  

致命错误......在'where id ='101''附近使用正确的语法... ... pro-status.php：12

Answer 1

删除逗号：

pass = :apass, where id = :aid"

应该是：

pass = :apass where id = :aid"

此外，永远不会存储纯文本密码！请使用 PHP built-in functions 来处理密码安全问题。如果您使用的PHP版本低于5.5，则可以使用password_hash() compatibility pack。 没有必要escape passwords 或在散列之前对它们使用任何其他清理机制。这样做更改密码并导致不必要的额外编码。

Answer 2

应该是

$sql = "update posts set status = :astatus, user = :auser, pass = :apass where id = :aid";

只需删除comma (,)

后的:apass即可

Answer 3

SQL语法问题请删除逗号并在$ sql上使用它。

$sql = "update posts set status = :astatus, user = :auser, pass = :apass where id = :aid";

Answer 4

在,

之前移除where

$sql = "update posts set status = :astatus, user = :auser, pass = :apass where id = :aid";