我正在使用JQuery验证插件,它对远程验证的用户名是独一无二的。JQuery远程验证不显示错误
该脚本返回“true”或“false”,您将从下面的代码中看到。但我不确定它为什么不显示错误......值得一提的是,显示了用户的其他错误。
jQuery验证规则:
User: {
required: true,
minlength: 6,
remote:{
url: 'scripts/userCheck.php',
type: "post"
}
}
jQuery验证消息:
User: {
required: "Please Enter a Username",
minlength: "Username must be more than 6 characters in Length",
remote: "User already exists"
}
userCheck.php:
<?php
/**
* Created by PhpStorm.
* User: nathanenglish5
* Date: 30/12/2015
* Time: 09:05
*/
include_once "init.php";
include_once "../resources/signup.class.php";
$signup = new signup($DB_con);
$return = "false";
$count = 0;
if (isset($_Post['User'])) {
$uid = $_Post['User'];
$sql = "SELECT COUNT(username) FROM username WHERE username = '$uid'";
$count = $signup->dataview($sql);
if($signup->dataview($sql)==0){
$return = "true";
}
}
?>
所有的数据视图类确实是返回一个数字。
任何帮助或指导将是伟大的!
哪里是'data'发布在远程? – AnkiiG
虽然看着萤火虫我可以看到在请求正文用户正在传递,而不需要数据部分? –