Question

我正在尝试使用下面的代码为jQuery验证插件添加一个新方法。我的目标是检查电子邮件地址是否已存在于数据库（MySQL）中。如果是，它将通知用户注册另一个电子邮件地址。不知何故，总是返回的结果是“已经收到电子邮件”。

这些是validate.js中的代码：

$(document).ready(function(){
 $.validator.addMethod("uniqueEmail", function(value, element) {
   $.ajax({
      type: "POST",
      url: "availability.php",
      data: value,
      success: function(exist)
      {
       if(exist>0) {
         return true;
       } return false;
      }
  });
 } ,"Email is already taken");
 $('#signup form').validate({
  rules: {
   firstname: {
    required: true,
    minlength:3
   },
   lastname: {
    required: true,
    minlength: 3
   },
   affiliation: {
    required: true,
   },
   occupation: {
    required: true,
   },
   email: {
    required: true,
    email: true,
    uniqueEmail: true
   },
   password: {
    minlength: 6,
    required: true
   },
   repassword: {
    equalTo: "#password"
   }
  },
  messages: {
   firstname: {
    minlength: "Your first name should be more than 3 characters"
   },
   lastname: {
    minlength: "Your last name should be more than 3 characters"
   },
  }, 
  success: function(label) {
   label.text('OK!').addClass('valid');
  }
 });
});

这些是我的php文件中的代码：

<?php

include_once $_SERVER['DOCUMENT_ROOT'] . '/braddclient/includes/magicquotes.inc.php';
include $_SERVER['DOCUMENT_ROOT'] . '/braddclient/includes/db.inc.php';

$email = strtolower($_POST['email']);
$email = mysqli_real_escape_string($link, $email);
$sql = "SELECT * FROM bradduser WHERE email='$email'";
$result = mysqli_query($link, $sql);
$num = mysqli_num_rows($result);

echo $num;

?>

Answer 1

您无法使用自定义方法完成此操作，因为您需要验证插件才能知道这是远程（或更重要的是，异步）请求。

幸运的是，它具有内置功能，可以提供帮助。您可以在此处使用remote代替自定义方法，因此请完全保留自定义方法并更改此内容：

uniqueEmail: true

为此：

remote: { url: "availability.php", type: "post" }

然后将错误消息添加到messages，如下所示：

email: { remote: "Email is already taken" }

同时更改您的PHP端以匹配，返回true或false，如下所示：

<?php
  include_once $_SERVER['DOCUMENT_ROOT'] . '/braddclient/includes/magicquotes.inc.php';
  include $_SERVER['DOCUMENT_ROOT'] . '/braddclient/includes/db.inc.php';

  $email = strtolower($_POST['email']);
  $email = mysqli_real_escape_string($link, $email);
  $sql = "SELECT * FROM bradduser WHERE email='$email'";
  $result = mysqli_query($link, $sql);
  if(mysqli_num_rows($result) > 0) {
    echo "false";  //validation fails, email in use
  } else {
    echo "true";   //validation passes
  }
?>

jQuery Validation插件addMethod

1 个答案: