Question

我正在使用codeigniter和AJAX使用jQuery创建一个登录表单。我的问题是如何显示从codeigniter返回的验证错误。如果用户正确登录，一切正常，但我在codeigniter中使用验证工具，如果登录失败，我想在页面上显示这些错误。如何将验证错误从控制器传递回ajax函数calll的.success？我的代码在下面..谢谢！

登录视图

<script type="text/javascript">

    $('#login_form').submit(function(e) {

        var username = $('#username').val();
        var password = $('#password').val();

        var loginData = "&username=" + username + "&password=" + password;

        $.ajax({
             type: "POST",
             url: "<?php echo site_url('login/validate') ?>",
             data: loginData,
             success: function(data) {

                   //My question is how to dislpay data if validation_errors() has something and do nothing if login properly worked.
             }
        });
    });

</scirpt>

<form id="login_form">
  Username: <input type="text" name="username" id="username" size="15" /><br />
  Password: <input type="password" name="passwort" id="password" size="15" /><br />
  <div align="center">
      <p><input type="submit" value="Login" /></p>
  </div>
 </form>

登录控制器

function validate() {

    foreach($_POST as $key => $value ) {
         $$key = trime(value);
    }

    $query = ..... call the model to validate in DB this works and will return true or false if it authenticates...

    if( $query ) {
         //redirect to the loggedin view..
         rediect('site/loggedin');
    }
    else {

         //here I want to pass back the validation errors to be shown on the current view. 
    }

}

Answer 1

jondavidjohn就在眼前。但是，要获得验证，您需要通过ajax：

返回validation_errors（）

您需要在某处放置验证错误：

HTML

<div class="errors"></div>

JQUERY

$.ajax({
    type: "POST",
    url: "<?php echo site_url('login/validate') ?>",
    data: loginData,
    success: function(data) {
        if (data == undefined) {
            window.location = "<?php echo site_url('controller/loggedin') ?>"; //javascript redirect by jondavidjohn
        }
        else {
            $('.errors').html(data); //how you get and set the errors to the html file
        }            
    }
});

PHP

function login() {
    if ($this->form_validation->run() != TRUE) {
      echo validation_errors(); // this returns the errors via ajax
    }
}

Answer 2

您必须通过javascript处理重定向，因此在ajax成功处理程序

中

$.ajax({
    type: "POST",
    url: "<?php echo site_url('login/validate') ?>",
    data: loginData,
    success: function(data) {
        if (data) {    //if the data returned is true (or 1)
            window.location = "<?php echo site_url('controller/loggedin') ?>"; //javascript redirect
        }
        else {         //if the data returned is not true (or 0)
            //show your error here
        }            
    }
});

并且在您的控制器中，您只需回显1或0，具体取决于登录的成功与否。如果$query是布尔值true / false，则只需回显它。

function validate() {

    foreach($_POST as $key => $value ) {
         $$key = trime(value);
    }

    $query = ..... call the model to validate in DB this works and will return true or false if it authenticates...

    echo $query;  //should be true/false which will be interpreted by the ajax success callback.

}

将验证从控制器传递回ajax函数

2 个答案: