Question

我有一个表单，其中包含用于电子邮件的输入文本框和用于名称的输入文本框。

当用户输入他/她的电子邮件ID时，应自动使用Ajax在名称输入文本框中提取并显示该电子邮件ID的正确名称（在数据库中）。

以下是我的代码，但是当我输入电子邮件ID并按下标签按钮时，没有显示任何名称（AJAX的工作方式）。有人能指出我出错的地方吗？谢谢。

我正在尝试使用Cakephp，jquery和ajax。

来自浏览器的源代码：

<script type="text/javascript">
//<![CDATA[
$(document).ready(function () {$("#MerryParentEmail").bind("keyup", function (event) 
{$.ajax({async:true, dataType:"html", success:function (data, textStatus)   
{$("#MerryParentName").html(data);}, type:"post", url:"\/merry_parents\/getname"});
    return false;});});
    //]]>
    </script>

view.ctp

<?php
echo $this->Session->flash();
echo $this->Form->create('MerryParent', array('default'=>false));

echo '<fieldset>';
echo '<legend>Parent Information</legend>';
echo $this->Form->input('MerryParent.email');
if (isset($name))
   echo $this->Form->input('MerryParent.name', array('label'=>'Parent/Guardian Name','value'=>$name));
else
   echo $this->Form->input('MerryParent.name', array('label'=>'Parent/Guardian Name'));

$this->Js->get('#MerryParentEmail');
$this->Js->event('keyup',
$this->Js->request(array(
'controller'=>'merry_parents',
        'action'=>'getname'),
            array('async'=>true,
                  'update'=>'#MerryParentName',
                  'method'=>'post')
     )
);

echo '</fieldset>';

echo $this->Form->end('Submit');

 ?>

merry_parents_controller.php

<?php
 class MerryParentsController extends AppController{
public $many_children_flag='false';
var $name='MerryParents';

function beforeFilter()//executed before any controller action logic
{   parent::beforeFilter();
    $this->Security->validatePost=false;  //set to false to completely skip the validation of POST request
        //parent::beforeFilter();
    if(isset($this->Security) && $this->RequestHandler->isAjax() && $this->action = 'getname'){
        $this->Security->enabled = false;
        $this->autoRender=false;
        }   
}

function getname(){
 if (!empty($this->data)){
    //var_dump($this->data);
    $merryparent_info=$this->MerryParent->getMerryParents($this->data['MerryParent']['email']);
    //print_r($merryparent_info);
    $name=$merryparent_info['MerryParent']['name'];
    $this->set('name',$name);
    //$this->render('/merry_parents/ajax_input');
    }
}
function view(){
    //$name='';
    //var_dump($this->data);
    //$this->getname();
}

merry_parent.php模型

function getMerryParents($field_value){
    if (is_int($field_value))
        $conditions=array('merryParent.id'=>$field_value);
    else
        $conditions=array('merryParent.email'=>$field_value);

    //debug($conditions);

    $merryparent_info=$this->find('first',array(
                                'conditions'=>$conditions,
                                'recursive'=>-1   //fetches merry_parents table data only not the associated data
                                ));

    //debug($merryparent_info);
    return $merryparent_info;
}

Answer 1

您必须将表单数据发送回控制器。将JS请求绑定到keyup事件时，您将通过CakePHP代理方法分别输入jQuery jQuery。并且没有自动绑定到您的表单数据，因此您必须自己管理它。

然而CakePHP中有一个方便的参数！尝试

$this->Js->get('#MerryParentEmail');
$this->Js->event('keyup',
$this->Js->request(array(
'controller'=>'merry_parents',
        'action'=>'getname'),
            array('async'=>true,
                  'update'=>'#MerryParentName',
                  'dataExpression'=>true,
                  'data'=>$this->Js->serializeForm(array('inline'=>true)),
                  'method'=>'post')
     )
);

cakephp js helper ajax：email id不在输入文本框中显示名称

1 个答案: