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我有一个关于 cakePHP 的问题。我在我的视图中创建了两个下拉列表。当用户更改一个列表中的值时，我希望第二个更改。目前，我的工作方式如下：当用户从列表框一中选择时，会触发点击事件。这会触发一个 jQuery ajax 函数，该函数从我的控制器中调用一个函数。这一切都很好，但是我如何异步（或查看）重新渲染我的控件？我知道我可以将数组序列化为 json，然后在 javascript 中重新创建控件，但似乎应该有一种更“CakePHP”的方式。这不就是渲染的目的吗？任何帮助都会很棒。这是我到目前为止的代码：

jQuery：

function changeRole(getId){

$.ajax({
    type: 'POST',
    url: 'ResponsibilitiesRoles/getCurrentResp',
    data:  { roleId: getId },
    cache: false,
    dataType: 'HTML',
    beforeSend: function(){
    },
    success: function (html){

    },
    error: function(XMLHttpRequest, textStatus, errorThrown) {

    }

});

看法：

<?php


echo 'Roles:';
    echo'<select name="myOptions" multiple="multiple">';
    foreach ($rolesResponsibility as $role) {
       echo' <option onclick="changeRole(this.value);"  value="';  echo $role["ResponsibilitiesRole"]["role_id"]; echo '">'; echo $role["r"]["role_name"]; echo '</option>';
    }
    echo '</select>';

echo 'Responsbility:';
echo'<select name="myOptionsResp" multiple="multiple">';
foreach ($respResponsibility as $responsibility) {
    echo' <option  value="';  echo $responsibility["responsibility"]["id"]; echo '">'; echo $responsibility["responsibility"]["responsibility_name"]; echo '</option>';
}
echo '</select>';

?>

控制器功能：

公共函数 getCurrentResp(){ $getId = $this->request->data['roleId'];

$responsibilityResp = $this->ResponsibilitiesRole->find('all',
    array("fields" => array('role.role_name','ResponsibilitiesRole.role_id','responsibility.*'),'joins' => array(
        array(
            'table' => 'responsibilities',
            'alias' => 'responsibility',
            'type' => 'left',
            'foreignKey' => false,
            'conditions'=> array('ResponsibilitiesRole.responsibility_id = responsibility.id')
        ),
        array(
            'table' => 'roles',
            'alias' => 'role',
            'type' => 'left',
            'foreignKey' => false,
            'conditions'=> array('ResponsibilitiesRole.role_id = role.id')
        )

    ),
        'conditions' => array ('ResponsibilitiesRole.role_id' => $getId),

    ));
$this->set('respResponsibility', $responsibilityResp);

//do something here to cause the control to be rendered, without have to refresh the whole page       

}
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1 回答 1

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  • js 事件是在选择标签上触发的更改，而不是单击
  • 您可以使用表单助手来构建您的表单。
  • 注意按照 cakephp 的方式命名。

因为你的代码有点混乱，我会做其他简单的例子：

Country hasMany City
User belongsTo Country
User belongsTo City

ModelName/TableName (fields)
Country/countries (id, name, ....) 
City/cities (id, country_id, name, ....)
User/users (id, country_id, city_id, name, ....)

查看/用户/add.ctp

<?php
    echo $this->Form->create('User');
    echo $this->Form->input('country_id');
    echo $this->Form->input('city_id');
    echo $this->Form->input('name');
    echo $this->Form->end('Submit');

    $this->Js->get('#UserCountryId')->event('change',
        $this->Js->request(
            array('controller' => 'countries', 'action' => 'get_cities'),
                array(
                    'update' => '#UserCityId',
                    'async' => true,
                    'type' => 'json',
                    'dataExpression' => true,
                    'evalScripts' => true,
                    'data' => $this->Js->serializeForm(array('isForm' => false, 'inline' => true)),
            )
        )
    );
    echo $this->Js->writeBuffer();

?>

用户控制器.php / 添加：

public function add(){
    ...
    ...
    // populate selects with options
    $this->set('countries', $this->User->Country->find('list'));
    $this->set('cities', $this->User->City->find('list'));
}

CountryController.php / get_cities：

public function get_cities(){
    Configure::write('debug', 0);
    $cities = array();
    if(isset($this->request->query['data']['User']['country_id'])){
        $cities = $this->Country->City->find('list', array(
                  'conditions' => array('City.country_id' => $this->request->query['data']['User']['country_id'])
        ));
    }
    $this->set('cities', $cities);
}

查看/Cities/get_cities.ctp：

<?php 
    if(!empty($cities)){
        foreach ($cities as $id => $name) {
?>
<option value="<?php echo $id; ?>"><?php echo $name; ?></option>
<?php           
        }
    }
?>
于 2013-10-28T20:45:51.663 回答