我创建了一个Dependent下拉列表,现在我想在更新页面上获取这些值。我该怎么办?
我创建了2个下拉列表 - 第一个客户和第二个员工 在更新页面,我得到了客户的价值,但我没有得到员工的价值(因为它是依赖下拉)
表格
<?php
//First drop-down
echo $form->field($model, 'client')->dropDownList($Client,
['prompt'=>'-Select Client-',
'onchange'=>'
$.post
(
"'.urldecode(
Yii::$app->urlManager->createUrl
('leads/lists&id=')).'"+$(this).val(), function( data )
{
$( "select#staff_id" ).html( data );
});
']); ?>
// depend dropdown
<?php echo $form->field($model, 'staff')
->dropDownList
(
['prompt'=>'-Choose a Sub Category-'],
['id'=>'staff_id','value'=>$Staff]
);
?>
控制器
public function actionLists($id)
{
$sql = "select * from staff where client='$id' ";
//exit;
$models = Staff::findBySql($sql)->asArray()->all();
//echo "<pre>";print_r($model);exit;
if(sizeof($models) >0){
echo "<option>-Choose a Sub Category-</option>";
foreach($models as $model){
echo "<option value='".$model['id']."'>".$model['fname']."</option>";
}
}
else{
echo "<option>-Choose a Sub Category-</option><option></option>";
}
}
答案 0 :(得分:1)
首先将$ modelsStaff变量添加到您的创建和更新操作中,如下所示:
<?
public function actionCreate()
{
$modelsStaff=null;
$model = new model();
if ($model->load(Yii::$app->request->post()) && $model->save())
{
return $this->redirect(['view', 'id' => $model->id]);
}
else
{
return $this->render('create', [ 'model' => $model,'modelsStaff'=>$modelsStaff]);
}
}
public function actionUpdate($id)
{
$model = $this->findModel($id);
if ($model->load(Yii::$app->request->post()) && $model->save())
{
return $this->redirect(['view', 'id' => $model->id]);
}
else
{
$sql = "select * from staff where client='$model->client'";
$modelsStaff = Staff::findBySql($sql)->asArray()->all();
return $this->render('update', [ 'model' => $model,'modelsStaff'=>$modelsStaff]);
}
}
?>
在您的更新操作中,使用$ model-&gt;客户端查找所有员工并获取此客户端下的所有员工并更新您的视图
<?php
//First drop-down
echo $form->field($model, 'client')->dropDownList($Client,
['prompt'=>'-Select Client-',
'onchange'=>'
$.post
(
"'.urldecode(
Yii::$app->urlManager->createUrl
('leads/lists?id=')).'"+$(this).val(), function( data ) //<---
{
$( "select#staff_id" ).html( data );
});
']); ?>
// depend dropdown
<?php echo $form->field($model, 'staff')->dropDownList
($modelsStaff,
['prompt'=>'-Choose a Sub Category-'],
['id'=>'staff_id','value'=>$Staff]
);
?>
答案 1 :(得分:0)
您必须在控制器中创建单独的功能(如示例):
public function actionLists($id)
{
$posts = \common\models\Post::find()
->where(['category_id' => $id])
->orderBy('id DESC')
->all();
if (!empty($posts)) {
$option = '<option>-Select Option-</option>';
foreach($posts as $post) {
$options .= "<option value='".$post->id."'>".$post->title."</option>";
}
return $options;
} else {
return "<option>-</option>";
}
}
和查看文件中的示例(示例):
use yii\helpers\ArrayHelper;
$dataCategory=ArrayHelper::map(\common\models\Category::find()->asArray()->all(), 'id', 'name');
echo $form->field($model, 'category_id')->dropDownList($dataCategory,
['prompt'=>'-Choose a Category-',
'onchange'=>'
$.post( "'.Yii::$app->urlManager->createUrl('post/lists?id=').'"+$(this).val(), function( data ) {
$( "select#title" ).html( data );
});
']);
$dataPost=ArrayHelper::map(\common\models\Post::find()->asArray()->all(), 'id', 'title');
echo $form->field($model, 'title')
->dropDownList(
$dataPost,
['id'=>'title']
);
这来自Yii文档:https://www.yiiframework.com/wiki/723/creating-a-dependent-dropdown-from-scratch-in-yii2