Question

我有一个像这样的json：

{"response":{"count":7,"list":[{"kdTeacher":"001","nmTeacher":"Eulis Irma"},{"kdTeacher":"002","nmTeacher":"Ni Wayan"},{"kdTeacher":"003","nmTeacher":"Dwi Widi"}]},"metaData":{"message":"OK","code":200}}

我希望教师名称和教师代码显示在下拉列表中。

在我的控制器中这样：

$model=new Teacher;

$paises = file_get_contents(Yii::getPathOfAlias('webroot.assets') .
   DIRECTORY_SEPARATOR . "teacher.json");
$jsondecode = CJSON::decode($paises, true);

$temp = array();

foreach ($jsondecode as $key => $value) {
        $temp[] = $value;
}

    $this->render('techer', array(
        'data'=>$temp,
        'model'=>$model,
    ));
}

并在视野中

<div class="row">
    <?php echo $form->labelEx($model,'name'); ?>
    <?php echo $form->dropDownList($model,'nama',CHtml::listData($data,'kodeTeacher','nmTeacher')); ?>
    <?php echo $form->error($model,'name'); ?>
</div>

ID

Answer 1

问题是您无法获取教师数据，因为您对foreach循环使用了错误的数据。教师代码和名称存储在已解码的json的['response']['list']索引处。看看：

$model=new Teacher;

$paises = file_get_contents(Yii::getPathOfAlias('webroot.assets') .
   DIRECTORY_SEPARATOR . "teacher.json");

$decodedJson = json_decode($paises, true);
$temp = array();

//check if list of teachers is available
if(isset($decodedJson['response']) && isset($decodedJson['response']['list']))
{
    //use foreach on ['response']['list'] index - here are teachers data stored
    foreach($decodedJson['response']['list'] as $teacher)
        $temp[$teacher['kdTeacher']] = $teacher['nmTeacher'];
}

$this->render('techer', array(
    'data'=>$temp,
    'model'=>$model,
));

从JSON获取教师数据的工作示例：CLICK!!!

在视图中（现在您不需要使用CHtml::listData方法）：

<div class="row">
    <?php echo $form->labelEx($model,'name'); ?>
    <?php echo $form->dropDownList($model,'nama',$data); ?>
    <?php echo $form->error($model,'name'); ?>
</div>

Yii Json到Array下拉列表

1 个答案: