Question

我有一个链接，当它按下它通过渲染ajax从控制器请求一个页面，以前我曾经只传递id但现在我想传递一个额外的参数到控制器，我如何实现这一点，这就是我试过的

这是仅将一个参数传递给控制器的链接

Html::a('click me', ['#'],
['value' => Url::to('checktruck?id='.$model->id), //this is where the param is passed 
'id' => 'perform']);

这是期望2个参数的控制器代码：

public function actionChecktruck($id,$category)  //it expects 2 parameters from above link
{
     $truckdetails = Truck::find()->where(['id' =>$id])->one();

    if (Yii::$app->request->post()) {
      $checklistperform = new TodoTruckChecklist();
        $truck = Truck::find()->where(['id'=>$id])->one();
        $checklistperform->truck_id=$id;
        $checklistperform->registered_by=Yii::$app->user->identity->id;
        $checklistperform->save();
        $truck->update();

        var_dump($checklistperform->getErrors());
        //var_dump($truck->getErrors());
    }
    else {
        $truckcategory = Checklist::find()->where(['truck_category'=>$truckdetails->truck_category])->andWhere(['checklist_category'=>$category])->all();
        return $this->renderAjax('truckyard/_checklistform', [
            'truckcategory' => $truckcategory,'truckvalue'=>$id,
        ]);

    }

}

这是我在发布请求期间依赖于上述控制器的另一个按钮的jquery代码

$("#postbutn").click(function(e) {

   $.post("checktruck?id="+truckid,  //here i would like to pass 2 params
                {checked:checked,unchecked:unchecked,truckid:truckid}
            )
 }

这是没有帖子时的jquery代码

如何在链接中传递额外参数，甚至传递控制器的$ .post请求

Answer 1

首先，当您使用JQuery ajax提交表单时，无需为链接设置值

Html::a('click me', ['#'],['id' => 'perform']);

使用此ID您可以按如下方式提交请求

$this->registerJs("$('#perform').click(function(event){

event.preventDefault(); // to avoid default click event of anchor tag

$.ajax({
    url: '".yii\helpers\Url::to(["your url here","id"=>$id,"param2"=>param2])."',        
    success: function (data) {
           // you response here
        },
      });
});");

没有必要提及方法属性为'POST'，你想通过GET方法发送

最后在您的控制器中，您需要接受以下参数

public function actionChecktruck()  //it expects 2 parameters from above link
{
     $id = Yii::$app->request->queryParams['id'];
     $param2 = Yii::$app->request->queryParams['param2'];

     $truckdetails = Truck::find()->where(['id' =>$id])->one();

    if (Yii::$app->request->post()) {
      $checklistperform = new TodoTruckChecklist();
        $truck = Truck::find()->where(['id'=>$id])->one();
        $checklistperform->truck_id=$id;
        $checklistperform->registered_by=Yii::$app->user->identity->id;
        $checklistperform->save();
        $truck->update();

        var_dump($checklistperform->getErrors());
        //var_dump($truck->getErrors());
    }
    else {
        $truckcategory = Checklist::find()->where(['truck_category'=>$truckdetails->truck_category])->andWhere(['checklist_category'=>$category])->all();
        return $this->renderAjax('truckyard/_checklistform', [
            'truckcategory' => $truckcategory,'truckvalue'=>$id,
        ]);

    }

}

Answer 2

试试这个，在视图文件

中

$this->registerJs("$('#postbutn').click(function(){
$.ajax({
    url: '".yii\helpers\Url::to(["u r URL"])."',        
    method: 'POST',       
    data: {id:id, truckid:truckid },
    success: function (data) {
        },
      });
});");

使用jquery将yii2 ajax请求中的两个参数传递给控制器

2 个答案: