Question

我试图通过JS函数插入多个记录.JS函数调用执行插入的另一个YII控制器的动作。

我正在从MySql表读取数据，对数据执行某些操作并将更改的数据保存到数组post_data。后来我将这个数组转换为url格式并将参数作为参数传递给JS函数。此函数具有ajax来调用另一个控制器来执行插入。

我想插入2个表。插入第一个contact_details表，然后插入工程师表，所以我希望这个ajax调用以串行方式运行。即，我希望下一次插入仅在上一次插入完成后才开始。

我的代码是：

<script>

     function call_migration_api(post_data)
    {
      //console.log( 'post data in js func = '+post_data); 

      //var db_data = post_data.serialize();//tried this, but not working.
     var db_data = post_data;

     $.ajax({
          type: "POST",
          url:"http://localhost/amica_migration/rapport/chs/Migrationapi/createengineer",
          data: db_data,
          success: function(server_response)
          {
             console.log(server_response)
          }//end of success

     });//end of $.ajax


   }//end of function.

</script>

<?php

   //reading all values in $post_data array.

   foreach ( $post_data as $key => $value) 
    {
        $post_items[] = $key . '=' . $value;
    }
    $post_string = implode ('&', $post_items);

    echo "<script>call_migration_api('$post_string');</script>";// Call to AJAX function.



?>

迁移API代码：

<?php

      public function actionCreateEngineer()
      {
            $engineer_model=new Engineer();
    $contactdetails_model=new ContactDetails();

            //****** RETRIVING CONTACT DETAILS ***********
            $contactdetails_model->address_line_1= $_POST['address_line_1'];
    $contactdetails_model->town= $_POST['town'];
    $contactdetails_model->postcode_s= $_POST['postcode_s'];
    $contactdetails_model->postcode_e= $_POST['postcode_e'];
    $contactdetails_model->telephone= $_POST['telephone'];
    $contactdetails_model->email= $_POST['email'];

        //********** SAVING CONTACT ***************
    if($contactdetails_model->save())
    {
        echo "<br>Contact Model Saved";

        $engineer_model->contact_details_id=$contactdetails_model->id;
        $engineer_model->delivery_contact_details_id=$contactdetails_model->id;

                    //****** RETRIVING ENGINEER DETAILS ***********
        $engineer_model->first_name=$_POST['first_name'];
        $engineer_model->last_name=$_POST['last_name'];
        $engineer_model->active=$_POST['active'];
        $engineer_model->id=$_POST['engg_id'];


            //****** SAVING ENGINEER DETAILS ***********    
        if($engineer_model->save())
        {
            echo "<br>Engineer saved";
        }
        else
        {
            echo "<br>Engineer not saved******** ENGG ERROR = ";
            print_r($engineer_model->getErrors());
        }


    }//end of if contact save.
    else
    {
        echo "<br>Contacts NOT SAVED ///////// CONTACT ERROR = ";
        print_r($contactdetails_model->getErrors());
    }

      }//end of create engineer.

?>

通过AJAX插入多个数据库

0 个答案: