Question

在表格work_details中，我有7列(id, project, work_description, percentage, time_in, time_out, fk)。现在我想将arraylist和fk保存到表格中。

我试过编码，但我知道这不是正确的方法。

  public void addWorkDetails(ArrayList<SearchResults> listItems, long id)
    {
        JSONObject object= new JSONObject();

        for(int i=0;i<listItems.size();i++)
        {
            try
            {
                object.put("Count : "+String.valueOf(i + 1),listItems.get(i));
            }catch(JSONException e)
            {
                e.printStackTrace();
            }
        }

        class AddWorkDetails extends AsyncTask<String, Void, String> {
            ProgressDialog loading;

            @Override
            protected void onPreExecute() {
                super.onPreExecute();
                loading = ProgressDialog.show(WorkDetailsTable.this, "Please Wait",null, true, true);
            }

            @Override
            protected void onPostExecute(String s) {
                super.onPostExecute(s);
                loading.dismiss();
                Toast.makeText(getApplicationContext(),s,Toast.LENGTH_LONG).show();
            }

            @Override
            protected String doInBackground(String... params) {
                HashMap<String, String> data = new HashMap<String,String>();
               // what should I put here ?
                RequestHandler rh=new RequestHandler();
                String result = rh.sendPostRequest(Config.ADD_WORKDETAILS,data);
                return  result;
            }
        }

        AddWorkDetails ru = new AddWorkDetails();
        ru.execute(listItems,id);
    }

腓

<?php

   if($_SERVER['REQUEST_METHOD']=='POST'){

   $list[]=$_POST['listItems'];
   $id=$_POST['id'];

   foreach($list as $value){
   $value=mysqli_real_escape_string($val);

    $sql="INSERT INTO work_details (project, work_description, percentage, timeIn, timeOut, id) VALUES ('$val', '$id')";

  //Importing our db connection script
        require_once('dbConnect.php');

        //Executing query to database
        if(mysqli_query($con,$sql)){
            echo ' Added Successfully';
        }else{
            echo 'Could Not Add Data';
        }

        //Closing the database 
        mysqli_close($con);
    }
?>

注意 listItems 持有project,work_description,percentage,time_in and time_out。

Answer 1

您的SQL查询不正确：

$sql="INSERT INTO work_details (project, work_description, percentage, timeIn, timeOut, id) VALUES ('$val', '$id')";

您需要设置6列，并且只提供2个值。

不是迭代$list并对每个值执行查询，而是应该在迭代时构造查询，并在完成时仅在结束时执行一次。例如：

foreach (...) {
    $sql = $sql . "'$val', ";
    ...
}
$sql = "INSERT INTO work_details (project, work_description, percentage, timeIn, timeOut, id) VALUES (" . $sql . "'$id')";

这只是一个想法..我不是一个PHP家伙，所以我写的内容存在错误。

编辑：这假定列表中的顺序始终相同。

将ArrayList从android发送到php MySQL

1 个答案: