值总是返回0,它应返回插入的行数

Question

我有两张桌子,一张是Information另一张桌子WorkForce

信息

劳动力

该TWF column在劳动力来获得id的信息,其假设回报1,但它的返回值0.如果id在信息是 5,twf应该是5了.

首先我使用它a来表示所选行的数量,然后将参数添加a到addWorkForce.在toast始终显示0.

  a=addInformation(name,weather,date2,status,first1[1],last1[1]);
  Toast.makeText(getApplicationContext(),a+"",Toast.LENGTH_LONG).show();
  addWorkForce(Sub, NoP, NoH,a);

addInformation函数

  public long addInformation( final String name, final String weather, final String date2, final String status, final String timeIn, final String timeOut)
    {
        class AddInfo 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>();
                data.put(Config.KEY_USER_NAME,name);
                data.put(Config.KEY_WEATHER,weather);
                data.put(Config.KEY_DATE,date2);
                data.put(Config.KEY_STATUS,status);
                data.put(Config.KEY_TIMEIN,timeIn);
                data.put(Config.KEY_TIMEOUT,timeOut);
                RequestHandler rh=new RequestHandler();
                String result = rh.sendPostRequest(Config.ADD_INFORMATION,data);
                return  result;
            }
        }

         AddInfo ru = new AddInfo();
         ru.execute(name,weather,date2,status,timeIn,timeOut);

        return 0;

    }

有人可以帮我解决问题吗？

addInformation.php

<?php 

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

        //Getting values
        $name = $_POST['name'];
        $weather = $_POST['weather'];
        $date = $_POST['date'];
                $status = $_POST['status'];
                $timeIn = $_POST['timeIn'];
                $timeOut = $_POST['timeOut'];

        //Creating an sql query
        $sql = "INSERT INTO information(name, weather, date, status, time_in, time_out) VALUES ('$name','$weather','$date', '$status', '$timeIn', '$timeOut')";

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

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

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

这就是我使用代码将行插入sqlite的方法,它可以工作

 a = ts.insertTimeSheet(name, weather, date2, status, first1[1], last1[1]);

时间表

 public long insertTimeSheet(String name, String weather, String date, String status, String TimeIn, String TimeOut) {
        database = dbHelper.getWritableDatabase();
        ContentValues values = new ContentValues();
        values.put(MyDatabaseHelper.Name, name);
        values.put(MyDatabaseHelper.Weather, weather);
        values.put(MyDatabaseHelper.Date, date);
        values.put(MyDatabaseHelper.Status, status);
        values.put(MyDatabaseHelper.TimeIn_Info, TimeIn);
        values.put(MyDatabaseHelper.TimeOut_Info, TimeOut);
        return database.insert(MyDatabaseHelper.TABLE_INFO, null, values); // if the id in Information is 5, twf display 5 too

Answer 1

根据您的代码添加信息:

a=addInformation(name,weather,date2,status,first1[1],last1[1]);

在这里,您调用addInformation主线程并将返回值保存到变量a.但问题是,你的信息添加实际执行中进行doInBackground的AsyncTask,实际上做的工作线程任务(异步调用),所以你总是会得到一个= 0的默认值(作为长).

您需要做的是将代码移动到以下onPostExecute功能AsyncTask:

public void addInformation( final String name, final String weather, final String date2, final String status, final String timeIn, final String timeOut)
    {
        class AddInfo 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();
                addWorkForce(Sub, NoP, NoH,a);
            }

            @Override
            protected String doInBackground(String... params) {

                HashMap<String, String> data = new HashMap<String,String>();
                data.put(Config.KEY_USER_NAME,name);
                data.put(Config.KEY_WEATHER,weather);
                data.put(Config.KEY_DATE,date2);
                data.put(Config.KEY_STATUS,status);
                data.put(Config.KEY_TIMEIN,timeIn);
                data.put(Config.KEY_TIMEOUT,timeOut);
                RequestHandler rh=new RequestHandler();
                String result = rh.sendPostRequest(Config.ADD_INFORMATION,data);
                return  result;
            }
        }

         AddInfo ru = new AddInfo();
         ru.execute(name,weather,date2,status,timeIn,timeOut);

    }

还有另一种使用回调的方法.实现一个接口,在完成AsyncTask后,您可以发送回调来完成另一项工作.