将php值存储到java

 addInformation(status, timeIn);
 lastID =? // what should I write in order to get the last ID ? 
  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, Long.parseLong(s));
               // addWorkDetails(results, Long.parseLong(s));
            }
            @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);
    }

<?php 
    if($_SERVER['REQUEST_METHOD']=='POST'){
        //Getting values
        $status = $_POST['status'];
        $timeIn = $_POST['timeIn'];

        //Creating an sql query
        $sql = "INSERT INTO information(status, time_in) VALUES ('$status', '$timeIn')";
        //Importing our db connection script
        require_once('dbConnect.php');
        //Executing query to database
        if(mysqli_query($con,$sql)){
            echo 'Information Added Successfully';
            $insertId=mysql_insert_id();
            echo json_encode($insertId);
        }else{
            echo 'Could Not Add Information';
        }
        //Closing the database 
        mysqli_close($con);
    }
?>

根据注释，我正在创建新的答案，因为我的两个答案基本上都是正确的。我做了一个例子，使用android（apachecommons 4.5.1）和php5.6。这两个版本（4.5.1和5.6）都不是要求，只是我现在使用的版本。

示例假设您有一个名为information的mysql表，其中包含字段status、time_in，另一个字段标记为AUTO_INCREMENT。

---

Java部件

在原始：：doInBackground（字符串…params）函数中，您可以使用

HttpClient client = HttpClients.createDefault();
HttpPost httpPost = new HttpPost("http://localhost/so/sendrequest/addInformation.php");
try{
   List<NameValuePair> data = new ArrayList<NameValuePair>(2);
   data.add(new BasicNameValuePair("status", "ok"));
   data.add(new BasicNameValuePair("timein", "12:55"));
   httpPost.setEntity(new UrlEncodedFormEntity(data));
   String response = EntityUtils.toString(client.execute(httpPost).getEntity());
   System.out.println(response); //here you have your insertid
}catch(ClientProtocolException e){
   // TODO Auto-generated catch block
}catch(IOException e){
   // TODO Auto-generated catch block
}

基于本地Java

import java.io.IOException;
import java.util.ArrayList;
import java.util.List;

和apachecommons，应该包含在android库中（如果没有，请下载链接）

import org.apache.http.NameValuePair;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.HttpClient;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.impl.client.HttpClients;
import org.apache.http.message.BasicNameValuePair;
import org.apache.http.util.EntityUtils;

---

PHP部分

addInformation.php

<?php
class AddInformation
{
   function response(){
      /** @var mysqli $con */
      require_once('dbConnect.php'); //$con = new mysqli('127.0.0.1', 'root', '', 'so');
      $status = $con->real_escape_string($_POST['status']);
      $timein = $con->real_escape_string($_POST['timein']);
      $con->query("INSERT INTO information (status, time_in) VALUES ('$status', '$timein')");
      echo $con->insert_id;
   }
}
$ai = new AddInformation();
$ai->response();

作为第一行，

 echo 'Information Added Successfully';

会破坏你的json格式，我建议你像一样格式化所需的json

{"status": "ok", "message": "Information Added Successfully", "data": {"lastid": 1234}}

之后，阅读如何使用

String result = rh.sendPostRequest(Config.ADD_INFORMATION, data);

关于如何在Java中将String转换为JSONObject。最令人满意的方法是为json响应准备java类，例如

class AddInformationResponse
{
   public String status;
   public String messsage;
   public ObjectNode data;
}

您可以在众多教程中的一篇中阅读如何将json映射到java对象。

在ADD_INFORMATION脚本中，您所需要做的就是在insert语句之后获取数据。因此，我会使用类似于以下的东西：

<?php 
    if($_SERVER['REQUEST_METHOD']=='POST'){
        //Getting values
        $status = $_POST['status'];
        $timeIn = $_POST['timeIn'];

        //Creating an sql query
        $sql = "INSERT INTO information(status, time_in) VALUES ('$status', '$timeIn')";
        //Importing our db connection script
        require_once('dbConnect.php');
        //Executing query to database
        if(mysqli_query($con,$sql)){
            $sql = "SELECT * FROM INFORMATION ORDER BY time_in DESC LIMIT 1";
            echo $sql;
            $insertId=mysql_insert_id();
            echo json_encode($insertId);
        }else{
            echo 'Could Not Add Information';
        }
        //Closing the database 
        mysqli_close($con);
    }
?>

实际上，我所做的是对刚刚插入INSERT语句中的数据创建一个SELECT语句。

希望这能有所帮助。