Http發布請求數據未傳遞

Question

我正在嘗試將一個簡單的post方法插入數據庫中。以下是指出我的錯誤的代碼。

這是用於發送發布請求的android代碼。我不知道出了什么問題，但未傳遞需要插入的參數，並且在數據庫中插入了空字符串。

爪哇

 String add = "myName";
 BufferedReader reader = null;
 String text = "";
 Log.d("K2NK", "Connecting....");
 URL url;
 String response = "";
 try {
        url = new URL("http://10.13.210.114/myDatabase/insert.php");
        URLConnection conn =  url.openConnection();

        conn.setDoOutput(true);

        OutputStreamWriter wr = new OutputStreamWriter(conn.getOutputStream());

        wr.write(add);
        wr.flush();

        //get the server response
        reader = new BufferedReader((new InputStreamReader(conn.getInputStream())));
        StringBuilder sb = new StringBuilder();
        String line = null;

        //read server response
        while ((line = reader.readLine()) != null) {
            //append server response in string
            sb.append(line + "\n");
        }
        text = sb.toString();
        Log.d("K2NK", text);

    }catch (IOException e) {
        e.printStackTrace();
    } finally {
        try {

            if(reader!=null){
            reader.close();}
        } catch (IOException e) {
            e.printStackTrace();
        }
      }

PHP腳本如下將數據插入數據庫

的PHP

<?php

$name = ($_POST['name']);
$con = new mysqli("localhost", "root", "","syncadapter");

if($con->connect_error){
   die("Connection failed: ".$con->connect_error);
 }

//sql to delete record
$sql="INSERT INTO table_student (name) values ('$name')";

if($con->query($sql)===TRUE){
        echo json_encode("insert successful: name:".$name);
 } else {
        echo json_encode("insert failed");
 }
?>

Answer 1

您傳遞的字符串應為：

String add = "name=myname";

它試圖找到一個不存在的名字命名變量后。