在PHP中成功返回查询成功到java - Returning success on query successfully in PHP to java

Returning success on query successfully in PHP to java

本文关键字：成功 java 查询返回 PHP | 更新日期: 2023-09-27

我需要检查用户名是否已经存在于数据库中。如果是，那么只登录。登录后，我需要一个响应从php文件发送到java类说成功。当我使用Json访问我的java文件中的结果数组时，虽然用户名已经存在于数据库中，但它给出了NULL。

帮忙吗?

我使用下面的Php和java代码

<?php
include("db_config.php");
$myusername = $_GET['username'];
$mypassword = $_GET['password'];

$sql="SELECT id FROM tablename WHERE username='$myusername' and  password='$mypassword'";    
echo $sql;
$result=mysql_query($sql);
$row=mysql_fetch_array($result);
$active=$row['active'];
$count=mysql_num_rows($result);
if($count==1)
{
$result['login'] == 'success';
}
else 
{
$result['login'] == 'failed';
}
echo json_encode($result);

这是我的Java代码

btnLogin.addActionListener(new ActionListener() 
    {
        public void actionPerformed(ActionEvent arg0)
        {
            username= textFieldUserName.getText().toString();
            password = textFieldPassword.getText().toString();
            //  discount = textFieldDiscount.getText().toString();
            List<NameValuePair> params = new ArrayList<NameValuePair>();


            params.add(new BasicNameValuePair("username",username));
            params.add(new BasicNameValuePair("password",password ));

            JSONObject json = jParser.makeHttpRequest(url_all_products, "GET", params);

            try {           
                status = json.getString(TAG_SUCCESS);
                System.out.println(" status is "+status);
                if(status.equalsIgnoreCase("success"))
                {
                    System.out.println("Login success");
                }   

            } catch (JSONException e) {
                e.printStackTrace();
            }           
        }
    });

,我试过了如果(response.getStatusLine () .getStatusCode () = = 200)如果从服务器返回OK则返回200。

[Edit: Code ReFormation]

使用java

JSONObject json = (JSONObject)jParser.makeHttpRequest(url_all_products, "GET", params);
try {           
   //status = json.getString(TAG_SUCCESS);
   String status = (String) json.get("login");
   System.out.println(" status is "+status);
   if(status.equalsIgnoreCase("success"))
   {
       System.out.println("Login success");
   }   
 }