尝试将Android中捕获的用户名和密码传递给PHP查询MYSQL数据库答案

【问题标题】：Try to pass the username and password captured in Android to PHP to query MYSQL database尝试将Android中捕获的用户名和密码传递给PHP查询MYSQL数据库
【发布时间】：2015-01-24 12:25:01
【问题描述】：

代码查询用户表（“姓名”），检索用户输入的用户名（唯一）对应的中文姓名。

在 DEBUG 模式下，发现 $_POST['username'] 中的用户名在传递给 PHP 时为空。我可以在DEBUG模式下检索中文名称。请帮助找出参数传递中的问题。

向 PHP 发送 POST 请求的 Android 部分。

public class SigninActivity  extends AsyncTask<String,Void,String> {

private TextView statusField,roleField;
private Context context;
private int byGetOrPost = 0; 
//flag 0 means get and 1 means post.(By default it is get.)
public SigninActivity(Context context,TextView statusField,
TextView roleField,int flag) {
  this.context = context;
  this.statusField = statusField;
  this.roleField = roleField;
  byGetOrPost = flag;
}

protected void onPreExecute(){

}
@Override
protected String doInBackground(String... arg0) {
  if(byGetOrPost == 0){ //means by Get Method
     try{
        String username = (String)arg0[0];
        String password = (String)arg0[1];
        String link = "http://www.example.com/login.php?username="
        +username+"&password="+password;
        URL url = new URL(link);
        HttpClient client = new DefaultHttpClient();
        HttpGet request = new HttpGet();
        request.setURI(new URI(link));
        HttpResponse response = client.execute(request);
        BufferedReader in = new BufferedReader
       (new InputStreamReader(response.getEntity().getContent()));

       StringBuffer sb = new StringBuffer("");
       String line="";
       while ((line = in.readLine()) != null) {
          sb.append(line);
          break;
        }
        in.close();
        return sb.toString();
  }catch(Exception e){
     return new String("Exception: " + e.getMessage());
  }
  }
  else{
     try{
        String username = (String)arg0[0];
        String password = (String)arg0[1];
        String link="http://www.example.com/login.php";
        String data  = URLEncoder.encode("username", "UTF-8") 
        + "=" + URLEncoder.encode(username, "UTF-8");
        data += "&" + URLEncoder.encode("password", "UTF-8") 
        + "=" + URLEncoder.encode(password, "UTF-8");
        URL url = new URL(link);
        URLConnection conn = url.openConnection(); 
        conn.setDoOutput(true); 
        OutputStreamWriter wr = new OutputStreamWriter
        (conn.getOutputStream()); 
        wr.write( data ); 
        wr.flush(); 
        BufferedReader reader = new BufferedReader
        (new InputStreamReader(conn.getInputStream()));
        StringBuilder sb = new StringBuilder();
        String line = null;
        // Read Server Response
        while((line = reader.readLine()) != null)
        {
           sb.append(line);
           break;
        }
        return sb.toString();
     }catch(Exception e){
        return new String("Exception: " + e.getMessage());
     }
  }
   }
@Override
protected void onPostExecute(String result){
  this.statusField.setText("Login Successful");
  this.roleField.setText(result);
   }
  }


// Code in PHP

$con = mysqli_connect($mysql_hostname, $mysql_user, $mysql_password) or     die("Could not connect database");



if (mysqli_connect_errno($con))
{
   echo "Failed to connect to MySQL: " . mysqli_connect_error();
} 


mysqli_select_db($con, $mysql_database) or die("Could not select database");


if (DEBUG) {
    echo "Connected to database $mysql_database";
}


mysqli_set_charset($con, 'utf8');

$username = mysqli_real_escape_string($con, $_POST['username']);
$password = mysqli_real_escape_string($con, $_POST['password']);

if (DEBUG) {
    $username = "b";
} 

$qry1 = "SELECT c_f_name FROM name where username = \"$username\" ";

$result = mysqli_query($con, $qry1);
if(DEBUG) {
echo "This is the username: ";
echo $username;

if ($result) {
    echo $username;
//successful query
if (mysqli_num_rows($result) > 0) {
    echo "query successful - 
    ";
    echo mysqli_num_rows($result);
 } else {
    echo "query successful-0 rows
    ";
}
} else {
// query not successful
echo "Query unsuccessful";
}

}

$row = mysqli_fetch_array($result);
//Since there is only 1 record that match, just retrieve the first and only record
$data = $row[0];
if($data){
echo $data;
}

mysqli_close($con);
?>

【问题讨论】：

在 php 文件中使用 $_REQUEST 而不是 $_POST
缺乏 Http 的基础知识：内容类型
Ravi - 在 PHP 中更改为 $_REQUEST，代码现在可以运行。非常感谢。

标签： php android mysql parameter-passing

【解决方案1】：

您正在通过 url 传递您的用户名/密码

http://www.example.com/login.php?username="+username+"&password="+password

因此使用 $_GET['username'] 或 $_REQUEST['username'] 请求您的值。

【讨论】：

【解决方案2】：

您正在传递 GET 变量并尝试从 POST 方法中检索变量。这不起作用，您还必须使用 $GET['variable_name'] 或 $REQUEST['variable_name'] 从该 url 检索数据。

这是一个简单的规则，如果您通过 get 或 post 方法传递数据，那么您也必须以相同的方法检索该数据。

【讨论】：