$ _POST數組是空的，當我嘗試將數據發送到myregister.php

我正在啓動一個Android應用程序。對於登錄我需要發送數據庫的數據，但是當我嘗試使用$ _POST數組後，它似乎是空的（我試圖打印響應，我認爲這是我的問題）。

這裏是我的應用程序裏面的javacode：

private String register (String username, String password, String number) { 

    String reg_url = "myDomain/register.php"; 


    try { 
     URL url = new URL(reg_url); 
     HttpURLConnection httpURLConnection = (HttpURLConnection) url.openConnection(); 
     httpURLConnection.setRequestMethod("POST"); 
     httpURLConnection.setDoOutput(true); 
     httpURLConnection.setRequestProperty("Accept-Charset", "UTF-8"); 
     httpURLConnection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded"); 

     OutputStream OS = httpURLConnection.getOutputStream(); 

     BufferedWriter bufferedWriter = new BufferedWriter(new OutputStreamWriter(OS, "UTF-8")); 

     String data = URLEncoder.encode("username", "UTF-8") + " = " + URLEncoder.encode(username, "UTF-8") + "&" + 
         URLEncoder.encode("password", "UTF-8") + " = " + URLEncoder.encode(password, "UTF-8") + "&" + 
         URLEncoder.encode("number", "UTF-8") + " = " + URLEncoder.encode(number, "UTF-8"); 
     bufferedWriter.write(data); 

     bufferedWriter.flush(); 
     bufferedWriter.close(); 
     OS.close(); 
     InputStream IS = httpURLConnection.getInputStream(); 

     BufferedReader bufferedReader= new BufferedReader(new InputStreamReader(IS,"iso-8859-1")); 
     String response = ""; 
     String line; 

     while ((line = bufferedReader.readLine())!=null) response += line ; 

     Log.i("Response", response); 
     IS.close(); 
     bufferedReader.close(); 

     if (!response.toLowerCase().contains("fail")) 
      return Language.registered; 
     else 
      return Language.aProblemOccurred; 



    } catch (UnsupportedEncodingException e) { 
     e.printStackTrace(); 
     return Language.aProblemOccurred; 
    } catch (IOException e) { 
     e.printStackTrace(); 
     return Language.aProblemOccurred; 
    } 

}

這是註冊的簡單的PHP代碼：

<?php 

    require "init.php"; 

    $username = $_POST["username"]; 
    $password = $_POST["password"]; 
    $number = $_POST["number"]; 
    $mpt  = "1"; 

    $sql_query = "insert into Users values('$mpt' , '$number' , '$username' , '$password' , '$number', '$mpt' , '$mpt' , '$mpt' , '$mpt' , '$mpt' , '$mpt' , '$mpt');"; 

    if (mysqli_query ($res , $sql_query)) 
    { 
     echo "<h3> Data Insert Success...".$number.$password.$username.$_POST["password"]."<h3>"; 
    } 
    else 
    { 
     echo "Data Insert fail: Error:".mysqli_error($res).$number.$password.$username; 
    } 

?>

有人可以幫我????

來源

2016-05-21 Luigi Blu

你的腳本是[SQL注入攻擊]的風險（http://stackoverflow.com/問題/ 60174 /如何防止SQL注入在PHP）看看發生了什麼[小Bobby表]（http://bobby-tables.com/）甚至 [如果你是逃避投入，它不安全！]（http://stackoverflow.com/questions/5741187/sql-injection-that-gets-around-mysql-real-escape-string）你應該使用參數化查詢 – RiggsFolly

你最好顯示使用'init.php'的內容以及 – RiggsFolly

1）我如何使用參數化查詢？ 2）無論如何，我該如何解決我的$ _POST的問題？ –

我其實不知道java我知道的基礎知識，但我沒有在3年內使用它，所以我不知道你是如何從用戶從post方法獲取值到php。但是，您可以使用url作爲get方法來獲取來自用戶的輸入。

"myDomain/register.php?username=abc&password=qwqw1w1"

然後在PHP中獲取它的功能。或者你可以先採取用戶輸入一個變量，然後把他們用POST方法PHP，應該工作

$username = $_GET["username"]; 
$password = $_GET["password"]; 
$number = $_GET["number"];

來源

2016-05-21 15:58:15 Sumit

這樣，它似乎從瀏覽器的工作，但仍然不能從應用程序工作:(（（ –

你是否嘗試在應用程序??將用戶名的價值存儲在一個不同的變量，然後通過url中的變量註釋你的代碼 – Sumit

$ _POST數組是空的，當我嘗試將數據發送到myregister.php

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