2015-09-02 34 views
0
<?php 
     echo "Hello test <br>"; 
     $userid = $_GET['userid']; 
     echo "your user userid is " . $userid . " &"; 
     $salt = $_GET['salt']; 
     echo " your user salt is " . $salt; 
     // Query for finding the data from db 
     // Issue in query 
     $sql = "SELECT * FROM test.test where id=" .$userid AND "salt=".$salt; 
     echo "<br>" . $sql; 
     $result = $conn->query($sql); 
     if (!empty($result)) 
     {   
      echo "<br>Result Found"; 
     } 
     else 
     { 
      echo "<br> Invalid link !"; 
     } 
     // } 
?> 

我的查詢無法正常工作。如果我減少我的查詢id=".$userid它工作正常,但如果我添加其餘部分不工作。問題。不工作

+0

至少使用的代碼高亮編輯器。這將避免你偶然發現像這樣的簡單語法錯誤。 –

回答

1

$sql行應該是:

$sql = "SELECT * FROM sauberlux_com.tbl_b2cuser where id=".$userid." AND salt=".$salt; 
+0

我也嘗試過這個,但它的工作不正常。 – Atif

+0

問題是當我運行這個查詢它給了我$結果==錯誤,但它應該是真實的。 – Atif

0
$sql = "SELECT * FROM sauberlux_com.tbl_b2cuser where id = $userid AND salt = $salt"; 

試試吧。你

0
$sql = "SELECT * FROM sauberlux_com.tbl_b2cuser where id = $userid AND salt = '$salt' "; 

也可以指here SQL注入