查询中的问题。不工作

时间:2015-09-02 10:41:15

标签: mysql

<?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 !";
        }
        // }
?>

我的查询无法正常运行。如果我将查询减少到id=".$userid它可以正常工作,但如果我添加剩余部分则无效。

3 个答案:

答案 0 :(得分:1)

$sql行应为:

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

答案 1 :(得分:0)

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

试试吧。

答案 2 :(得分:0)

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

您也可以参考here进行sql注入