Question

我是php的初学者。我正在尝试创建一个Web应用程序，其中用户得分会针对他回答的每个正确答案进行更新。脚本如下

<?php
$username="surendra";
$useranswer=$_REQUEST['option'];
$qno=$_REQUEST['a'];

  $con=mysqli_connect("localhost","root","","gk");

//检查连接

  include('connection.php');
    $sql1="SELECT * FROM questions  WHERE  qid=$qno";
    $result=mysql_query($sql1);
while ($row = mysql_fetch_array($result))
{
    $correctanswer=$row['answer'];

}

   include('connection.php');
     $sql1="SELECT * FROM contest  WHERE  username LIKE '".$username."'";
     $result=mysql_query($sql1);
while ($row = mysql_fetch_array($result))
{
    $points=$row['points'];

}   

    if ($username=$correctanswer)
      {
         $con=mysqli_connect("localhost","root","","gk");
          // Check connection
       if (mysqli_connect_errno())
         {
          echo "Failed to connect to MySQL: " . mysqli_connect_error();
          }

        mysqli_query($con,"UPDATE contest SET points=$points+5
         WHERE username =$username");

          mysqli_close($con);
         }

  ?>

更新功能不起作用请帮助我吗？

Answer 1

尝试将$username括在单引号中：

mysqli_query($con,"UPDATE contest SET points=$points+5 WHERE username = '$username'");

if ($username==$correctanswer)和其他人一样。

然后我建议你学习如何绑定params并使用预处理语句。阅读here

Answer 2

您错过了=标志

if ($username=$correctanswer)
           --^

尝试：

if ($username == $correctanswer)

此外，还需要引用$username：

mysqli_query($con,"UPDATE contest SET points = $points + 5 WHERE username ='".$username."'");

Answer 3

您的代码中有两次include('connection.php');。

if ($username=$correctanswer)应为==。

考虑使用mysqli_ functions或PDO并保护您的代码免受SQL注入。

更新脚本不起作用？

3 个答案: