dongliang1996 2015-03-26 16:58
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MYSQL PHP查询不插入数据

This code inserts data from a form into a database named "test" and table named "signup". The insert query is not working:

   <?php
    echo"working2";
    if(isset($_POST['submit']))
    {
    echo"woerking";
    $conn = mysql_connect("localhost", "root", "", "test") or die("error connecting");
    echo"woerking1";
        $flag=0;
        $Rno=$_POST['rno'];
        echo$Rno;
        $fname=$_POST['fname'];
        echo $fname;
            $lname=$_POST['lname'];
        $email=$_POST['uemail'];
            $pswd=$_POST['pswd'];
        $repswd=$_POST['repswd'];

        //$choice="select Rollno from signup where Rollno='$Rno' ";
        //$chk=mysql_query($choice);
            //echo $chk;
        //while($row=mysql_fetch_array($chk))
        //{
            //echo "<font color='red'>ROLL-NUMBER ALREADY EXIST!!!!</font>";
            //$flag=1;
        //}

        if($flag==0)
    {
       $query="INSERT INTO `signup`(`Rollno`, `fname`, `lname`, `email-id`,    `password`, `retype`)     VALUES($Rno,'$fname','$lname','$email','$pswd','$repswd')";
*##//query not working##* 
            $tem=mysql_query($query);
                    if(isset($tem))
            {
                echo"REGISTERED SUCCESSFULLY";
            }
        }
    }
    ?>
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2条回答 默认 最新

  • dongyong3554 2015-03-26 17:46
    关注

    Try this change to start:

    $tem=mysql_query($query) or die(mysql_error());
    

    That will tell you the problem with the query. Unfortunately it won't tell you about the SQL injection vulnerabilities in your code - but please fix them anyway!

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
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