Question

此代码将表单中的数据插入名为“test”的数据库和名为“signup”的表中。插入查询无效：

   <?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";
            }
        }
    }
    ?>

Answer 1

尝试此更改开始：

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

这将告诉您查询的问题。不幸的是，它不会告诉你代码中的SQL注入漏洞 - 但无论如何请修复它们！

Answer 2

$query="INSERT INTO `signup`(`Rollno`,`fname`,`lname`,`email-id`,`password`,`retype`)VALUES('".$Rno."','".$fname."','".$lname."','".$email."','".$pswd."','".$repswd."')";

尝试这样..

MYSQL PHP查询不插入数据

2 个答案: