Undefined index form error

Question

i read all answer about undefined index error but not help full for me because i'm already using isset function to check plz how to slove this problem..

<?php
$con=mysqli_connect("localhost","root","","contact");
if (mysqli_connect_errno())
{
    echo "failed".mysqli_connect_error();
    }

checking for submited data

if(isset($_POST['submit']))
        {
        $name=$_POST['name']; 
        $website=$_POST['website']; 
        $gender=$_POST['gender']; 
        $comment=$_POST['comment']; 
        }
        $sql="insert into form(name,website,gender,comment) Values('$_POST[name]','$_POST[website]','$_POST[gender]','$_POST[comment]')";

        if(!mysqli_query($con,$sql))
            {
                die('error:'.mysqli_error($con));
            }
        else "added";

        mysqli_close($con);
        ?>
    <html>
        <body>
            <form method=post action="<?php echo $_SERVER['PHP_SELF']; ?>">
                Name: <input type="text" name="name"><br>
                E-mail: <input type="text" name="email"><br>
                Website: <input type="text" name="website"><br>
                <input type="radio" name="gender" value="female">Female
                <input type="radio" name="gender" value="male">Male<br>
                Comment: <textarea name="comment" rows="5" cols="40"></textarea>
                 <input type=submit name="submit"><br>
            </form>
        </body>

these errors comes

Notice: Undefined index: name in H:\Wamp\Xamp\htdocs\form.php on line 15

Notice: Undefined index: website in H:\Wamp\Xamp\htdocs\form.php on line 15

Notice: Undefined index: gender in H:\Wamp\Xamp\htdocs\form.php on line 15

Notice: Undefined index: comment in H:\Wamp\Xamp\htdocs\form.php on line 15

possible duplicate of PHP: "Notice: Undefined variable" and "Notice: Undefined index" — Rikesh
– Rikesh, Commented Apr 8, 2014 at 4:25
Take the closing bracket of isset($_POST['submit']) and place it in between mysqli_close($con); and ?> — Vagabond
– Vagabond, Commented Apr 8, 2014 at 4:29

Kraang Prime · Accepted Answer · 2014-04-08 04:35:29Z

Please try the following corrected code :

if(isset($_POST['submit']))
        {
        $name=isset($_POST['name']) ? $_POST['name'] : ''; 
        $website=isset($_POST['website']) ? $_POST['website'] : ''; 
        $gender=isset($_POST['gender']) ? $_POST['gender'] : ''; 
        $comment=isset($_POST['comment']) ? $_POST['comment'] : ''; 
        $sql="insert into form(name,website,gender,comment) Values('$name','$website','$gender','$comment')";

        // Open the database connection here
        // aka, mysqli_connect()


        if(!mysqli_query($con,$sql))
            {
                die('error:'.mysqli_error($con));
            }
        else "added";

        mysqli_close($con);
        }
        ?>
    <html>
        <body>
            <form method=post action="<?php echo $_SERVER['PHP_SELF']; ?>">
                Name: <input type="text" name="name"><br>
                E-mail: <input type="text" name="email"><br>
                Website: <input type="text" name="website"><br>
                <input type="radio" name="gender" value="female">Female
                <input type="radio" name="gender" value="male">Male<br>
                Comment: <textarea name="comment" rows="5" cols="40"></textarea>
                 <input type=submit name="submit"><br>
            </form>
        </body>

What I did is added validation to check if those fields are set, and if so, then set the value, if not, then set the variable (aka $name) to ''. You should probably add some further validation in the event of required fields being = '' (equal to blank).

I also adjusted your query to not use the $_POST vars, instead it uses the variables that you are assigning the $_POST values to, so you know they exist for sure.

And lastly, I moved the mysql connection code and query itself into the if(isset(submit)) statement so it does not try to process those on regular page load where the form has not been submitted yet.

Jaykumar Patel · Accepted Answer · 2014-04-08 04:27:31Z

1

Update this insert query,

$sql="insert into form(name,website,gender,comment) values('". $name ."','". $website ."','". $gender ."','". $comment ."')";

Hope this help you!

answered Apr 8, 2014 at 4:27

Jaykumar Patel

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Comments

Jenz · Accepted Answer · 2014-04-08 04:33:12Z

0

Change

if(isset($_POST['submit']))
        {
        $name=$_POST['name']; 
        $website=$_POST['website']; 
        $gender=$_POST['gender']; 
        $comment=$_POST['comment']; 
        }
        $sql="insert into form(name,website,gender,comment) Values('$_POST[name]','$_POST[website]','$_POST[gender]','$_POST[comment]')";
        if(!mysqli_query($con,$sql))
        {
            die('error:'.mysqli_error($con));
        }
       else "added";

       mysqli_close($con);
       ?>

to

if(isset($_POST['submit']))
{
        $name=$_POST['name']; 
        $website=$_POST['website']; 
        $gender=$_POST['gender']; 
        $comment=$_POST['comment']; 
        $sql="insert into form(name,website,gender,comment) values ('$name','$website','$gender','$comment')";
        if(!mysqli_query($con,$sql))
        {
            die('error:'.mysqli_error($con));
        }
        else "added";
        mysqli_close($con);

  }?>

edited Apr 8, 2014 at 4:33

answered Apr 8, 2014 at 4:25

Jenz

8,3718 gold badges50 silver badges83 bronze badges

2 Comments

Jaykumar Patel Over a year ago

Try @user1153551 answer

Rikesh Over a year ago

@Zu007 - take your complete query code inside if condition not just a sql statement.

Zu007 · Accepted Answer · 2014-04-08 04:34:37Z

0

than friends problem is slove What wrong?

if(isset($_POST['submit']))
        {
        $name=$_POST['name']; 
        $website=$_POST['website']; 
        $gender=$_POST['gender']; 
        $comment=$_POST['comment']; 


        $sql="insert into form(name,website,gender,comment) Values('". $name . "','" . $website . "','" . $gender . "','" . $comment . "')";

//these also in if(isset()) Block

if(!mysqli_query($con,$sql))
            {
                die('error:'.mysqli_error($con));
            }
        else "added";
        }

thanx to all

answered Apr 8, 2014 at 4:34

Zu007

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