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Dhaneja
Dhaneja

Reputation: 27

How to upload a radio button value entered in an html form

I have made the form so that it gets the selected radio value button and it gets passed to the php section and to the database.But in the database it shows as "on" no matter what the selection is.

I have no idea where I have gone wrong

HTML form:

<form action="Database.php" name="register" method="post">
<div>
    First Name   <input type="text"  name="fname"/><br>
    Last Name    <input type="text"  name="lname"/><br>
    Email        <input type="email" name="email"/><br>
    Contact No.  <input type="text"  name="num"/><br>
    Gender <br>  <input type="radio" name="g1" value="Male"/>Male

                     <input type="radio" name="g1" value="Female"/>Female
            <br>
            <br>
</form>

PHP:

$fname      = $conn->real_escape_string($_POST['fname']);
$lname      = $conn->real_escape_string($_POST['lname']);
$email      = $conn->real_escape_string($_POST['email']);
$cnumber    = $conn->real_escape_string($_POST['num']);
$gender     = $conn->real_escape_string($_POST['g1']);

$sql="INSERT INTO data (fname,  lname, email, cnumber, gender) 

VALUES ('".$fname. "','".$lname."','".$email."', '".$cnumber."', '".$gender."')";

I expected output to be male/female but it says "on"

Upvotes: 0

Views: 759

Answers (1)

Trevor Ackermann
Trevor Ackermann

Reputation: 176

Using your form I have used a small code and found this works

<form action="Database.php" name="register" method="post">
    <div>
        First Name   <input type="text"  name="fname"/><br>
        Last Name    <input type="text"  name="lname"/><br>
        Email        <input type="email" name="email"/><br>
        Contact No.  <input type="text"  name="num"/><br>
        Gender <br>  <input type="radio" name="g1" value="Male"/>Male

        <input type="radio" name="g1" value="Female"/>Female
        <br>
        <input type="submit" style="min-width:100%" align="center" name='submit' value="SUBMIT">
        <br>
</form>

Your PHP side should be this:

if(!isset($_POST['submit']))
{
//This page should not be accessed directly. Need to submit the form.
    echo "Submitted";
}
$fname = $_POST['fname'];
$lname = $_POST['lname'];
$g1 = $_POST['g1'];   
               $sql = "INSERT INTO etrack.test SET
                    fname = '".$fname."',
                    lname = '".$lname."',
                    g1 = '".$g1."'";

    if ($conn->query($sql) === TRUE) {
        echo "";
    } else {
        echo "Error: " . $sql . "<br>" . $conn->error;
        echo("Error description: " . mysqli_error($con));
        echo "Error 1";
    }

I am getting the required results

Upvotes: 1

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