PHP: how to change url of php page on pressing html button?

Question

I want to go to other php page on pressing html button of same site. I have tried it by using header function but it was not working.
Here is the simple html code:

<input type="button" name="Insert_Ad" value="Post and ad">

If user clicks this button it will take it to another php page For Example with URL adddress 'http://localhost/Product.php/Product.php'

Here is the code of that PHP page to which I want to go on pressing button

   <html>
    <head>
        <meta charset="UTF-8">
        <title>Insert form data</title>
    </head>
    <body>
    <form method="post" action ="Product.php" id="contact-form">

        <input type="text" name="product_category" placeholder="product_category"  required />



        <input type="text" name="product_name"  placeholder="product_name" required />

        <textarea id = "address" name="product_description" placeholder="product_description"  required /></textarea>


        <input type="text" name="product_image1"  placeholder="product_image1" required />


        <input type="text" name="product_image2"  placeholder="product_image2" required />

        <input type="text" name="product_image3"  placeholder="product_image3" required />

        <input type="text" name="product_image4"  placeholder="product_image4" required />

        <input type="text" name="product_image5"  placeholder="product_image5" required />


        <div class="btn-group" role="group">
            <input type="submit" class="btn btn-default" name="Add" value="Enter the box" style="margin-top: 15px; margin-right: 15px; border-radius: 4px;">

        </div>

    </form>

    <?php
$servername = "localhost";
$username = "root";
$password = "zz224466";
$dbname = "zain";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

if(isset($_POST['Add']))
{
    $product_category = $_POST['product_category'];
    $product_name = ($_POST['product_name']);
    $product_description = ($_POST['product_description']);
    $product_image1 = ($_POST['product_image1']);
    $product_image2 = ($_POST['product_image2']);
    $product_image3 = ($_POST['product_image3']);
    $product_image4 = ($_POST['product_image4']);
    $product_image5 = ($_POST['product_image5']);


    $sql = "INSERT INTO zain.product (product_category,product_name,product_description,product_image1,product_image2,product_image3,product_image4,product_image5)
VALUES ('$product_category','$product_name','$product_description','$product_image1','$product_image2','$product_image3','$product_image4','$product_image5')";
mysqli_query($conn,$sql);
    if (mysqli_query($conn,$sql))
    {
        echo "New record created successfully";
    }
    else
    {
        echo "Error: " . $sql . "<br>" . $conn->error;
    }
    $conn->close();
}
?>


    </body>
   </html>

Kindly help me how to do that

On pressing HTML button I want it will take user to php page — Zain Farooq, Jan 23 '16 at 13:11
i think, what you are looking for is ajax https://developer.mozilla.org/en-US/docs/AJAX/Getting_Started. Or do you just want a link, looking like a button? ... PHP is only serverside, clientside its html and js, so you can use a link to go to another url or you can load new content via ajax — Jarlik Stepsto, Jan 23 '16 at 13:11
actually it should be possible to do it over isset post request. could you provide the error you are getting or is simply nothing running? — Philipp Braun, Jan 23 '16 at 13:13
When I use isset function it does not show any error but does not work too — Zain Farooq, Jan 23 '16 at 13:15
can you create a fiddle with your code and error (or provide a link to your page) .... see also: http://stackoverflow.com/questions/2906582/how-to-create-an-html-button-that-acts-like-a-link — Jarlik Stepsto, Jan 23 '16 at 13:16
change echo "New record created successfully"; to header("location:abc.php"); — Sahil Manchal, Jan 23 '16 at 13:17

score 1 · Accepted Answer · answered Jan 23 '16 at 13:15

1

This is the simplest way:
<input type="button" name="Insert_Ad" value="Post and ad" onclick="location.href='your_url_here'">

Or you can use a form to submit to the given url:
<form action="your_url_here"> <input type="submit" name="Insert_Ad" value="Post and ad"> </form>

answered Jan 23 '16 at 13:15

trank

952
11
8

Ok it was my mistake.. I find it working... Thankyou very much – Zain Farooq Jan 23 '16 at 13:21

Brian Mosigisi · Answer 2 · 2017-04-04T15:34:35.500

I would recommend using an anchor tag. If you want to give it the appearance of a button, then use CSS to style it into a button. For instance, if you were using twitter bootstrap, this would involve giving it a class of btn

<a href="/Product.php" class='btn'>Post an ad</a>

If you insist on using a button, then you might want to wrap it around a HTML form, where the action attribute points to the page you would like to navigate to. The method has to be GET. If it is the only button in the form, then it will default to being the submit button even in the absence of a type="submit" attribute.

<form action="/Product.php" method="GET">
    <input type="button" name="Insert_Ad" value="Post and ad">
</form>

PHP: how to change url of php page on pressing html button?

2 Answers2