CODEWITHSUNDEEP
phphtml
Yujian
Yujian

Reputation: 169

How to pass value of current PHP file to another PHP file by clicking a button?

I'm trying to link one php file to another one. The first file, which is a query result of one mysql command, contains introduction of all the products.

enter image description here

When the "Details" button is clicked, I want the page to be redirected to another php file, which accepts the query result (ID, name) from the first file and output the detailed information.

Now I'm stuck at the "button" step. Could anyone help me to link the two files?

<div><button href = "Product_details.php" type="submit" name="Details" class="btn btn-default">Details</button></div>

Upvotes: 0

Views: 2377

Answers (1)

Yassine CHABLI
Yassine CHABLI

Reputation: 3724

you can use the $_GET variable like : 

<div><button href = "Product_details.php?data1=value2&data2=value2" type="submit" name="Details" class="btn btn-default">Details</button></div>

value1 and value2 are supposed as two variables available in the current php file which the button exist named $data1 and $data2

it's goning to be something like :

<div><button href = "Product_details.php?data1=<?php echo $data1?>&data2=<?echo $data2?>" type="submit" name="Details" class="btn btn-default">Details</button></div>

and in the Product_details.php , you can get you data as : 

$data1 = $_GET['data1'];
$data2 = $_GET['data2'];

Upvotes: 5

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