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You are attempting to pass JavaScript variables to PHP using a hidden input in a form. However, your code is unable to retrieve the value of $_POST['hidden1'] into $salarieid. This is because you cannot directly pass variable values from JavaScript to PHP.
PHP executes on the server-side, and it does not have knowledge of client-side operations. To transfer variables to PHP, you must employ alternative mechanisms, such as submitting the form using the GET or POST methods.
A more effective approach would be to submit the form and process the variable using PHP. Here is a revised code example:
<!DOCTYPE html> <html> <head> <title>My Test Form</title> </head> <body> <form method="POST"> <p>Please, choose the salary id to proceed result:</p> <p> <label for="salarieids">SalarieID:</label> <?php $query = "SELECT * FROM salarie"; $result = mysql_query($query); if ($result) : ?> <select>
This code creates a form with a drop-down menu populated with salary IDs retrieved from a database. When the user selects an ID and submits the form, the selected value is posted to the PHP script. The PHP script then uses the posted value to query the database and display the corresponding information.
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