PHP variable to URL in address bar

Question

I have the code:

echo('
 <div class="row">
    <div class="col-sm-2" > 
    <form method="POST" action="plan.php">
<p>Selectati data raportului:<br>
 <select name="SelectDR" onchange="form.submit()">
 <option value="">Select...</option>
');

It is a form from which I select a value.

$option = isset($_POST['SelectDR']) ? $_POST['SelectDR'] : false;
if ($option){
    $queryTXT = "SELECT * FROM ".$tabel." WHERE DataRaport='".$option."'";
...

The records of a SQLtable are filtered by the value selected from the form.

Now, what I want is to display in the url plan.php?DataRaport& variable php

I've tried to put in the form url (<form method="POST" action="plan.php?DataRaport'.$option.'">), but that doesn't show the variable.

If you want to get the DataReport from the URL, you would need to use GET — Phiter, Feb 19 '16 at 11:33

score 5 · Answer 1 · answered Feb 19 '16 at 11:34

5

You need GET not POST

<form method="GET">

answered Feb 19 '16 at 11:34

Ben Rhys-Lewis

3,118
8
34
45

score 0 · Answer 2 · answered Feb 19 '16 at 11:39

First of all. If you want to echo this HTML every time, just use it as HTML:

<div class="row">
<div class="col-sm-2" > 
<form method="POST" action="plan.php">
     <p>Selectati data raportului:<br>
     <select name="SelectDR" onchange="form.submit()">
     <option value="">Select...</option>
     </select>
</form>

<?php //your php stuff

To get the value from the URL, you need to change one thing:

<form method="POST" action="plan.php?DataReport=<?php echo $option?>">

Then you would get this variable as:

$dataReport = $_GET['DataReport'];

This would return the option value as POST and the DataReport as GET.

You can change the form method to GET if you wish, but you would then need to use $_GET to pick the selected option.

PHP variable to URL in address bar

2 Answers2