PHP: Sending Variable from Dropdown to URL

Question

I kindly ask for your input in how to send a variable via URL to an other PHP site:

  <form action="deny.php" method="get">
  <div align="left"></div>
  <p><span class="style13" style="height: 23px;">
    <select name="deny" class="style28" style="width: 320px">
      <option value="Select" selected="selected">Select</option>
      <option value="Price">Too expensive</option>
      <option value="Agency">Other Agency</option>
    </select>
    </span></p>
  <p><span class="style13" style="height: 23px;">
  <input type="hidden" name="id" value=<? echo $id; ?> />
  </span> </p>
  <p>
  <? echo '<td><a href="deny.php?submit='.$id.'&deny='.$_GET['deny'].'">Send Feedback</a></td>'; ?>
  </p>
</form>

$id is correct, but $deny is empty

I even tried with $deny (instead of $_GET['deny']) and $_POST[deny] - but $deny is always empty. (can be controlled in the link)

Thanks for your suggestions!

BR,

Stefan

you can't use a link to submit (without js) the form, change <a> to <input type="submit"..> — donald123
– donald123, Commented Dec 19, 2014 at 9:07

Ajay Patel · Accepted Answer · 2014-12-19 12:04:19Z

Replace your code with below one:

<script type="text/javascript" src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<form action="deny.php" method="get">
  <div align="left"></div>
  <p><span class="style13" style="height: 23px;">
    <select name="deny" class="style28" style="width: 320px">
      <option value="Select" selected="selected">Select</option>
      <option value="Price">Too expensive</option>
      <option value="Agency">Other Agency</option>
    </select>
    </span></p>
  <p><span class="style13" style="height: 23px;">
  <input type="hidden" name="id" value="<?php echo $id; ?>" />
  </span> </p>
  <p>
  <a href="#" id="feedbackLink">Send Feedback</a>
  </p>
</form>
<script type="text/javascript">
    $( document ).ready(function() {
        $('select[name="deny"]').change(function(){
            var link = 'deny.php?submit=<?php echo $id; ?>&deny=';
            $('a#feedbackLink').attr('href', link + $(this).val());
        });
        $('select[name="deny"]').trigger('change');
    });
</script>

You can't use $_GET, $_POST, etc. variables without submitting form.

Hope this will resolved your problem.

Quentin · Accepted Answer · 2014-12-19 09:08:08Z

1

$_GET['deny'] won't be populated until you submit the form to the current script.

Don't try to use PHP to construct the URL. Let the browser do it. Just put a submit button in the form and have appropriate inputs and action. That's what forms are for.

answered Dec 19, 2014 at 9:08

Quentin

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Comments

Amit Kumar Sahu · Accepted Answer · 2016-04-12 10:13:50Z

1

<form action="deny.php" method="get">
  <div align="left"></div>
  <p><span class="style13" style="height: 23px;">
    <select name="deny" class="style28" style="width: 320px">
      <option value="Select" selected="selected">Select</option>
      <option value="Price">Too expensive</option>
      <option value="Agency">Other Agency</option>
    </select>
    </span></p>
  <p><span class="style13" style="height: 23px;">

  </span> </p>
  <p>
  <input type="submit" value="submit">
  </p>
</form>

edited Apr 12, 2016 at 10:13

answered Dec 19, 2014 at 9:13

Amit Kumar Sahu

4956 silver badges16 bronze badges

2 Comments

Bosstone Over a year ago

Thank you for your comments and suggestions. I tried it with this echo "<input type='submit' name='submit' value='" . $id . "'>"; now everything will be passed, but the button name is the value of the id instead of the name "send feedback"... :-)

Amit Kumar Sahu Over a year ago

which text you want to display on button?

Mirceac21 · Accepted Answer · 2014-12-19 09:28:21Z

<form action="deny.php" method="get">
    <div align="left"></div>
    <p>
        <span class="style13" style="height: 23px;">

            <select name="deny" class="style28" style="width: 320px">
                <option value="Select" selected="selected">Select</option>
                <option value="Price">Too expensive</option>
                <option value="Agency">Other Agency</option>
            </select>
        </span>
    </p>
    <p>
         <span class="style13" style="height: 23px;">
              <input type="hidden" name="id" value="<?php echo $id; ?>" /> 
         </span>
    </p>
    <p>
         <input type="submit" value="Send Feedback" value="submit" />
    </p>
</form>

 <?php
      if (isset($_GET)) {
          var_dump($_GET);
      }
 ?>

Bosstone · Accepted Answer · 2014-12-19 09:39:56Z

0

<form action="/deny.php?id=<? $id ?>" method="get">

<input name="submit" type="submit" value="Feedback" />

Did the trick :-)

Thanks,

Stefan

answered Dec 19, 2014 at 9:39

Bosstone

1891 silver badge12 bronze badges

Comments

Raham · Accepted Answer · 2014-12-19 10:18:55Z

0

Try this one,it has been tested.

<form action="your_page.php" method="GET"> <!--you may use POST for security as well -->
      <select name="deny">
        <option value="">Select Option</option>
        <option value="Price">Too expensive</option>
        <option value="Agency">Other Agency</option>
      </select>
      <!-- you msut have $id = something -->
      <input type="hidden" name="id" value="<?php echo $id; ?>" /> 
      <input type="submit" value="Send Feedback" value="submit" />
  </form>

 <?php
      echo "<pre>";print_r(($_GET));
 ?>

answered Dec 19, 2014 at 10:18

Raham

4,9613 gold badges26 silver badges29 bronze badges

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