I need to simply pass a form variable into a URL variable. I suspect that it's something easy to do but I'm having a hard time finding clear steps (that aren't tons of code) online anywhere.

Here's my current form code

<form id="zip_search" method="post" action="dealers.php">
    <label for="zipfield"><a href="dealers.php">Find a Dealer</a></label>
    <input name="tZip" type="text" id="zipfield" value="ZIP CODE" onblur="if(this.value=='') this.value='ZIP CODE';" onfocus="if(this.value=='ZIP CODE') this.value='';" />
    <input type="image" value="Submit" class="submitbutton" src="/images/submit_button.gif" />
  </form>

And all I need is for it to send the browser to something like this:

.php?zip=55118

Thank you in advance for any help.

---

Update to question

Thanks to Drew and Anton's responses here's an update. Changing the input name attribute to match the URL var name (tZip to zip) along with changing POST to GET did the trick but for some reason it's adding two additional URL variables as well (&x=0&y=0). I'm guessing this is something incorrect with my PHP code as I'm not a PHP wizard by any stretch. Here's all the code:

PHP Function

<?php
function processForm() {
    $zipCode = $_GET['zip'];
    $url = "dealers.php?zip=" . $zipCode;
    header("Location: $url");
    exit;
}
?>

Form

<form id="zip_search" method="get" action="dealers.php">
    <label for="zipfield"><a href="dealers.php">Find a Dealer</a></label>
    <input name="zip" type="text" id="zipfield" value="ZIP CODE" onblur="if(this.value=='') this.value='ZIP CODE';" onfocus="if(this.value=='ZIP CODE') this.value='';" />
    <input type="image" value="Submit" class="submitbutton" src="/images/submit_button.gif" />
  </form>

URL Output Example

.php?zip=12345&x=0&y=0

---

Additional Related Question

How is this working if processForm() is only defined but not called anywhere else. It seems to me that the processForm() function should be in the action attribute in the opening form element. Any insight? Thanks in advance.

I need to simply pass a form variable into a URL variable. I suspect that it's something easy to do but I'm having a hard time finding clear steps (that aren't tons of code) online anywhere.

Here's my current form code

<form id="zip_search" method="post" action="dealers.php">
    <label for="zipfield"><a href="dealers.php">Find a Dealer</a></label>
    <input name="tZip" type="text" id="zipfield" value="ZIP CODE" onblur="if(this.value=='') this.value='ZIP CODE';" onfocus="if(this.value=='ZIP CODE') this.value='';" />
    <input type="image" value="Submit" class="submitbutton" src="/images/submit_button.gif" />
  </form>

And all I need is for it to send the browser to something like this:

http://www.mydomain./dealers.php?zip=55118

Thank you in advance for any help.

---

Update to question

Thanks to Drew and Anton's responses here's an update. Changing the input name attribute to match the URL var name (tZip to zip) along with changing POST to GET did the trick but for some reason it's adding two additional URL variables as well (&x=0&y=0). I'm guessing this is something incorrect with my PHP code as I'm not a PHP wizard by any stretch. Here's all the code:

PHP Function

<?php
function processForm() {
    $zipCode = $_GET['zip'];
    $url = "dealers.php?zip=" . $zipCode;
    header("Location: $url");
    exit;
}
?>

Form

<form id="zip_search" method="get" action="dealers.php">
    <label for="zipfield"><a href="dealers.php">Find a Dealer</a></label>
    <input name="zip" type="text" id="zipfield" value="ZIP CODE" onblur="if(this.value=='') this.value='ZIP CODE';" onfocus="if(this.value=='ZIP CODE') this.value='';" />
    <input type="image" value="Submit" class="submitbutton" src="/images/submit_button.gif" />
  </form>

URL Output Example

http://www.domain./dealers.php?zip=12345&x=0&y=0

---

Additional Related Question

How is this working if processForm() is only defined but not called anywhere else. It seems to me that the processForm() function should be in the action attribute in the opening form element. Any insight? Thanks in advance.

• php
• javascript
• forms
• url
• variables

Share Improve this question Follow edited Dec 1, 2010 at 17:33 Brian Larson asked Nov 30, 2010 at 19:56 Brian LarsonBrian Larson 4511 gold badge11 silver badge88 bronze badges 1

• Others have suggested the solution for you. To get the essence of this I'd suggest you o follow this course cs75.tv/2009/fall – Anton S Commented Nov 30, 2010 at 22:24

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2 Answers 2

Sorted by: Reset to default 4

Change the form method to "get"

You will need to change the form method from POST to GET and also rename the text input from tZip to zip otherwise your URL will look like this:

http://www.mydomain./dealers.php?tZip=55118

instead of

http://www.mydomain./dealers.php?zip=55118