help with a php form please

[monkeymade]

ok, I'm sure this is something simple I'm doing wrong, it always is lol.

 

I'm making a form, instead of a normal submit button, I am using an image.  The code for the submit button is this:

 

<input type="image" src="images/savedata.jpg" border="0" name="submit" value="test">

 

Then, I try and retrieve the value of it.  I have more than one form on the page, and I do it this way so I know which form was filled out:

 

<?PHP
$submit = $_REQUEST['submit'];
echo "the value of submit is $submit";
?>

 

This should echo out the word "test" since thats the value I assigned to the button (or so I'd hoped), but instead it comes out blank.

 

I've also tried

 

$submit = $_POST['submit'];

 

Thinking that maybe request just wasn't working in this instance, but still with no luck.  Anyone know what I am doing wrong here?  Or if this can even be done?  Thanks for your help!

[bpops]

I've personally never used an image for a submit button, so can't help you there, but you could always use a hidden variable to accomplish this easily:

 

<input type="hidden" name="hiddenSubmit" value="test">

[revraz]

The input type should be "submit".  Maybe change the graphic with CSS.

[bpops]

But the type="image" should work for submitting form, it's just the value he's having trouble getting.

[redarrow]

let see the whole form then...

[redarrow]

try this please..

 

<INPUT TYPE="image" SRC="images/savedata.jpg" HEIGHT="30" WIDTH="173" BORDER="0" ALT="Submit Form">

 

 

 

 

[monkeymade]

nope RedArrow, that didn't do it either, just scrunched up my image lol.

 

I don't know if it will help at all, but this is the page I am using it on:

 

http://admin.dadnladlawnservice.com/customermaint.php

 

There are 5 images at the bottom, each will be a "submit button" and each will do different things, this is why I am not using a hidden field, because its basically just 1 form, with 5 different submit buttons.  A hidden field would get entered with all 5, and not tell me which one was pushed.

 

I haven't added the security part yet, so its open to anyone to look at it, if you can help with this that would be great!

[redarrow]

you can not name the image submit button example and try!

 

working example with image as submit button

<?php
$name=$_POST['name'];
if($name){
echo $name;
}
?>


<form method="POST" action=" ">
<br>
<br>
your name please
<br><br>
<input type="text" name="name">
<br><br>
<INPUT TYPE="image" SRC="http://z.about.com/d/paranormal/1/0/4/S/myrtle_ghost.jpg" HEIGHT="10%" WIDTH="10%" BORDER="0" ALT="Submit Form">
</form>

[redarrow]

your have to use the isset around a form condition.........

<?php

$name=$_POST['name'];

if(isset($name)){

echo $name;
}

?>

[monkeymade]

ahh the isset, that might work for what I want to do.  Thanks guys

[Jessica]

Actually,

<?php
if(isset($_POST['name'])){
   $name=$_POST['name'];
   echo $name;
}
?>

 

is more correct and will not trigger a warning.