type=image value problems

[Pepe]

Hello php coders,

I am creating a form that uses images instead of buttons to submit information and I found some problems: type=image does not pass the value.

I am creating this system to delete lines from a shopping cart, therefore every "delete" button has a different value but I am not able to get it once the form is processed.

Exmaple:

<table>
<form name="formulario" action="pruebas.php" method="post">
<tr>
<td>
<input name="delete" type="IMAGE" value="7" src="a.gif" width="70" height="26" border="0">
</td>
</tr>
</form>
</table>

once this form is submitted I echo $delete and nothing appears on the screen, I could use $delete_x and get the some screen coordinates.

I got a workaraound but it looks a little complex:

I inserted the value of the button inside the name tag like

name="delete_id"

while (list ($clave, $val) = each ($HTTP_POST_VARS)) {
if (ereg("x", $clave) > 0){
$clave2_array = explode("_", $clave);
$clave2 = $clave2_array[1];
echo "The value is" . $clave2 . " and the button is" . $clave2_array[0];
} // if
} //while

this way I get the value passed.

This problem only happens on IE, Mozilla treats the type=image as a type=submit.

Did you find any other solution to this?

Thanks

[bad76]

Hi
you can use an onclick() function that fill an hidden called "delete" with right value...

[Pepe]

Originally posted by bad76
Hi
you can use an onclick() function that fill an hidden called "delete" with right value...

Using onclick() means javascript is enabled in the browser, I want my shopping cart to be JS FREE.

[bad76]

Just use an hidden input called "delete" 😃
for each image

[EDIT]
anyway i personally don't support this guys.
I think JS not dangerous for user.

(when else i support no cookie...)