<FORM> HELP!!!!!!!!!!!!!!!!!

[Anon]

I nave a form with multiple <input type=image> buttons, I want to echo the value of the button chosen when the user pushes the button

******************Example*************
/////////////////////form.html

<form method=post action=script.php>
<input type=image src=images/button.gif name=button_1 value=2>

<input type=image src=images/button.gif name=button_2 value=3>

<input type=image src=images/button.gif name=button_3 value=4>

<input type=image src=images/button.gif name=button_4 value=5>
</form>

///////////////////script.php

<?
if ($button_1 && $button_1 == 2)
{
echo 'button_1';
}
elseif ($button_2 && $button_2 == 3)
{
echo 'button_2';
}
elseif ($button_3 && $button_3 == 4)
{
echo 'button_3';
}
elseif ($button_4 && $button_4 == 5)
{
echo 'button_5';
}
?>
******************Example End**********

If I choose a button the variable is not passed to the script.php how do you assign variables to <input type=image>?

Thanks

[Anon]

Hey,

Several things I would do.

First, it looks as though you don't have register_globals enabled in your php.ini file. You can go and enable them but, for security purposes, I advise you against it. On the other hand, you can use the following to refer to form variables from a submitted form. Instead of $button_1, use $HTTP_POST_VARS['button_1'] (I took note of the fact you used the POST method, replace POST with GET if you choose to switch to GET in your HTML form).

If this doesn't work, tune up these things:

Put quotes around all of your attributes. I don't know about PHP, but JavaScripthates non-quoted attributes.

I don't quite understand the logic in your if statements. If your trying to make sure that $button_1 is defined AND that it's value is equal to 2, your syntax is a bit off. PHP has a special function to check is something is defined or not. It's isset().

PHP symnol for AND is &, not &&.

Also, picture this scenario. If $button_1 is not defined, then PHP won't like the part after the && because $button_1 is not defined so it doesn't have the variable to test for equality. This would make your code a bit more complex, but try nesting another IF that checks for equality inside the IF that checks for definition. You'll be able to avoid this problem.

I know all of this can be a little confusing, so I won't mind if you reply asking me to convert your code to code based on what I wrote. I'm pretty busy so it might take me am hour till I get to it, but I'll do it.

Good Luck, elenev

[Anon]

these 2 code snippets work the same, isset is not needed

if (isset($button_1) && $button_1 == 2)
{
}

if(($button_1) && $button_1 == 2)
{
}

Also, I would not use the example I supplied in my application, I use much more complex methods that are object generated, if I used my code for examples in this forum very few would understand it and I would not get any replies

[mrskhris]

chris, there are some specifics in using <input type=image> control. Let's say you have this tag in your form:
<input type="image" src="foo.gif" name="my_image" value="2">

When you submit your form, you'll get 2 variables:
my_image_x and my_image_y, which are coordinates of the point where user clicked. But there will be no variable my_image...