PHP and Buttons in HTML

[mag1]

Hi, I've got a form with a number of input fields on it. I have a javascript function that checks the fields on the click of a submit button. I've also create a hidden input field name=DBConnect this value is set to false on entering the javascript function and to True providing all is OK. Please see below code that I was hoping would work but isn't, any suggestions as to how I should go about doing this? Is it possible?

<button onClick='CheckFields()'
<?php if($DBConnect=="false")
{?> type=button name="Check"<?php}
else if($DBConnect=="True"){?>type=submit name="submit"<?php}?>> Enter Complaint</button>

Thanks

[coopster]

Welcome to WebmasterWorld, mag1.

PHP processing happens at the server-side, not on the client. PHP can write out your code for you, but all PHP processing happens at the server. Maybe if you explain a bit more of what you are trying to accomplish, we can help you overcome the issue.

[mag1]

Hi,

What I'm trying to do is, based on the result of the CheckFields function i.e. Value of DBConnect, I want to determine the type of button. If DBConnect = false then I want the button type to be button, if it's true I want the button type to be submit. I'd like to do some Client side checking of fields before going to the server side.

Does this expain it any better?

Thanks.

[coopster]

OK, I think I follow. Have you checked out the JavaScript Jumpstart -- Form Validation threads in the Javascript Library [webmasterworld.com]? I think they explain how to do what you are looking for.

[mcibor]

To do on client checking use javascript, not PHP

in the html:

<form name="test" id="test" action="index.html" method="POST"><button onclick="java_check(); return false;">Submit</button></form>

in javascript

function java_check()
{
//here you check everything, if fine

document.test.submit();
}

Hope it helps you somehow!
Best regards
Michal Cibor