PHP echo out jQuery in IF statement - php

Question

EDIT: What a TW*T. Sorry everyone for wasting your time. Just missed a Google jQuery link on one F'in page. Whoops.
Hi, i have a div containing 3 forms. These should be the only thing on the screen on page load. When any of them are submitted, a graph gets shown below. What i'm trying to do is within the PHP IF statement is make the div disappear that contains the forms. Sound simple?
This is my code:
if($_GET['submit1']){
echo "<script type='text/javascript'>$('#options').css('display','none');</script>";
However, when i do submit one of the forms (therefore a $_GET has occurred) the div is still there??
EDIT:
If i try people answer on one line i get this:
Parse error: syntax error, unexpected '(', expecting T_VARIABLE or '$'
But if i put in people's multiline answers, no error, but div still shows!

Answer 1

Why do you want to hide the forms with JavaScript?
Simply do it with PHP:
<?php
if(!isset($_GET['submit1'])) {
?>
//<form> your form HTML here
<?php
} else {
?>
<p>Your submitted data: <?php print_r($_GET); ?></p>
<?php
}
?>
So your forms are only shown if you have NOT submitted one of them. You might have to adjust the parameters if you have multiple forms, for example
if(!isset($_GET['submit1']) && !isset($_GET['submit2'])) {
Edit: If you want to keep your forms after submitting but only hide it, you could do it that way:
<?php
$formsVisible = !isset($_GET['submit1']));
$formsDisplay = $formsVisible ? 'block' : 'none';
?>
<form style="display:<?php echo $formsDisplay; ?>">
<!-- ... --->
</form>

Answer 2

You need to add a DOM ready event:
if($_GET['submit1']) {
echo '<script type="text/javascript">' . "\n";
echo ' $(function() {' . "\n";
echo ' $("#options").hide();' . "\n";
echo ' });' . "\n";
echo '</script>' . "\n";
}

Answer 3

edit
You are getting a parse error because you're using double quotes, so the php parser is reading the dollar sign as a php variable. I would switch to a single quote php syntax to make your life easier:
if($_GET['submit1']){
echo '<script type="text/javascript">
$(function(){
$("#options").css("display","none");
});
</script>';
}
Be sure to wrap your jquery code in the onLoad function $(function(){};