PHP echo out jQuery in IF statement

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!

Can you edit the answer with the parse error you're receiving? — JakeParis
– JakeParis, Commented Jan 24, 2011 at 14:39

Alex · Accepted Answer · 2011-01-24 14:42:21Z

3

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>

edited Jan 24, 2011 at 14:42

answered Jan 24, 2011 at 14:35

Alex

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2 Comments

benhowdle89 Over a year ago

The reason i did it with javascript is so i could add a button to slideToggle() the form back in/out with jQuery

benhowdle89 Over a year ago

Looks good, i'm just a little confused. I have forms, then on same page, but below the form, the PHP if $_GET. So i'm not sure how to fit your answer in

Stephen · Accepted Answer · 2011-01-24 14:34:03Z

1

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";
}

edited Jan 24, 2011 at 14:34

answered Jan 24, 2011 at 14:28

Stephen

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7 Comments

benhowdle89 Over a year ago

echo "<script type='text/javascript'>$(document).ready(function(){$('#options').hide();});</script>"; produces an error: Parse error: syntax error, unexpected '(', expecting T_VARIABLE or '$'

Stephen Over a year ago

Copy and paste my example verbatim.

Stephen Over a year ago

In your pasted example, you did not include the terminating bracket for the if block. }. Maybe you are adding it with my example, but it occurs later in your script, and that is what is breaking.

benhowdle89 Over a year ago

Hi Stephen, i have copied & pasted correctly, i get no error, but the div still shows!?

Stephen Over a year ago

Question: If you hardcode the script (outside the if block) so that it is supposed to hide the div every time, does it work?

|

JakeParis · Accepted Answer · 2011-01-24 14:42:21Z

0

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(){};

answered Jan 24, 2011 at 14:42

JakeParis

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