Javascript SyntaxError: Unexpected identifier

Question

I wrote some small script, but since I did it with Google help I need some advice. After putting this into Chrome console, I'm getting this error:

SyntaxError: Unexpected identifier

Here is my code:

jQuery(document).ready(function(){ 
    if jQuery("#colorbox").css("display", "none");) {
        jQuery("#close-news").css("display", "none");
    } 
    else {
        jQuery("#close-news").css("display", "visible");
    }
});

Any ideas what's wrong?

If you get a SyntaxError, that means you have a SyntaxError, and need to take a close look at your code, and correct the invalid JavaScript. — user1106925
– user1106925, Commented Jul 25, 2012 at 15:32
the script is works now, but i dont see any changes on my page. Was it correct to to use .css on colorbox id, while it has .style in code — Avdept
– Avdept, Commented Jul 25, 2012 at 15:36

Alex · Accepted Answer · 2012-07-25 15:31:13Z

2

change:

if jQuery("#colorbox").css("display", "none");)

to:

if (jQuery("#colorbox").css("display") == 'none')

answered Jul 25, 2012 at 15:31

Alex

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Alex Dn · Accepted Answer · 2012-07-25 15:32:16Z

1

if (jQuery("#colorbox").css("display", "none")) {
    jQuery("#close-news").css("display", "none");
} 
else {
    jQuery("#close-news").css("display", "visible");
}

answered Jul 25, 2012 at 15:32

Alex Dn

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Comments

epascarello · Accepted Answer · 2012-07-25 15:46:15Z

0

Another way to write it with the syntax errors fixed:

jQuery(function () {
    if (jQuery("#colorbox").is(":visible")) {
        jQuery("#close-news").css("display", "none");
    } else {
        jQuery("#close-news").css("display", "visible");
    }
});

and even simpler with toggle

jQuery(function () {
    var isVisible = jQuery("#colorbox").is(":visible")) {
    jQuery("#close-news").toggle(!isVisible);        
});

answered Jul 25, 2012 at 15:46

epascarello

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