Jquery .show() not working in firefox

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 4  2072

南旧

I have a problem with the jquery show function in firefox.

Situation: I have loaded a iframe with an aspx page in a popup (fancybox). In that aspx page I have a button

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醉话见心

2021-02-14 02:07
I had a similar problem. I did set this in my css:
```
body { display: none }
```
and tried to change this in jQuery:
```
$("body").show();
```
which did not work in Firefox.

After I changed the HTML to
```
<body style="display: none">
```
and removing the part from CSS I was able to use jQuery show(); flawless

It seems that the inheritance of css rules does not follow the same routine in Firefox as in IE / Chrome.
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不知归路

2021-02-14 02:25

It may be just a side-effect of a js error. It's always worth hitting F12 and seeing if there are any js errors in the console. I had a similar issue where I couldn't figure out why .show() wasn't working on an input that had a style of 'display:none' in Firefox and it turned out I had a badly formed selector which was missing a closing ] and threw an error you don't see, until you open Firebug. Chrome was more tolerant showed the button while Firefox caught the bad selector.

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醉酒成梦

2021-02-14 02:28
I had a similar problem and my solution was to do
```
$("#psharebutton").css("display", "inline-block");
```
instead of
```
$("#psharebutton").show();
```
A jquery "show" is identical to setting the display style attribute to "block". I found that I needed "inline-block" for firefox and chrome (I didn't need this for IE). I believe (but have not confirmed) that jquery is smart enough to know what the previous display value is and use that, but in my case I started out with display set to "none" so there was no previous value.
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青春惊慌失措

2021-02-14 02:29

Just write down like $("#psharebutton").attr('display','') or $("#psharebutton").attr('display','block').

function dolinkedin(url, title, summary, source) {
        $(".smbutton").addClass("buttonDisabled");
        $("#linklinkedin").removeClass("buttonDisabled");
        $("#psharebutton").attr('display','');

        parent.sizeFancybox();
    }

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