set offset lead to show() method failed

[mrbbljc@…]

$( "#ContactBtn" ).bind( "click", function(evt) {
    var offset = $(this).offset(),
        newOff = {};
    newOff.left = offset.left - 250;
    newOff.top = offset.top - 350;//
    $( "#ContactPanel" ).offset( newOff ).show();
});

when I click ContactBtn,first it show,but when I use ("#ContactBtn").hide() It must click twice to show the div panel; I'm not sure it's a bug; But,

$( "#ContactBtn" ).bind( "click", function(evt) {
    var x, y;
    if( typeof window.event != 'undefined' ){
        x = evt.clientX;
        y = evt.clientY;
    } else {
        x = evt.pageX;
        y = evt.pageY;
    }

    $( "#ContactPanel" ).css( "top", y - 350 + "px" )
        .css( "left", x - 250 + "px" )
        .show();
});

It's work; I'm sorry my poor English;

[snover]

Please provide a reduced test case on jsFiddle demonstrating this issue. Also, please tells us what browser you experienced this issue in.

[anonymous]

test url: http://jsfiddle.net/sPMGM/1/ I found that I click the button div every time,the offset() method's final result different. I use IE8,and set overflow hidden,so it disappear;

[mrbbljc@…]

Test url: http://jsfiddle.net/sPMGM/1/

[jitter]

Thanks for reporting back with and providing a test case!

This isn't a bug in jQuery. The  offset() documentation states

jQuery does not support getting the offset coordinates of hidden elements.

which explains your problems. That it works the first time is only a "coincidence" because at that time top/left are 0. Just change .ContactPanel top/left to something different from 0px and then it also won't work on the first click.

The correct way to go about this is to use .css()

 working test case based on your code