set offset lead to show() method failed

[[email protected]…]

$( "#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;

[[email protected]…]

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