Use javascript to obtain the client resolution

Source: Internet
Author: User
Tags relative visibility
I made a webpage navigation drop-down menu, which can be quickly generated using firework. I only need to copy the code to the page. However, different resolutions on the user end lead to the movement of the drop-down menu position, how to keep the correct display under the menu of 800 and 1024, modify a function in the js file, y only needs to change the x coordinate variable.
Function MM_showMenu (menu, x, y, child, imgname ){
If (! Window. mmWroteMenu) return;
MM_clearTimeout ();
If (menu ){
Var obj = FIND (imgname) | document. images [imgname] | document. links [imgname] | document. anchors [imgname];
If (window. screen. width = 800 & window. screen. height = 600) // used for 800*600 resolution settings
{
X = movexbyserver POS (X-100, obj );
Y = moveYbySlicePos (y, obj );
}
If (window. screen. width = 1024 & window. screen. height = 768) // sets the resolution.
{
X = moveXbySlicePos (x, obj );
Y = moveYbySlicePos (y, obj );
}
}
If (document. layers ){
If (menu ){
Var l = menu. menuLayer | menu;
L. top = l. left = 1;
HideActiveMenus ();
If (this. visibility) l = this;
Window. ActiveMenu = l;
} Else {
Var l = child;
}
If (! L) return;
For (var I = 0; I <l. layers. length; I ++) {if (! L. layers [I]. isHilite) l. layers [I]. visibility = "inherit ";
If (l.layers+ I }.doc ument. layers. length> 0) MM_showMenu (null, "relative", "relative", l. layers [I]);
}
If (l. parentLayer ){
If (x! = "Relative") l. parentLayer. left = x | window. pageX | 0;
If (l. parentLayer. left + l. clip. width> window. innerWidth) l. parentLayer. left-= (l. parentLayer. left + l. clip. width-window. innerWidth );
If (y! = "Relative") l. parentLayer. top = y | window. pageY | 0;
If (l. parentLayer. isContainer ){
L. Menu. xOffset = window. pageXOffset;
L. Menu. yOffset = window. pageYOffset;
L. parentLayer. clip. width = window. ActiveMenu. clip. width + 2;
L. parentLayer. clip. height = window. ActiveMenu. clip. height + 2;
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