Gadget vista error var with flyout
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by Jiinn
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Published on 2010-02-09T00:20:24Z
Indexed on
2010/04/11
21:03 UTC
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windows-desktop-gadgets
hello !
i have a problem with a variable for the flyout :
var friendsUser = "";
var friendsMdp = "";
System.Gadget.Settings.write("variableName", variableName);
System.Gadget.settingsUI = "Settings.html";
System.Gadget.onSettingsClosed = SettingsClosed;
System.Gadget.Flyout.visible = SettingsClosed;
function SettingsClosed() {
variableName = System.Gadget.Settings.read("variableName");
friendsUser = System.Gadget.Settings.read("friendUser");
friendsMdp = System.Gadget.Settings.read("friendMdp");
setContentText();
}
function flyFriends()
{
System.Gadget.Flyout.file = 'friends.htm';
System.Gadget.Flyout.show = true ;
var flyoutDiv = System.Gadget.Flyout.document.parentWindow;
flyoutDiv.gMyVar = friendsUser;
flyoutDiv.gMyVar2 = friendsMdp;
}
If i use this my flyout var is undefined , and if i write : var friendsUser = "test"; i have Test in var and after use setting i have nothing ... if i write var in flyoutDiv before System.Gadget.Flyout.show = true ; gadget bug .
my settings dont have a problem, but the refresh of the var ...
have you a idea ?
thank you for all !
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