взгляните на этот пример
class Person
{
public int ID { get; set; }
public string FirstName { get; set; }
public string LastName { get; set; }
public string Phone { get; set; }
}
class Pet
{
public string Name { get; set; }
public Person Owner { get; set; }
}
public static void LeftOuterJoinExample()
{
Person magnus = new Person {ID = 1, FirstName = "Magnus", LastName = "Hedlund"};
Person terry = new Person {ID = 2, FirstName = "Terry", LastName = "Adams"};
Person charlotte = new Person {ID = 3, FirstName = "Charlotte", LastName = "Weiss"};
Person arlene = new Person {ID = 4, FirstName = "Arlene", LastName = "Huff"};
Pet barley = new Pet {Name = "Barley", Owner = terry};
Pet boots = new Pet {Name = "Boots", Owner = terry};
Pet whiskers = new Pet {Name = "Whiskers", Owner = charlotte};
Pet bluemoon = new Pet {Name = "Blue Moon", Owner = terry};
Pet daisy = new Pet {Name = "Daisy", Owner = magnus};
// Create two lists.
List people = new List {magnus, terry, charlotte, arlene};
List pets = new List {barley, boots, whiskers, bluemoon, daisy};
var query = from person in people
where person.ID == 4
join pet in pets on person equals pet.Owner into personpets
from petOrNull in personpets.DefaultIfEmpty()
select new { Person=person, Pet = petOrNull};
foreach (var v in query )
{
Console.WriteLine("{0,-15}{1}", v.Person.FirstName + ":", (v.Pet == null ? "Does not Exist" : v.Pet.Name));
}
}
// This code produces the following output:
//
// Magnus: Daisy
// Terry: Barley
// Terry: Boots
// Terry: Blue Moon
// Charlotte: Whiskers
// Arlene:
, теперь вы можете include elements from the left
, даже если этот элемент has no matches in the right
, в нашем случае мы восстановили Arlene
, даже если он не имеет соответствия в правой
здесь ссылка
Таким образом, Вы хотите переменную в функции, доступной за пределами функции? Почему бы не возвратить результаты переменной в функции? var x = function returnX { var x = 0; return x; }
идея...
<html xmlns="http://www.w3.org/1999/xhtml">
<head id="Head1" runat="server">
<title></title>
<script type="text/javascript">
var offsetfrommouse = [10, -20];
var displayduration = 0;
var obj_selected = 0;
function makeObj(address) {
var trailimage = [address, 50, 50];
document.write('<img id="trailimageid" src="' + trailimage[0] + '" border="0" style=" position: absolute; visibility:visible; left: 0px; top: 0px; width: ' + trailimage[1] + 'px; height: ' + trailimage[2] + 'px">');
obj_selected = 1;
return trailimage;
}
function truebody() {
return (!window.opera && document.compatMode && document.compatMode != "BackCompat") ? document.documentElement : document.body;
}
function hidetrail() {
var x = document.getElementById("trailimageid").style;
x.visibility = "hidden";
document.onmousemove = "";
}
function followmouse(e) {
var xcoord = offsetfrommouse[0];
var ycoord = offsetfrommouse[1];
var x = document.getElementById("trailimageid").style;
if (typeof e != "undefined") {
xcoord += e.pageX;
ycoord += e.pageY;
}
else if (typeof window.event != "undefined") {
xcoord += truebody().scrollLeft + event.clientX;
ycoord += truebody().scrollTop + event.clientY;
}
var docwidth = 1395;
var docheight = 676;
if (xcoord + trailimage[1] + 3 > docwidth || ycoord + trailimage[2] > docheight) {
x.display = "none";
alert("inja");
}
else
x.display = "";
x.left = xcoord + "px";
x.top = ycoord + "px";
}
if (obj_selected = 1) {
alert("obj_selected = true");
document.onmousemove = followmouse;
if (displayduration > 0)
setTimeout("hidetrail()", displayduration * 1000);
}
</script>
</head>
<body>
<form id="form1" runat="server">
<img alt="" id="house" src="Pictures/sides/right.gif" style="z-index: 1; left: 372px; top: 219px; position: absolute; height: 138px; width: 120px" onclick="javascript:makeObj('Pictures/sides/sides-not-clicked.gif');" />
</form>
</body>
</html>
я не протестировал это, но если бы Ваш код работал до того небольшого изменения, то это должно работать.