Nach etwas mehr Arbeit, bekam ich diese Arbeit. Ich bin mir immer noch nicht sicher, warum die localData-Variable benötigt wird, da die Kendo-Baumansicht sie anscheinend nicht verwendet, sobald ein Knoten bereits dort ist. Hier ist meine Lösung sowieso:
<div id="treeview"> </div>
<script>
var serviceRoot = "http://<local name>:58754/api/";
var localData;
$(document).ready(function() {
var homogeneous = new kendo.data.HierarchicalDataSource({
transport: {
read: function (options) {
if (typeof options.data.ID != 'undefined') {
var id = options.data.ID;
var data = getNextLevel(localData, id);
if (data) {
options.success(data);
} else {
var currentnode = get(localData, id);
if (currentnode.Level == 1) {
$.ajax({
url: serviceRoot + "tree",
data: 'ID=' + currentnode.ID + '&Level=' + currentnode.Level,
type: "Get",
success: function (result) {
setTimeout(function() {
var res = result;
addToLocalData(localData, res, currentnode.ID);
options.success(res);
}, 1000);
},
error: function (result) {
options.error(result);
}
});
} else {
if (currentnode.Level == 2) {
$.ajax({
url: serviceRoot + "tree",
data: 'ID=' + currentnode.ID + '&Level=' + currentnode.Level,
type: "Get",
success: function (result) {
setTimeout(function() {
var res = result;
addToLocalData(localData, res, currentnode.ID);
options.success(res);
}, 1000);
},
error: function (result) {
options.error(result);
}
});
}
}
}
}
else {
$.ajax({
url: serviceRoot + "tree",
data: 'ID='+ null +'&Level='+ null,
type: "Get",
success: function (result) {
setTimeout(function() {
options.success(result);
}, 1000);
localData = result;
},
error: function (result) {
options.error(result);
}
});
}
}
},
schema: {
model: {
id: "ID",
hasChildren: "HasChildren"
}
}
});
$("#treeview").kendoTreeView({
dataSource: homogeneous,
dataTextField: "Name"
});
});
//Checks if nodes are already in the tree and returns it if it does
function getNextLevel(data, id) {
if (!id) {
return data;
} else {
for (var i = 0; i < data.length; i++) {
if (data[i].ID == id) {
return data[i].Items;
} else if (data[i].Items) {
for (var j = 0; j < data[i].Items.length; j++) {
if (data[i].Items[j].ID == id) {
return data[i].Items[j].Items;
}
}
}
}
}
}
//Get Tree object for a given ID
function get(data, id) {
if (id) {
for (var i = 0; i < data.length; i++) {
if (data[i].ID == id) {
return data[i];
}
else if (data[i].Items) {
for (var j = 0; j < data[i].Items.length; j++) {
if (data[i].Items[j].ID == id) {
return data[i].Items[j];
}
}
}
}
}
return null;
}
//Add newly read nodes to cached tree
function addToLocalData(localdata, data, id) {
if (!id) {
return localdata;
} else {
for (var i = 0; i < localdata.length; i++) {
if (localdata[i].ID == id) {
localdata[i].Items = data;
return;
} else {
if (localdata[i].Items) {
for (var j = 0; j < localdata[i].Items.length; j++) {
if (localdata[i].Items[j].ID == id) {
localdata[i].Items[j].Items = data;
return;
}
}
}
}
}
}
}
</script>
ich eine gespeicherte Prozedur unter Verwendung von Werten von 3 Tabellen in ein Baumobjekt zu lesen. Hier ist der Code für das Baumobjekt:
public class Tree
{
public Guid ID { get; set; }
public string Name { get; set; }
public bool HasChildren { get; set; }
public int Level { get; set; }
public IEnumerable<Tree> Items { get; set; }
}
Und mein Stored Procedure:
ALTER PROCEDURE [dbo].[GetTreeItems]
@ID uniqueidentifier, @CurrentLevel int
AS BEGIN SET NOCOUNT ON;
if @CurrentLevel is null
select IDStation as ID, StationName as Name, null as IDParent, 1 as [Level] ,
case when (select COUNT(*) from Unit where Unit.IDStation = Station.IDStation) > 0 then 1 else 0 end as HasChildren
from Station
order by [Level], Name
--union
else if @CurrentLevel = 1
select IDUnit as ID, UnitName as Name, Station.IDStation as IDParent, 2 as [Level],
case when (select COUNT(*) from Component where Component.IDUnit = Unit.IDUnit) > 0 then 1 else 0 end as HasChildren
from Unit inner join Station on Station.IDStation = Unit.IDStation
where Station.IDStation = @ID
order by [Level], Name
--union
if @CurrentLevel = 2
select IDComponent as ID, ComponentName as Name, Unit.IDUnit as IDParent,
3 as [Level], 0 as HasChildren
from Component inner join Unit on unit.IDUnit = Component.IDUnit
where Unit.IDUnit = @ID
order by [Level], Name
END
Dank Alexander! Dies gibt einen guten Ausgangspunkt. – prthrokz
Könnten Sie bitte Ihre Lösung posten? Ich arbeite seit Tagen an diesem jsBin-Vorschlag und kann immer noch nicht das zweite Level zur Arbeit bringen. Bitte. – Amanda