[英]Building json from tabular hierarchical data
假設我有一些數據,如下所示:
{
"Menu": {
"aaa": "aaa",
"bbb": {
"ccc": "ccc",
"ddd": "ddd"
},
"eee": "eee"
}
}
我可以通過這種關系方式將這種類型的分層數據保存到數據庫中:
http://i.stack.imgur.com/lmuq1.jpg
樣本清單:
List<MenuItem> menuItems = new List<MenuItem>();
menuItems.Add(new MenuItem() { SiteMenuId = 1, ParentId = null, MenuName = "Menu", Url = null, SiteId = 1 });
menuItems.Add(new MenuItem() { SiteMenuId = 2, ParentId = 1, MenuName = "aaa", Url = "aaa", SiteId = 1 });
menuItems.Add(new MenuItem() { SiteMenuId = 3, ParentId = 1, MenuName = "bbb", Url = null, SiteId = 1 });
menuItems.Add(new MenuItem() { SiteMenuId = 4, ParentId = 3, MenuName = "ccc", Url = "ccc", SiteId = 1 });
menuItems.Add(new MenuItem() { SiteMenuId = 5, ParentId = 3, MenuName = "ddd", Url = "ddd", SiteId = 1 });
menuItems.Add(new MenuItem() { SiteMenuId = 6, ParentId = 1, MenuName = "eee", Url = "eee", SiteId = 1 });
因此,當我從db獲得作為MenuItem對象列表的關系數據時,如何將其轉換回json?
public partial class MenuItem
{
public int SiteMenuId { get; set; }
public int SiteId { get; set; }
public string MenuName { get; set; }
public string Url { get; set; }
public Nullable<int> ParentId { get; set; }
public int CreatedUser { get; set; }
public System.DateTime CreatedDate { get; set; }
public Nullable<int> ModifiedUser { get; set; }
public Nullable<System.DateTime> ModifiedDate { get; set; }
}
我必須使用Dictionary或ExpandoObject或其他東西嗎? 我想擁有與開始時完全相同的格式。
您可以為此創建KeyValuePair對象:
KeyValuePair<string, List<Object>> toExport = new KeyValuePair<int, int>("Menu", new List<Object>());
然后,您可以添加元素,如下所示:
toExport.Value.Add(new KeyValuePair<string, string>("aaa", "aaa"));
要向其中添加復合內容,您可以執行以下操作:
KeyValuePair<string, List<Object>> bbb = new KeyValuePair<string, List<Object>>("bbb", new List<Object>());
bbb.Value.Add(new KeyValuePair<string, string>("ccc", "ccc"));
bbb.Value.Add(new KeyValuePair<string, string>("ddd", "ddd"));
toExport.Value.Add(bbb);
構建對象后,可以使用NewtonSoft的JsonConvert.SerializeObject方法。
您還可以創建一個幫助器類來幫助您。
編輯:動態數據的創建。
public class DynamicKeyValueBuilder {
private KeyValuePair<string, List<Object>> toExport;
public DynamicKeyValueBuilder(string mainKey) {
toExport = new KeyValuePair<string, List<Object>>(mainKey, new List<Object>());
}
public string getJSON() {
return JsonConvert.SerializeObject(this.toExport);
}
private KeyValuePair<string, List<Object>> searchParent(List<string> path) {
KeyValuePair<string, List<Object>> temp = (KeyValuePair<string, List<Object>>)this.toExport;
int index = 0;
while (index < path.Count) {
try {
temp = (KeyValuePair<string, List<Object>>)temp.First(item => item.Key == path.ElementAt(index)); //throws exception if value is not list or the element was not found
index++;
} catch (Exception exception) {
//handle exceptions
return null;
}
}
return temp;
}
//If value == null, we create a list
public boolean addElement(List<string> path, string key, string value) {
KeyValuePair<string, Object> parent = this.searchParent(path);
//failure
if (parent == null) {
return false;
}
parent.Value.Add((value == null) ? (new KeyValuePair<string, List<Object>>(key, new List<Object>())) : (new KeyValuePair<string, string>(key, value)));
return true;
}
}
代碼未經測試,如果遇到錯誤,請告訴我,而不僅僅是拒絕投票,我相信我會在這里提供幫助。
您可以像這樣實例化該類:
DynamicKeyValueBuilder myBuilder = new DynamicKeyValueBuilder("Menu");
當您打算添加新的<string, string>
元素時,可以這樣進行:
myBuilder.Add(new List<string>(new string[] {"Menu"}), "aaa", "aaa");
當您打算添加新的<string, List<Object>>
元素時,可以這樣進行:
myBuilder.Add(new List<string>(new string[] {"Menu"}), "bbb", null);
當您打算在內部列表中添加內容時,可以這樣進行:
myBuilder.Add(new List<string>(new string[] {"Menu", "bbb"}), "ccc", "ccc");
使用Json.net ,我們可以編寫一個自定義轉換器,以從MenuItem
列表生成所需的json。
注意:我省略了轉換器的閱讀器部分,以使其簡潔(因為它實際上與問題無關),但是其邏輯與編寫器部分相似。
class MenuItemJsonConverter : JsonConverter
{
public override bool CanConvert(Type objectType)
{
return objectType == typeof (MenuItemCollection) || objectType==typeof(List<MenuItem>);
}
public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
{
throw new NotImplementedException();
}
public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
{
var map=new Dictionary<int,JObject>();
var collection = (List<MenuItem>) value;
var root=new JObject();
var nestedItems=collection.GroupBy(i => i.ParentId).ToLookup(g=>g.Key); //or we can simply check for item.Url==null but I believe this approach is more flexible
foreach (var item in collection)
{
if (item.ParentId == null)
{
var firstObj=new JObject();
root.Add(item.MenuName,firstObj);
map.Add(item.SiteMenuId,firstObj);
continue;
}
var parent = map[item.ParentId.Value];
if (!nestedItems.Contains(item.SiteMenuId))
{
parent.Add(item.MenuName,item.Url);
continue;
}
var jObj = new JObject();
parent.Add(item.MenuName, jObj);
map.Add(item.SiteMenuId, jObj);
}
writer.WriteRaw(root.ToString());
}
}
這是直接的用法示例:
var menuItems = new List<MenuItem>();
menuItems.Add(new MenuItem() { SiteMenuId = 1, ParentId = null, MenuName = "Menu", Url = null, SiteId = 1 });
menuItems.Add(new MenuItem() { SiteMenuId = 2, ParentId = 1, MenuName = "aaa", Url = "aaa", SiteId = 1 });
menuItems.Add(new MenuItem() { SiteMenuId = 3, ParentId = 1, MenuName = "bbb", Url = null, SiteId = 1 });
menuItems.Add(new MenuItem() { SiteMenuId = 4, ParentId = 3, MenuName = "ccc", Url = "ccc", SiteId = 1 });
menuItems.Add(new MenuItem() { SiteMenuId = 5, ParentId = 3, MenuName = "ddd", Url = "ddd", SiteId = 1 });
menuItems.Add(new MenuItem() { SiteMenuId = 6, ParentId = 1, MenuName = "eee", Url = "eee", SiteId = 1 });
var json = JsonConvert.SerializeObject(menuItems,Formatting.Indented,new MenuItemJsonConverter());
或者我們可以從List<>
派生,並用[JsonConverter(typeof(MenuItemJsonConverter))]
裝飾它,以方便使用:
[JsonConverter(typeof(MenuItemJsonConverter))]
class MenuItemCollection : List<MenuItem>
{
}
然后像這樣簡單地使用它:
var menuItems = new MenuItemCollection();
menuItems.Add(new MenuItem() { SiteMenuId = 1, ParentId = null, MenuName = "Menu", Url = null, SiteId = 1 });
menuItems.Add(new MenuItem() { SiteMenuId = 2, ParentId = 1, MenuName = "aaa", Url = "aaa", SiteId = 1 });
menuItems.Add(new MenuItem() { SiteMenuId = 3, ParentId = 1, MenuName = "bbb", Url = null, SiteId = 1 });
menuItems.Add(new MenuItem() { SiteMenuId = 4, ParentId = 3, MenuName = "ccc", Url = "ccc", SiteId = 1 });
menuItems.Add(new MenuItem() { SiteMenuId = 5, ParentId = 3, MenuName = "ddd", Url = "ddd", SiteId = 1 });
menuItems.Add(new MenuItem() { SiteMenuId = 6, ParentId = 1, MenuName = "eee", Url = "eee", SiteId = 1 });
var json = JsonConvert.SerializeObject(menuItems,Formatting.Indented);
如果可以使用NewtonSoft,請使用來反序列化JSON內容,並查看生成的對象的外觀。 然后,創建一個與反序列化的結果結構匹配的類。 逆向工程...
使用此方法:
var obj = JsonConvert.DeserializeObject("{ "menu": { "aaa": "aaa"......} }");
讓我知道你的發現。
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