从平面列表创建递归对象
本文关键字:递归 对象 创建 列表 平面 | 更新日期: 2023-09-27 18:09:04
如下图所示:
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 });
是否可以将此数据结构转换为可以序列化为以下json的格式:
{
"Menu": {
"aaa": "aaa",
"bbb": {
"ccc": "ccc",
"ddd": "ddd"
},
"eee": "eee"
}
}
如果答案是肯定的,我该怎么做呢?
你可以考虑使用Json.NET。
Json。. NET将任何IDictionary
序列化为JSON键/值对对象-但转换为Dictionary<string, object>
然后序列化将是有问题的,因为。NET字典是无序的,并且您(可能)希望在序列化为JSON时保留 MenuItem
对象的相对顺序。因此,使用LINQ到JSON手动转换为JObject
对象树是有意义的,因为JSON。. NET保留对象属性的顺序。
你可以这样做:
public static string CreateJsonFromMenuItems(IList<MenuItem> menuItems)
{
return new JObject
(
menuItems.ToTree(
m => (int?)m.SiteMenuId,
m => m.ParentId, m => new JProperty(m.MenuName, m.Url),
(parent, child) =>
{
if (parent.Value == null || parent.Value.Type == JTokenType.Null)
parent.Value = new JObject();
else if (parent.Value.Type != JTokenType.Object)
throw new InvalidOperationException("MenuItem has both URL and children");
child.MoveTo((JObject)parent.Value);
})
).ToString();
}
(注意,如果MenuItem
同时具有非空Url
和子集合,则该方法会抛出异常)
它使用以下扩展方法:
public static class JsonExtensions
{
public static void MoveTo(this JToken token, JObject newParent)
{
if (newParent == null)
throw new ArgumentNullException();
var toMove = token.AncestorsAndSelf().OfType<JProperty>().First(); // Throws an exception if no parent property found.
if (toMove.Parent != null)
toMove.Remove();
newParent.Add(toMove);
}
}
public static class RecursiveEnumerableExtensions
{
static bool ContainsNonNullKey<TKey, TValue>(this IDictionary<TKey, TValue> dictionary, TKey key)
{
if (dictionary == null)
throw new ArgumentNullException();
return key == null ? false : dictionary.ContainsKey(key); // Dictionary<int?, X> throws on ContainsKey(null)
}
public static IEnumerable<TResult> ToTree<TInput, TKey, TResult>(
this IEnumerable<TInput> collection,
Func<TInput, TKey> idSelector,
Func<TInput, TKey> parentIdSelector,
Func<TInput, TResult> nodeSelector,
Action<TResult, TResult> addMethod)
{
if (collection == null || idSelector == null || parentIdSelector == null || nodeSelector == null || addMethod == null)
throw new ArgumentNullException();
var list = collection.ToList(); // Prevent multiple enumerations of the incoming enumerable.
var dict = list.ToDictionary(i => idSelector(i), i => nodeSelector(i));
foreach (var input in list.Where(i => dict.ContainsNonNullKey(parentIdSelector(i))))
{
addMethod(dict[parentIdSelector(input)], dict[idSelector(input)]);
}
return list.Where(i => !dict.ContainsNonNullKey(parentIdSelector(i))).Select(i => dict[idSelector(i)]);
}
}
工作小提琴。
您可以尝试创建一个这样的类
public class MenuItemTr
{
public MenuItemTr
{
this.MenuItems= new List <MenuItem>
}
public int SiteMenuId {get; set;}
public int ParentId {get; set;}
public string MenuName {get; set;}
public string Url {get; set;}
public int SiteId {get; set;}
public List <MenuItemTr> MenuItems {get; set;}
}
,然后解析为树
var MenuItem = menuItems.GenerateTree(c => c.SiteMenuId, c => c.ParentId);
和使用这个解决方案从这个线程Nice &将List of items转换为Tree的通用方法
public static IEnumerable<TreeItem<T>> GenerateTree<T, K>(
this IEnumerable<T> collection,
Func<T, K> id_selector,
Func<T, K> parent_id_selector,
K root_id = default(K))
{
foreach (var c in collection.Where(c => parent_id_selector(c).Equals(root_id)))
{
yield return new TreeItem<T>
{
Item = c,
Children = collection.GenerateTree(id_selector, parent_id_selector, id_selector(c))
};
}
}
}