Question

我有一个datagridview，它绑定到dataBase（我有bindingSource和bindingNavigator）

所以，如果我想显示所有表格（在我的情况下是livre），我写下这段代码：

query = from x in ctx.livre
        select x;
livreBindingSource.DataSource = query.ToList();

我在dataGridView中添加了一个名为Hierarchy的列，以便添加一些不在表livre中的信息（包含在列表中）。

所以我想将此列表绑定到该列层次结构

示例（没有列Hierarchy）：

query = from x in ctx.livre
        select x;
livreBindingSource.DataSource = query.ToList();


id | name

1 | name1

2  | name2

...

示例（使用列层次结构）：

query = from x in ctx.livre
        select x;
livreBindingSource.DataSource = query.ToList();
//Some code here for binding list<string> to column Hierarchy

id | name | Hierarchy

1 | name1 | 1 is for name1

2  | name2 | this is second

...

我该怎么做？谢谢！

Answer 1

更改你的选择以在linq中添加列？

像：

select new { x.id, x.name, Hierarchy = (x.id == 1 ? "1 is for name1" : "this is second") };

如果层次结构数据在另一个列表中，您可以加入它然后引用

var result = from y in ctx.Livre
                     join h in hierList on y.id equals h.hierId
                     select new { y.id, y.name, Hierarchy = h.hierString };

好的，在这里添加一些代码来显示扁平的层次结构

public class Rayon
    {
        public int idLocation { get; set; }
        public string LocationName { get; set; }
        public List<Rayon> idParentLocation { get; set; }
    }

    public class Livre
    {
        public int id { get; set; }
        public string name { get; set; }
        public string author { get; set; }
        public List<Rayon> Location { get; set; }
    }

List<Rayon> library = new List<Rayon>();
        library.Add(new Rayon() { idLocation = 1, idParentLocation = null, LocationName = "BookCase1" });
        library.Add(new Rayon() { idLocation = 10, idParentLocation = new List<Rayon>() { library.Find(i => i.LocationName.Equals("BookCase1")) }, LocationName = "1Shelf1" });
        library.Add(new Rayon() { idLocation = 11, idParentLocation = new List<Rayon>() { library.Find(i => i.LocationName.Equals("BookCase1")) }, LocationName = "1Shelf2" });
        library.Add(new Rayon() { idLocation = 12, idParentLocation = new List<Rayon>() { library.Find(i => i.LocationName.Equals("BookCase1")) }, LocationName = "1Shelf3" });
        library.Add(new Rayon() { idLocation = 2, idParentLocation = null, LocationName = "BookCase2" });
        library.Add(new Rayon() { idLocation = 20, idParentLocation = new List<Rayon>() { library.Find(i => i.LocationName.Equals("BookCase2")) }, LocationName = "2Shelf1" });
        library.Add(new Rayon() { idLocation = 21, idParentLocation = new List<Rayon>() { library.Find(i => i.LocationName.Equals("BookCase2")) }, LocationName = "2Shelf2" });
        library.Add(new Rayon() { idLocation = 22, idParentLocation = new List<Rayon>() { library.Find(i => i.LocationName.Equals("BookCase2")) }, LocationName = "2Shelf3" });
        library.Add(new Rayon() { idLocation = 3, idParentLocation = null, LocationName = "BookCase3" });
        library.Add(new Rayon() { idLocation = 30, idParentLocation = new List<Rayon>() { library.Find(i => i.LocationName.Equals("BookCase3")) }, LocationName = "3Shelf1" });
        library.Add(new Rayon() { idLocation = 31, idParentLocation = new List<Rayon>() { library.Find(i => i.LocationName.Equals("BookCase3")) }, LocationName = "3Shelf2" });
        library.Add(new Rayon() { idLocation = 32, idParentLocation = new List<Rayon>() { library.Find(i => i.LocationName.Equals("BookCase3")) }, LocationName = "3Shelf3" });

        List<Livre> bkList = new List<Livre>();
        bkList.Add(new Livre() { id = 1, name = "Catch-22", author = "Heller", Location = new List<Rayon>() { library.Find(i => i.idLocation == 30) } });

        var test = bkList.SelectMany(b => b.Location.SelectMany(c => c.idParentLocation.Select(p => new { id = b.id, name = b.name, author = b.author, hierarchy = p.LocationName + "->" + c.LocationName + "->" + b.name })));

按列表填充一个Datagridview列

1 个答案: