我需要一种方法来对来自数据库的大量(大约120或更多值)数据进行分类。没有什么想到的,所以我决定在这里问。我需要C#
或SQL
解决方案(或至少提示)。
当前需要排序的数据在一列中,但我对任何建议持开放态度。下面是一个数据示例:
- A2-11 (Main branch that contains all A2 from below)
- A2-113
- A2-114
- A2-115
- .
- .
- .
- F-L-5 (Main branch)
- F-L-55 (Another branch in there)
- F-L-56 (Contains all values below)
- F-L-566
- F-L-567
- .
- .
- .
先谢谢!!
可以显示为小树图:
TreeExample
答案 0 :(得分:0)
也许使用Hierarchyid? 大卫
答案 1 :(得分:0)
尝试自定义OrderBy
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Data;
namespace ConsoleApplication50
{
class Program
{
static void Main(string[] args)
{
DataTable dt = new DataTable();
dt.Columns.Add("Category", typeof(string));
string[] data = { "Main-A2-11", "Main-A2-113", "Main-A2-114", "Main-A2-115",
"Third-F-L-56", "Third-F-L-566", "Third-F-L-567",
"Second-F-L-5", "Second-F-L-55"
};
foreach (string d in data)
{
dt.Rows.Add(new object[] { d });
}
dt = dt.AsEnumerable().GroupBy(x => x.Field<string>("Category").Contains("-") ?
x.Field<string>("Category").Substring(0,x.Field<string>("Category").IndexOf("-")) :
x.Field<string>("Category"))
.Select(x => x.OrderBy(y => new SortRecord(y,"Category"))).SelectMany(x => x).CopyToDataTable();
}
}
public class SortRecord : IComparer<SortRecord >, IComparable<SortRecord>
{
string[] rowSplitArray;
int length = 0;
public SortRecord(DataRow x, string name)
{
rowSplitArray = x.Field<string>(name).Split(new char[] { '-' });
length = rowSplitArray.Length;
}
public int Compare(SortRecord rowA, SortRecord rowB)
{
int len = Math.Min(rowA.length, rowB.length);
for (int i = 0; i < len; i++)
{
if (rowA.rowSplitArray[i] != rowB.rowSplitArray[i])
return rowA.rowSplitArray[i].CompareTo(rowB.rowSplitArray[i]);
}
return rowA.length.CompareTo(rowB.length);
}
public int CompareTo(SortRecord row)
{
return Compare(this, row);
}
}
}
答案 2 :(得分:0)
实现结果有多种方法,但如果您还需要决定性能,请按照本文进行操作。What are the options for storing hierarchical data in a relational database?
你当然可以在sql中使用hierarchyid。 https://docs.microsoft.com/en-us/sql/relational-databases/hierarchical-data-sql-server
This可能对我有帮助。