Question

我有课程：

  public class Person
  {  
    public string Name{get;set;}   
    public bool IsMarried{get;set;}  
    public List<Child> Children { get; set; }
  }
  public class Child
  {
    public bool Greater5 { get; set; }
    public List<MyColor> ListColor { get; set; }
  }
  public class MyColor
  {
    public string History { get; set; }
    public string Type { get; set; }
  }

我想获取IsMarried = true且Children.Greater5 = true的所有人的列表字符串（MyColor.Type，它们不为空）。我可以通过以下方式实现这一目标：

  List<Person> list = personList.Where(l => l.IsMarried == true).ToList();
  List<Child> childList = list.SelectMany(c => c.Children).Where(test => test.Greater5 == true).ToList();
  List<string> wantedList= childList.SelectMany(test => test.ListColor.Where(c => !String.IsNullOrEmpty(c.Type)).Select(t => t.Type)).ToList();

但我想知道是否可以使用Linq实现更简单？

Answer 1

您可以链接SelectMany以拼合嵌套的子项。

List<string> wantedList = personList
    .Where(person => person.IsMarried)
    .SelectMany(person => person.Children
        .Where(child => child.Greater5)
        .SelectMany(child => child.ListColor
            .Where(color => !String.IsNullOrEmpty(color.Type))
            .Select(color => color.Type)))
    .ToList();

Answer 2

如果您看起来“更简单”，也可以使用查询语言：

List<string> wantedList = (from person in personList where person.IsMarried
                           from child in person.Children where child.Greater5 
                           from color in child.ListColor where !string.IsNullOrEmpty(color.Type)
                           select color.Type).ToList();

如何从列表列表列表中获取列表<string> C＃

2 个答案: