Question

有一些比这更快的方法来找到有条件的所有人吗？

if (!String.IsNullOrEmpty(name) && !String.IsNullOrEmpty(lastname) && !String.IsNullOrEmpty(phone))
{
      List<Person> newList = List.FindAll(s => s.Name == name && s.Surname == lastname && s.Phone == phone);
}
else if (!String.IsNullOrEmpty(name) && !String.IsNullOrEmpty(lastname))
{
      List<Person> newList = List.FindAll(s => s.Name == name && s.Surname == lastname);
}

等

Answer 1

您的版本可能是运行时最快的选项。您创建的List<T>.FindAll谓词效率很高，因为它检查的次数最少。

但是，您可以使用LINQ使代码更简单，更易于维护：

IEnumerable<Person> people = List; // Start with no filters

// Add filters - just chaining as needed
if (!string.IsNullOrWhitespace(name) && !string.IsNullOrWhitespace(lastname))
{
    people = people.Where(s => s.Name == name && s.Surname == lastname);
    if (!string.IsNullOrWhitespace(phone))
        people = people.Where(s => s.Phone == phone);
}

//... Add as many as you want

List<Person> newList = people.ToList(); // Evaluate at the end

这将更加可维护，并且可能“足够快”，因为过滤通常不会在紧密循环中完成。

Answer 2

我改写这种方式只是为了让它更容易阅读：

if (!String.IsNullOrEmpty(name) && !String.IsNullOrEmpty(lastname)) {
     if (!String.IsNullOrEmpty(phone))
     {
           List<Person> newList = List.FindAll(s => s.Name == name && s.Surname == lastname && s.Phone == phone);
     }
     else 
     {
          List<Person> newList = List.FindAll(s => s.Name == name && s.Surname == lastname);
     }
}

列出＆lt;＆gt; .FindAll，条件很少

2 个答案: