是否有任何简单的LINQ表达式将我的整个List<string>集合项连接到具有分隔符字符的单个字符串?
如果集合是自定义对象而不是字符串呢?假设我需要连接object。name。
是否有任何简单的LINQ表达式将我的整个List<string>集合项连接到具有分隔符字符的单个字符串?
如果集合是自定义对象而不是字符串呢?假设我需要连接object。name。
当前回答
我认为如果你在扩展方法中定义逻辑,代码将更易于阅读:
public static class EnumerableExtensions {
public static string Join<T>(this IEnumerable<T> self, string separator) {
return String.Join(separator, self.Select(e => e.ToString()).ToArray());
}
}
public class Person {
public string FirstName { get; set; }
public string LastName { get; set; }
public override string ToString() {
return string.Format("{0} {1}", FirstName, LastName);
}
}
// ...
List<Person> people = new List<Person>();
// ...
string fullNames = people.Join(", ");
string lastNames = people.Select(p => p.LastName).Join(", ");
其他回答
using System.Linq;
public class Person
{
string FirstName { get; set; }
string LastName { get; set; }
}
List<Person> persons = new List<Person>();
string listOfPersons = string.Join(",", persons.Select(p => p.FirstName));
我已经使用LINQ做到了这一点:
var oCSP = (from P in db.Products select new { P.ProductName });
string joinedString = string.Join(",", oCSP.Select(p => p.ProductName));
您可以使用聚合(Aggregate)将字符串连接成单个字符分隔的字符串,但如果集合为空,则会抛出无效操作异常(Invalid Operation Exception)。
可以将聚合函数与种子字符串一起使用。
var seed = string.Empty;
var seperator = ",";
var cars = new List<string>() { "Ford", "McLaren Senna", "Aston Martin Vanquish"};
var carAggregate = cars.Aggregate(seed,
(partialPhrase, word) => $"{partialPhrase}{seperator}{word}").TrimStart(',');
你可以用字符串。Join并不关心您是否传递给它一个空集合。
var seperator = ",";
var cars = new List<string>() { "Ford", "McLaren Senna", "Aston Martin Vanquish"};
var carJoin = string.Join(seperator, cars);
string result = String.Join(delimiter, list);
是充分的。
警告-严重性能问题
虽然这个答案确实产生了预期的结果,但与这里的其他答案相比,它的性能较差。在决定使用它时要非常谨慎
通过使用LINQ,这应该工作;
string delimiter = ",";
List<string> items = new List<string>() { "foo", "boo", "john", "doe" };
Console.WriteLine(items.Aggregate((i, j) => i + delimiter + j));
类描述:
public class Foo
{
public string Boo { get; set; }
}
用法:
class Program
{
static void Main(string[] args)
{
string delimiter = ",";
List<Foo> items = new List<Foo>() { new Foo { Boo = "ABC" }, new Foo { Boo = "DEF" },
new Foo { Boo = "GHI" }, new Foo { Boo = "JKL" } };
Console.WriteLine(items.Aggregate((i, j) => new Foo{Boo = (i.Boo + delimiter + j.Boo)}).Boo);
Console.ReadKey();
}
}
这是我最好的:)
items.Select(i => i.Boo).Aggregate((i, j) => i + delimiter + j)