如何拆分存储在 list<*> 中的字符串，然后使用 linq 样式或 lambda 插入到同一个列表中？

Question

I have Writen o code using linq

var result=from m in driver_list select(m.Email,m.LisancePlate)

我的代码输出是这样的

[e-mail];[e-mail]----34 KZ 7898 [e-mail];[e-mail]---- 34 TZK 6785

我想使用这样的结果进行新输出 [e-mail]---34 KZ 7898 [e-mail]---34 KZ 7898 [e-mail]---34 TZK 6785 [e-mail]---34 TZK 6785

Answer 1

从给定的输出中，我有一个列表，如下所示：

List<string> test = new List<string>()
{
    "[e-mail];[e-mail]----34 KZ 7898",
    "[e-mail];[e-mail]---- 34 TZK 6785"
};

以下内容将生成请求的输出：

var result = 
test
.Select 
(
    t =>new 
    {
        Emails =  t.Split(new string[] { "----" }, StringSplitOptions.None)[0],
        LicensePlate = t.Split(new string[] { "----" }, StringSplitOptions.None)[1]
    }
)
.Select 
(  
    t => new
    {
        Email = t.Emails.Split(';'),
        t.LicensePlate}
    )
.SelectMany 
( 
    t => t.Email,
    (x,t)=>new {Email = t,CarPlaket = x.LicensePlate)
)
.ToList();
foreach(var email in result)
    Console.WriteLine("{0}---{1}",email.Email,email.CarPlaket;)

上述解决方案不会将结果放在同一个列表中，但会创建一个新列表。

Answer 2

.Net的正则表达式引擎支持环顾四周的组捕获。所以它变得简单！

string input="[e-mail];[e-mail]----34 KZ 7898 [e-mail];[e-mail]---- 34 TZK 6785";
String[] emailValues=Regex.Matches(input,@"(w+@w+[.]w+)(?=.*?(-+.*?)(?=s*w+@|$))")
                          .Cast<Match>()
                          .Select(x=>x.Groups[1].Value+x.Groups[2].Value)
                          .ToArray();

这将适用于任意数量的电子邮件 ID

Answer 3

看起来你必须使用Regex，因为你在34 KZ 7898中有多个spaces。我已经尝试了这段代码，它按您的预期工作：

string input = "[e-mail];[e-mail]----34 KZ 7898 [e-mail];[e-mail]---- 34 TZK 6785";
var result = string.Join(" ", Regex.Split(input, "\s(?=\S+@)")
                             .Select(x=>{
                                var s = x.Split(new string[] {"----"}, StringSplitOptions.RemoveEmptyEntries);
                                return string.Join(" ",s[0].Split(';').Select(e => e + "----" + s[s.Length - 1]).ToArray());
                             }).ToArray());//No need ToArray() since .NET 4.0
//Output
[e-mail]---34 KZ 7898 [e-mail]---34 KZ 7898 [e-mail]---34 TZK 6785 [e-mail]---34 TZK 6785

Answer 4

var src_in_arac = from log in Src_Log_List.ToList()
                              join ml in mail_arac_sorgu on log.Node equals ml.TasitTanımaNo
                              join sr in surucu_list.Select(ss =>
                              {
                                  string SCKartNo = "";
                                  string AdSoyad = ss.AdSoyad;
                                  if (ss.SrcKardId != "")
                                      SCKartNo = "000" + (Convert.ToInt64(ss.SrcKardId).ToString("X"));
                                  return new { AdSoyad, SCKartNo };
                              })
                              on log.SCID equals sr.SCKartNo
                              into data
                              from ss in data.DefaultIfEmpty()
                              where log.Task == "GPSEvents"

                              select new { log.Speed, AdSoyad = (ss == null ? "İZİNSİZ KULLANIM" : ss.AdSoyad), ml.Plaka, log.OdoMeter, log.StandStill, log.Date, ml.Grup, ml.Email ,log.SC};

我的代码在上面。对不起，我的英语不好。邮件字段 ml.Email，我想拆分用";"连接的邮件。然后进入萨马字符串。