[英]How to add from an array of strings to a list, each word that has its letters in alphabetical order?
[英]How to create Alphabetical list with letters?
大家好,我试图实现按列的Alpha排序列表
上所示的画面
但是我的算法尚不清楚,也许有人可以在
string[] letters = new string[] { "A", "B", "C", "D", "E", "F", "G", "H", "I",
"J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X",
"Y", "Z", "Å", "Ä", "Ö", "0-9" };
int j = 0, s = 0, i = 1;
var fullServices = (from se in EntityBase.db.Services
orderby se.Name
select se).ToList();
int total = fullServices.Count;
var grouped = (from l in letters
select new ServiceInfo
{
Letter = l,
Services = EntityBase.db.Services.Where(se => se.Name.StartsWith(l)).ToList(),
Total = EntityBase.db.Services.Where(se => se.Name.StartsWith(l)).Count()
}).ToList();
Dictionary<int, List<ServiceInfo>> result = new Dictionary<int, List<ServiceInfo>>();
changecell:
List<ServiceInfo> item = new List<ServiceInfo>();
while (j < letters.Count())
{
letterchange:
List<Service> _services = new List<Service>();
while (s < total)
{
if ((s == (5 + (total % 5 > i ? 1 : 0)) * i))
{
item.Add(new ServiceInfo() { Letter = letters[j], Services = _services });
result.Add(i, item);
if (i == 6)
goto exit;
i++;
goto changecell;
}
//start render services
if (fullServices.ElementAt(s).Name.StartsWith(letters[j]))
{
_services.Add(fullServices.ElementAt(s));
s++;//increment service in list
}
else //letter switch
{
item.Add(new ServiceInfo() { Letter = letters[j], Services = _services });
j++;
goto letterchange;
}
}//end render services
}
exit:
return View(result);
在我的代码中,我看到缺少字母XYZÅÄÖ的字母,看起来像这样
这是呈现字典的代码
<% foreach (KeyValuePair<int, List<BL.Models.Infos.ServiceInfo>> col in Model)
{ %>
<ul class="col">
<% foreach (var item in col.Value)
{ %>
<% if (!item.Services.Any())
{%>
<li class="disabled">
<h1>
<%= item.Letter %></h1>
</li>
<%}
else
{ %>
<li>
<h1>
<a href="/service/info/<%= item.Letter %>"><%= item.Letter %></a>
</h1>
</li>
<% foreach (var service in item.Services)
{ %>
<li><a href="/service/info/<%= service.Name %>"><%= service.Name %></a></li>
<%}
}
}%>
</ul>
<%} %>
请帮忙...
好吧,您肯定是对的,代码不是特别清楚:)
我并没有真正遵循代码的主循环,但这是一个更简单的起点。 请注意,它不能正确地将0-9分组(目前仅处理0):老实说,我不确定解决该问题的最佳方法。 您可能要推迟处理,直到收到一些与任何正常字母都不匹配的条目为止。
using System;
using System.Linq;
class Test
{
static void Main(string[] args)
{
ShowGroups();
}
private static readonly char[] Letters =
"ABCDEFGHIJKLMNOPQRSTUVWXYZÅÄÖ0".ToCharArray();
// This is taking the place of EntityBase.db.Services
// for the purposes of the test program
public static string[] Services = { "Blogger", "Delicious",
"Digg", "Ebay", "Facebook", "Feed", "Flickr",
"Friendfeed", "Friendster", "Furl", "Google",
"Gosupermodel", "Lastfm", "Linkedin", "Livejournal",
"Magnolia", "Mixx", "Myspace", "NetOnBuy", "Netvibes",
"Newsvine", "Picasa", "Pownce", "Reddit", "Stumbleupon",
"Technorati", "Twitter", "Vimeo", "Webshots",
"Wordpress" };
public static void ShowGroups()
{
var groupedByLetter =
from letter in Letters
join service in Services on letter equals service[0] into grouped
select new { Letter = letter, Services = grouped };
// Demo of how to access the groups
foreach (var entry in groupedByLetter)
{
Console.WriteLine("=== {0} ===", entry.Letter);
foreach (var service in entry.Services)
{
Console.WriteLine (" {0}", service);
}
Console.WriteLine();
}
}
}
我不知道您打算如何将结果分成5个相等的列...
这可能还很遥远-我刚刚在记事本中完成了此操作(目前无法访问IDE)。 似乎您正在尝试填充Dictionary<int, List<ServiceInfo>> result
实例,并且我假设您的View
方法了解如何将结果布置在列中。
开始:
string[] letters = new string[] { "A", "B", "C", "D", "E", "F", "G", "H", "I", "J",
"K", "L", "M", "N", "O", "P", "Q", "R", "S", "T",
"U", "V", "W", "X", "Y", "Z", "Å", "Ä", "Ö", "0-9" };
var result = new Dictionary<int, List<ServiceInfo>>();
foreach (var letter in letters)
{
var services = (from se in EntityBase.db.Services
where se.Name.StartsWith(letter)
orderby se.Name
select select new ServiceInfo
{
Letter = letter,
Services = EntityBase.db.Services.Where(se => se.Name.StartsWith(letter)).ToList(),
Total = EntityBase.db.Services.Where(se => se.Name.StartsWith(letter)).Count()
}).ToList();
result.Add(i, services);
}
如果正确的话,它可能不是最快的实现,但可读性更高。
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