我有一个像这样的字符串 -
var roleDetails = "09A880C2-8732-408C-BA09-4AD6F0A65CE9^Z:WB:SELECT_DOWNLOAD:0000^Product Delivery - Download^1,24B11B23-1669-403F-A24D-74CE72DFD42A^Z:WB:TRAINING_SUBSCRIBER:0000^Training Subscriber^1,6A4A6543-DB9F-46F2-B3C9-62D69D28A0B6^Z:WB:LIC_MGR_HOME_REDL:0000^License Manager - Home use^1,76B3B165-0BB4-4E3E-B61F-0C0292342CE2^Account Admin^Account Admin^1,B3C0CE51-00EE-4A0A-B208-98653E21AE11^Z:WB:1BENTLEY_ISA_ADMIN:0000^Co-Administrator^1,CBA225BC-680C-4627-A4F6-BED401682816^ReadOnly^ReadOnly^1,D80CF5CF-CB6E-4424-9D8F-E29F96EBD4C9^Z:WB:MY_SELECT_CD:0000^Product Delivery - DVD^1,E0275936-FBBB-4775-97D3-9A7D19D3E1B4^Z:WB:LICENSE_MANAGER:0000^License Manager^1";
用“，”分隔它会返回 -
[0] "09A880C2-8732-408C-BA09-4AD6F0A65CE9^Z:WB:SELECT_DOWNLOAD:0000^Product Delivery - Download^1"
[1] "24B11B23-1669-403F-A24D-74CE72DFD42A^Z:WB:TRAINING_SUBSCRIBER:0000^Training Subscriber^1"
[2] "6A4A6543-DB9F-46F2-B3C9-62D69D28A0B6^Z:WB:LIC_MGR_HOME_REDL:0000^License Manager - Home use^1"
[3] "76B3B165-0BB4-4E3E-B61F-0C0292342CE2^Account Admin^Account Admin^1"
[4] "B3C0CE51-00EE-4A0A-B208-98653E21AE11^Z:WB:1BENTLEY_ISA_ADMIN:0000^Co-Administrator^1"
[5] "CBA225BC-680C-4627-A4F6-BED401682816^ReadOnly^ReadOnly^1"
[6] "D80CF5CF-CB6E-4424-9D8F-E29F96EBD4C9^Z:WB:MY_SELECT_CD:0000^Product Delivery - DVD^1"
[7] "E0275936-FBBB-4775-97D3-9A7D19D3E1B4^Z:WB:LICENSE_MANAGER:0000^License Manager^1"
所有元素都包含克拉（^）。所以用^符号进一步分割每个元素将返回四个元素。
但我想加入所有第一个元素，然后加入所有第二个元素，然后加入第三个，依此类推，得到如下结果 -
[0]: 09A880C2-8732-408C-BA09-4AD6F0A65CE9, 24B11B23-1669-403F-A24D-74CE72DFD42A, 6A4A6543-DB9F-46F2-B3C9-62D69D28A0B6, 76B3B165-0BB4-4E3E-B61F-0C0292342CE2, B3C0CE51-00EE-4A0A-B208-98653E21AE11, CBA225BC-680C-4627-A4F6-BED401682816, D80CF5CF-CB6E-4424-9D8F-E29F96EBD4C9, E0275936-FBBB-4775-97D3-9A7D19D3E1B4
[1]: Z:WB:SELECT_DOWNLOAD:0000,Z:WB:TRAINING_SUBSCRIBER:0000, Z:WB:LIC_MGR_HOME_REDL:0000,Account Admin, Z:WB:1BENTLEY_ISA_ADMIN:0000, ReadOnly, Z:WB:MY_SELECT_CD:0000, Z:WB:LICENSE_MANAGER
[2]: Product Delivery - Download, Training Subscriber, License Manager - Home use, Account Admin, Co-Administrator, ReadOnly, Product Delivery - DVD, License Manager
[3]: 1,1,1,1,1,1,1,1
实现这一目标的最快捷，最简单的方法是什么？
这是我到目前为止所尝试的 -
var rolearray = roleDetails.Split(',').Select(s => s.Split('^')).Select(a => new { RoleId = a[0], RoleNme = a[1], FriendlyName = a[2], IsUserInRole = a[3] });
但是这并没有以我需要的方式回归。但是我想加入所有[0]，然后加入所有[1]等等
解： 在比较解决方案并在循环中运行10次以查看性能后，我发现Jamiec建议的解决方案花费的时间更少。所以选择这个解决方案。
答案 0 :(得分：16)
Pure LINQ解决方案：
roleDetails.Split(',')
.SelectMany(x => x.Split('^').Select((str, idx) => new {str, idx}))
.GroupBy(x => x.idx)
.Select(grp => string.Join(", ", grp.Select(x => x.str)))
答案 1 :(得分：10)
最简单的方法就是：
var split = roleDetails.Split(',')
.Select(x => x.Split('^').ToArray())
.ToArray();
然后，您将访问像多维锯齿状数组
Console.WriteLine(split[0][0]);
// result: 09A880C2-8732-408C-BA09-4AD6F0A65CE9
实例：http://rextester.com/NEUVOR15080
如果你想要所有元素分组
Console.WriteLine(String.Join(",",split.Select(x => x[0])));
Console.WriteLine(String.Join(",",split.Select(x => x[1])));
Console.WriteLine(String.Join(",",split.Select(x => x[2])));
Console.WriteLine(String.Join(",",split.Select(x => x[3])));
答案 2 :(得分：5)
您可以在此处使用Linq的聚合和Zip扩展方法。
聚合：对结果中的每个元素执行指定的操作，同时向前传递结果。
Zip： Zip扩展方法对两个集合起作用。它将两个系列中的每个元素一起处理。
var roleDetails = "09A880C2-8732-408C-BA09-4AD6F0A65CE9^Z:WB:SELECT_DOWNLOAD:0000^Product Delivery - Download^1,24B11B23-1669-403F-A24D-74CE72DFD42A^Z:WB:TRAINING_SUBSCRIBER:0000^Training Subscriber^1,6A4A6543-DB9F-46F2-B3C9-62D69D28A0B6^Z:WB:LIC_MGR_HOME_REDL:0000^License Manager - Home use^1,76B3B165-0BB4-4E3E-B61F-0C0292342CE2^Account Admin^Account Admin^1,B3C0CE51-00EE-4A0A-B208-98653E21AE11^Z:WB:1BENTLEY_ISA_ADMIN:0000^Co-Administrator^1,CBA225BC-680C-4627-A4F6-BED401682816^ReadOnly^ReadOnly^1,D80CF5CF-CB6E-4424-9D8F-E29F96EBD4C9^Z:WB:MY_SELECT_CD:0000^Product Delivery - DVD^1,E0275936-FBBB-4775-97D3-9A7D19D3E1B4^Z:WB:LICENSE_MANAGER:0000^License Manager^1";
var rolearray = roleDetails.Split(',')
.Select(s => s.Split('^'))
.Aggregate((s1Array, s2Array) => s1Array.Zip(s2Array, (s1, s2) => s1 + "," + s2).ToArray());
答案 3 :(得分：1)
string roleDetails = "09A880C2-8732-408C-BA09-4AD6F0A65CE9^Z:WB:SELECT_DOWNLOAD:0000^Product Delivery - Download^1,24B11B23-1669-403F-A24D-74CE72DFD42A^Z:WB:TRAINING_SUBSCRIBER:0000^Training Subscriber^1,6A4A6543-DB9F-46F2-B3C9-62D69D28A0B6^Z:WB:LIC_MGR_HOME_REDL:0000^License Manager - Home use^1,76B3B165-0BB4-4E3E-B61F-0C0292342CE2^Account Admin^Account Admin^1,B3C0CE51-00EE-4A0A-B208-98653E21AE11^Z:WB:1BENTLEY_ISA_ADMIN:0000^Co-Administrator^1,CBA225BC-680C-4627-A4F6-BED401682816^ReadOnly^ReadOnly^1,D80CF5CF-CB6E-4424-9D8F-E29F96EBD4C9^Z:WB:MY_SELECT_CD:0000^Product Delivery - DVD^1,E0275936-FBBB-4775-97D3-9A7D19D3E1B4^Z:WB:LICENSE_MANAGER:0000^License Manager^1";
var RawItems = roleDetails.Split(',').Select(x=> x.Split('^'));
var Items1 = RawItems.Select(x=> x.ElementAt(0));
var Items2 = RawItems.Select(x=> x.ElementAt(1));
var Items3 = RawItems.Select(x=> x.ElementAt(2));
var Items4 = RawItems.Select(x=> x.ElementAt(3));
答案 4 :(得分：1)
如果你不喜欢LINQ解决方案，这里有一个没有的解决方案：
var result = new string[4];
var i = 0;
foreach(var line in roleDetails.Split(','))
foreach(var piece in line.Split('^'))
result[i++ % 4] += (i <= 4 ? "" : ",") + piece;
基本上，你使用一个计数器来分割逗号和插入符号，并使用一个计数器告诉我们要连接哪个数组元素，以及是否使用逗号分隔符。
如果您的初始字符串比此示例中的字符串大得多，请考虑首先创建一个StringBuilders数组，因为这些字符串在连接时效果更好：
var stringBuilders = new StringBuilder[4];
var result = new string[4];
var i = 0;
for (var i = 0; i < 4; i++)
stringBuilders[i] = new StringBuilder();
foreach(var line in roleDetails.Split(','))
foreach(var piece in line.Split('^'))
stringBuilders[i++ % 4].Append((i <= 4 ? "" : ",") + piece);
foreach (var stringBuilder in stringBuilders)
result[i++ % 4] = stringBuilder.ToString();
答案 5 :(得分：0)
另一个LINQ解决方案。但不像@ Pavel那样干净：
string a = "", b = "", c = "", d = "";
roleDetails.Split(',').ToList().ForEach(x =>
{
a += x.Split('^')[0] + ',';
b += x.Split('^')[1] + ',';
c += x.Split('^')[2] + ',';
d += x.Split('^')[3] + ',';
});
MessageBox.Show(a.Trim(','));
MessageBox.Show(b.Trim(','));
MessageBox.Show(c.Trim(','));
MessageBox.Show(d.Trim(','));
输出：
a = 09A880C2-8732-408C-BA09-4AD6F0A65CE9,24B11B23-1669-403F-A24D-74CE72DFD42A,6A4A6543-DB9F-46F2-B3C9-62D69D28A0B6,76B3B165-0BB4-4E3E-B61F-0C0292342CE2,B3C0CE51-00EE-4A0A-B208-98653E21AE11,CBA225BC-680C-4627-A4F6-BED401682816,D80CF5CF-CB6E-4424-9D8F-E29F96EBD4C9,E0275936-FBBB-4775-97D3-9A7D19D3E1B4
b = Z:WB:SELECT_DOWNLOAD:0000,Z:WB:TRAINING_SUBSCRIBER:0000,Z:WB:LIC_MGR_HOME_REDL:0000,Account Admin,Z:WB:1BENTLEY_ISA_ADMIN:0000,ReadOnly,Z:WB:MY_SELECT_CD:0000,Z:WB:LICENSE_MANAGER:0000
c = Product Delivery - Download,Training Subscriber,License Manager - Home use,Account Admin,Co-Administrator,ReadOnly,Product Delivery - DVD,License Manager
d = 1,1,1,1,1,1,1,1
答案 6 :(得分：0)
相当干净，快速......
var sets = new[]
{
new List<string>(),
new List<string>(),
new List<string>(),
new List<string>(),
};
foreach (var role in roleDetails.Split(','))
{
var details = role.Split('^');
sets[0].Add(details[0]);
sets[1].Add(details[1]);
sets[2].Add(details[2]);
sets[3].Add(details[3]);
}
var lines = sets.Select(set => string.Join(",", set)).ToArray();
...要理解的小坚果，并没有真正保存任何性能......
var ret = roleDetails.Split(',')
.Aggregate(seed: new { SBS = new[] { new StringBuilder(), new StringBuilder(),
new StringBuilder(), new StringBuilder(), },
Start = true },
func: (seed, role) =>
{
var details = role.Split('^');
if (seed.Start)
{
seed.SBS[0].Append(details[0]);
seed.SBS[1].Append(details[1]);
seed.SBS[2].Append(details[2]);
seed.SBS[3].Append(details[3]);
return new
{
seed.SBS,
Start = false,
};
}
else
{
seed.SBS[0].Append(',').Append(details[0]);
seed.SBS[1].Append(',').Append(details[1]);
seed.SBS[2].Append(',').Append(details[2]);
seed.SBS[3].Append(',').Append(details[3]);
return seed;
}
},
resultSelector: result => result.SBS.Select(sb => sb.ToString()).ToArray()
);
答案 7 :(得分：-2)
这样的事情应该有效。
var roleDetails = "09A880C2-8732-408C-BA09-4AD6F0A65CE9^Z:WB:SELECT_DOWNLOAD:0000^Product Delivery - Download^1,24B11B23-1669-403F-A24D-74CE72DFD42A^Z:WB:TRAINING_SUBSCRIBER:0000^Training Subscriber^1,6A4A6543-DB9F-46F2-B3C9-62D69D28A0B6^Z:WB:LIC_MGR_HOME_REDL:0000^License Manager - Home use^1,76B3B165-0BB4-4E3E-B61F-0C0292342CE2^Account Admin^Account Admin^1,B3C0CE51-00EE-4A0A-B208-98653E21AE11^Z:WB:1BENTLEY_ISA_ADMIN:0000^Co-Administrator^1,CBA225BC-680C-4627-A4F6-BED401682816^ReadOnly^ReadOnly^1,D80CF5CF-CB6E-4424-9D8F-E29F96EBD4C9^Z:WB:MY_SELECT_CD:0000^Product Delivery - DVD^1,E0275936-FBBB-4775-97D3-9A7D19D3E1B4^Z:WB:LICENSE_MANAGER:0000^License Manager^1";
var firstSplit=roleDetails.Split(',');
var final = new string[4];
foreach(string Str in firstSplit)
{
var secondSplit=Str.Split('^');
for (int i = 0; i < 4; i++)
{
final[i] += secondSplit[i] + ", ";
}
}
我会留给你删除最后一个逗号。
答案 8 :(得分：-2)
您可以在此使用Tuple
var roles = roleDetails.Split(',')
.Select(x => x.Split('^'))
.Where(x=>x.Length==4)
.Select(x=>
new Tuple<string, string, string, string>(x[0], x[1], x[2], x[3]))
.ToList();
var item1 = string.Join(",", roles.Select(x=>x.Item1).ToArray());
var item2 = string.Join(",", roles.Select(x => x.Item2).ToArray());
var item3 = string.Join(",", roles.Select(x => x.Item3).ToArray());
var item4 = string.Join(",", roles.Select(x => x.Item4).ToArray());
答案 9 :(得分：-3)
你的尝试试图在一行中做所有事情，这使你更难理解发生了什么。
您已经使用了所需的所有工具（Select()和Split()）。如果通过将所有代码分隔成单独的代码行来使代码更具可读性，那么找到自己的方式会更容易：
//Your data string
string myDataString = "...";
//Your data string, separated into a list of rows (each row is a string)
var myDataRows = myDataString.Split(',');
//Your data string, separated into a list of rows (each row is a STRING ARRAY)
var myDataRowsAsStringArrays = myDataRows.Select(row => row.Split('^'))
现在，您需要做的就是检索正确的数据。
var firstColumnValues = myDataRowsAsStringArrays.Select(row => row[0]);
var secondColumnValues = myDataRowsAsStringArrays.Select(row => row[1]);
var thirdColumnValues = myDataRowsAsStringArrays.Select(row => row[2]);
var fourthColumnValues = myDataRowsAsStringArrays.Select(row => row[3]);
如果你这样选择，你可以将这些值加入一个逗号分隔的字符串中：
var firstColumnString = String.Join(", ", firstColumnValues);
var secondColumnString = String.Join(", ", secondColumnValues);
var thirdColumnString = String.Join(", ", thirdColumnValues);
var fourthColumnString = String.Join(", ", fourthColumnValues);