因此，我导入ListBox的文件如下所示（每个换行是ListBox中的新项目）：< / p>
C2 CAP-00128G 238.385 205.944 0 0402_8
C1 CAP-00128G 236.260 204.844 0 0402_8 //Same as above 1 (C2)
C18 CAP-00129G 230.960 190.619 0 0402_4
C19 CAP-00129G 248.060 195.344 0 0402_4
C63 CAP-00129G 231.535 203.844 0 0402_4
C65 CAP-00129G 232.260 195.419 0 0402_4 //Same as above 3 (C18, C19, C63)
C71 CAP-00130G 240.151 195.123 0 0402_4 //1 Occurrence
C4 CAP-00151G 247.435 202.019 90 0402_3 //1 Occurrence
L1 IND-00007G 248.110 205.719 0 0402_4 //1 Occurrence
R30 RES-00455G 273.010 184.644 0 0402_8
R9 RES-00455G 236.260 203.794 0 0402_8
R29 RES-00455G 272.960 174.694 0 0402_8 //Same as above 2 (R30, R9)
我正在查看ListBox中的每个项目并抓住第二列值并查看它们出现的次数 （请参阅文本后面的数字 - ，即向下看列时的2,4,1,1,1,3 并将其输出到.txt文件。这是它的样子（这是.txt文件的第一部分，第二部分将在下面列出）。
( 1 0 1 1 0 0 0 0 "" "CAP-00128G" "" 2 0 0 0 0 0 ) // 2 = C2, C1
( 2 0 1 1 0 0 0 0 "" "CAP-00129G" "" 4 0 0 0 0 0 ) // 4 = C18, C19, C63, C65
( 3 0 1 1 0 0 0 0 "" "CAP-00130G" "" 1 0 0 0 0 0 ) // 1 = C71
( 4 0 1 1 0 0 0 0 "" "CAP-00151G" "" 1 0 0 0 0 0 ) // 1 = C4
( 5 0 1 1 0 0 0 0 "" "IND-00007G" "" 1 0 0 0 0 0 ) // 1 = L1
( 6 0 1 1 0 0 0 0 "" "RES-00455G" "" 3 0 0 0 0 0 ) // 3 = R30, R9, R29
将此代码用于上述文本：
List<string> partList = new List<string>();
// Builds original list first.
foreach (var item in placementOneListBox.Items) //This is getting all the lines in the ListBox and storing them in a List<string>.
{
// Creates a string of the items in the ListBox.
var newItems = item.ToString();
// Replaces any multiple spaces (tabs) with a single space.
newItems = Regex.Replace(newItems, @"\s+", " ");
// Splits each line by spaces.
var eachItem = newItems.Split(' ');
// Add each item to the List.
partList.Add(eachItem[1]);
}
Dictionary<string, int> result = new Dictionary<string, int>();
foreach (var item in partList)
{
if (result.ContainsKey(item))
result[item]++;
else
result.Add(item, 1);
}
foreach (var item in result)
{
sw.WriteLine("( " + deviceCount + " 0 1 1 0 0 0 0 \"\" \"" + item.Key + "\" \"\" " + item.Value + " 0 0 0 0 0 )");
deviceCount++;
}
使用相同的ListBox我打印出.txt文件的后半部分。文本文件的第二部分如下所示：
( 1 1 "DD" 1 238.385 205.944 0 1 2 0 0 "C2" "" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 2 1 "DD" 1 236.260 204.844 0 1 2 0 0 "C1" "" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 3 1 "DD" 1 230.960 190.619 0 1 2 0 0 "C18" "" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 4 1 "DD" 1 248.060 195.344 0 1 2 0 0 "C19" "" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 5 1 "DD" 1 231.535 203.844 0 1 2 0 0 "C63" "" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 6 1 "DD" 1 232.260 195.419 0 1 2 0 0 "C65" "" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 7 1 "DD" 1 240.151 195.123 0 1 2 0 0 "C71" "" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 8 1 "DD" 1 247.435 202.019 90 1 2 0 0 "C4" "" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 9 1 "DD" 1 248.110 205.719 0 1 2 0 0 "L1" "" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 10 1 "DD" 1 273.010 184.644 0 1 2 0 0 "R30" "" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 11 1 "DD" 1 236.260 203.794 0 1 2 0 0 "R9" "" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 12 1 "DD" 1 272.960 174.694 0 1 2 0 0 "R29" "" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
使用代码：
foreach (var item in placementOneListBox.Items)
{
var newItems = item.ToString();
newItems = Regex.Replace(newItems, @"\s+", " ");
var eachItem = newItems.Split(' ');
sw.WriteLine("( " + lineCount + " 0 \"FD\" 1 " + eachItem[2] + " " + eachItem[3]
+ " 0 0 0 0 0 \"" + eachItem[0] + "\" \"\" \"\" \"\" \"\" 0 0 0 "
+ lineCount + " 0 0 0 0 0 0 0 \"\" )");
lineCount++;
}
sw.WriteLine(")");
sw.WriteLine(")");
sw.Close();
文件的第二部分正在打印出来，除了我需要替换文件第二部分中“DD”之后发生的值文件第一部分中的 第一（数字）列 .. 第二部分的 正在打印文件的方式，我想根据它是否与值匹配打印出来。这就是我希望文件看起来像：
( 1 1 "DD" 1 238.385 205.944 0 1 2 0 0 "C2" "" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 2 1 "DD" 1 236.260 204.844 0 1 2 0 0 "C1" "" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 3 1 "DD" 2 230.960 190.619 0 1 2 0 0 "C18" "" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 4 1 "DD" 2 248.060 195.344 0 1 2 0 0 "C19" "" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 5 1 "DD" 2 231.535 203.844 0 1 2 0 0 "C63" "" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 6 1 "DD" 2 232.260 195.419 0 1 2 0 0 "C65" "" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 7 1 "DD" 3 240.151 195.123 0 1 2 0 0 "C71" "" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 8 1 "DD" 4 247.435 202.019 90 1 2 0 0 "C4" "" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 9 1 "DD" 5 248.110 205.719 0 1 2 0 0 "L1" "" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 10 1 "DD" 6 273.010 184.644 0 1 2 0 0 "R30" "" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 11 1 "DD" 6 236.260 203.794 0 1 2 0 0 "R9" "" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 12 1 "DD" 6 272.960 174.694 0 1 2 0 0 "R29" "" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
答案 0 :(得分：2)
使用LINQ：
可以简单地，相对容易地（或不是）解决问题
// Using the input file:
const string str = "C2 CAP-00128G 238.385 205.944 0 0402_8\n"
+ "C1 CAP-00128G 236.260 204.844 0 0402_8 #Same as above 1 (C2)\n"
+ "C18 CAP-00129G 230.960 190.619 0 0402_4 \n"
+ "C19 CAP-00129G 248.060 195.344 0 0402_4\n"
+ "C63 CAP-00129G 231.535 203.844 0 0402_4\n"
+ "C65 CAP-00129G 232.260 195.419 0 0402_4 #Same as above 3 (C18, C19, C63)\n"
+ "C71 CAP-00130G 240.151 195.123 0 0402_4 #1 Occurrence\n"
+ "C4 CAP-00151G 247.435 202.019 90 0402_3 #1 Occurrence\n"
+ "L1 IND-00007G 248.110 205.719 0 0402_4 #1 Occurrence\n"
+ "R30 RES-00455G 273.010 184.644 0 0402_8\n"
+ "R9 RES-00455G 236.260 203.794 0 0402_8\n"
+ "R29 RES-00455G 272.960 174.694 0 0402_8 #Same as above 2 (R30, R9)\n";
// And the reader (notice that by using `StringSplitOptions.RemoveEmptyEntries`
// you don't need your regular expression to normalize spacing):
var listLines = new List<string[]>();
using (var tr = new StringReader(str))
{
string l;
while ((l = tr.ReadLine()) != null)
listLines.Add(l.Split(new[] {' '},
StringSplitOptions.RemoveEmptyEntries));
}
// this is a cool trick when you need to build up a dictionary. The problem
// with using a dictionary in a case like this is that, unless you have some
// multidictionary implementation, you are going to be stuck with a lot of
// extra code that maintains the value collection (you know what I mean... the
// List<string[]> part of Dictionary<string, List<string[]>>.
var lookup = (from line in listLines group line by line[1]).ToArray();
// Part 1 -- The Select allows me to run a lambda function. In this case,
// I use a variation that provides me with a counter. Nothing really special
// here.
var part1 = lookup.Select(
(x, i) =>
string.Format(@"( {0} 0 1 1 0 0 0 0 """" ""{1}"" """" {2} 0 0 0 0 0 )",
i + 1, x.Key, x.Count()));
// And Part 2 - A little trickier. I'm using SelectMany to flatten my dictionary
// back out, but while doing so, I use an anonymous type to effectively decorate
// each line with the index of the group it belongs to. From then on it is again
// just a standard call to string.Format (which, if you aren't using it, please do.
// It's a lot more readable than string concatenations).
var part2 = lookup.SelectMany((x,i) => x.Select(y => new { Index = i, Item = y }))
.Select((x,i) => string.Format(
@"( {0} 1 ""DD"" {1} {2} {3} {4} 1 2 0 0 ""{5}"" """" """" """" 0 0 0 0 0 0 0 0 0 0 0 """" )",
i + 1, x.Index + 1, x.Item[2], x.Item[3], x.Item[4], x.Item[0] ));
// And so you can see your results in the console:
Console.WriteLine(string.Join("\n", part1));
Console.WriteLine();
Console.WriteLine(string.Join("\n", part2));
最后，输出是：
( 1 0 1 1 0 0 0 0 "" "CAP-00128G" "" 2 0 0 0 0 0 )
( 2 0 1 1 0 0 0 0 "" "CAP-00129G" "" 4 0 0 0 0 0 )
( 3 0 1 1 0 0 0 0 "" "CAP-00130G" "" 1 0 0 0 0 0 )
( 4 0 1 1 0 0 0 0 "" "CAP-00151G" "" 1 0 0 0 0 0 )
( 5 0 1 1 0 0 0 0 "" "IND-00007G" "" 1 0 0 0 0 0 )
( 6 0 1 1 0 0 0 0 "" "RES-00455G" "" 3 0 0 0 0 0 )
( 1 1 "DD" 1 238.385 205.944 0 1 2 0 0 "C2" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 2 1 "DD" 1 236.260 204.844 0 1 2 0 0 "C1" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 3 1 "DD" 2 230.960 190.619 0 1 2 0 0 "C18" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 4 1 "DD" 2 248.060 195.344 0 1 2 0 0 "C19" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 5 1 "DD" 2 231.535 203.844 0 1 2 0 0 "C63" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 6 1 "DD" 2 232.260 195.419 0 1 2 0 0 "C65" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 7 1 "DD" 3 240.151 195.123 0 1 2 0 0 "C71" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 8 1 "DD" 4 247.435 202.019 90 1 2 0 0 "C4" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 9 1 "DD" 5 248.110 205.719 0 1 2 0 0 "L1" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 10 1 "DD" 6 273.010 184.644 0 1 2 0 0 "R30" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 11 1 "DD" 6 236.260 203.794 0 1 2 0 0 "R9" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
( 12 1 "DD" 6 272.960 174.694 0 1 2 0 0 "R29" "" "" "" 0 0 0 0 0 0 0 0 0 0 0 "" )
我认为这正是你所追求的。
所以...希望这里有一些有趣的新事物可以熟悉：
- 更新 -
如果如引号中所述，您是从列表框中获取项目，则可以初始化listlines，如下所示：
var listLines = from l in ListBox.Items
select l.ToString().Split(new[] {' '},
StringSplitOptions.RemoveEmptyEntries);
您根本不再需要StringReader。现在，您可能必须使用ToString（）来确保每个项目确实是一个字符串，因为我不确定该集合实际上是什么类型。
答案 1 :(得分：0)
它看起来像是一个分隔文件。
如果您将此文本文件加载到DataTable，则可以使用DataTable提供的一些功能（如自定义视图和查询{{1}）来查询并输出该文本文件。方法。
您还可以轻松遍历表格中的所有行，以执行您需要的任何替换。
因此，我们的想法是将文件加载到表中，根据需要操作表，并将表的内容呈现为文本以供输出。