我有一個tsv如下所示的文件Time Object pmPdDrb pmPdcDlSrb00:45 EUtranCellFDD=GNL02294_7A_1 2588007 162600:45 EUtranCellFDD=GNL02294_7B_1 18550 3200:45 EUtranCellFDD=GNL02294_7C_1 26199 3800:45 EUtranCellFDD=GNL02294_9A_1 3857243 751是否可以像下面這樣將其轉換為 XML？<xmlnode> <Time>00:45</Time> <Object>EUtranCellFDD=GNL02294_7A_1</Object> <pmPdDrb>2588007</pmPdDrb> <pmPdcDlSrb>1626</pmPdcDlSrb></xmlnode>我試過下面的代碼：var reader = File.ReadAllLines(logFile);var xml1 = new XElement("TopElement",reader.Select(line => new XElement("Item", line.Split('\t').Select((column, index) => new XElement("Column" + index,column))) ));
1 回答
偶然的你
TA貢獻1841條經驗 獲得超3個贊
嘗試以下：
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.Xml.Linq;
using System.IO;
namespace ConsoleApplication110
{
class Program
{
const string INPUT_FILENAME = @"c:\temp\test.txt";
const string OUTPUT_FILENAME = @"c:\temp\test.xml";
static void Main(string[] args)
{
string header = "<xmlnodes></xmlnodes>";
XDocument doc = XDocument.Parse(header);
XElement xmlnodes = doc.Root;
StreamReader reader = new StreamReader(INPUT_FILENAME);
string line = "";
string[] columnNames = null;
int lineCount = 0;
while((line = reader.ReadLine()) != null)
{
line = line.Trim();
if (line.Length > 0)
{
string[] splitArray = line.Split(new char[] { '\t', ' '}, StringSplitOptions.RemoveEmptyEntries);
if (++lineCount == 1)
{
columnNames = splitArray;
}
else
{
XElement newNode = new XElement("xmlnode");
xmlnodes.Add(newNode);
for(int i = 0; i < splitArray.Length; i++)
{
XElement xColumn = new XElement(columnNames[i], splitArray[i]);
newNode.Add(xColumn);
}
}
}
}
doc.Save(OUTPUT_FILENAME);
}
}
}
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