将xml文件读入datagridview修改以后保存成新的xml文件怎么做？-.NET技术/C#

------回答---------

------其他回答（10分）---------

DataSet ds; load:

ds = new DataSet(); ds.ReadXml(filepath);

this.dataGridView1.DataSource = ds.Tables[0]; ……………… Save：

ds.AcceptChanges(); ds.WriteXml(filepath); ------其他回答（10分）---------

// 将xml文件转换为DataSet

public static DataSet ConvertXMLFileToDataSet(string xmlFile) {

StringReader stream = null; XmlTextReader reader = null; try {

XmlDocument xmld = new XmlDocument(); xmld.Load(xmlFile);

DataSet xmlDS = new DataSet(); stream = new StringReader(xmld.InnerXml); reader = new XmlTextReader(stream); xmlDS.ReadXml(reader); return xmlDS; }

catch (System.Exception ex) {

throw ex; } finally

{

if (reader != null) reader.Close(); } }

// 将DataSet转换为xml文件

public static void ConvertDataSetToXMLFile(DataSet xmlDS, string xmlFile)

{

MemoryStream stream = null; XmlTextWriter writer = null; try {

stream = new MemoryStream();

writer = new XmlTextWriter(stream, Encoding.Unicode); xmlDS.WriteXml(writer);

int count = (int)stream.Length; byte[] arr = new byte[count]; stream.Seek(0, SeekOrigin.Begin); stream.Read(arr, 0, count);

UnicodeEncoding utf = new UnicodeEncoding(); StreamWriter sw = new StreamWriter(xmlFile);

sw.WriteLine( \ version=\\ \ encoding=\\ \ sw.WriteLine(utf.GetString(arr, 0, arr.Length).Trim()); sw.Close(); }

catch (System.Exception ex) {

throw ex; } finally {

if (writer != null) writer.Close(); } }

分别调用以上2个函数就可以了