Question

目前我正在使用单个XML文件中的数据表中的所有记录创建XML文件，我想使用linq分别为每个记录创建XML文件。

DataTable dtTest = new DataTable();
dtTest.Columns.Add("Name");
dtTest.Columns.Add("NickName");
dtTest.Columns.Add("Code");
dtTest.Columns.Add("reference");

dtTest.Rows.Add("Yash", "POPs", "Vapi", "None1");
dtTest.Rows.Add("shilpa", "shilpa", "valsad", "None2");
dtTest.Rows.Add("Dinesh", "dinu", "pune", "None3");
dtTest.Rows.Add("rahul", "mady", "pardi", "None4");

XDocument xmlDoc = new XDocument(
    new XElement(
        "File",
        from fields in dtTest.AsEnumerable()
        select new XElement(
            "company_details",
            new XElement(
                "company",
                new XElement(
                    "Name",
                    fields.Field<string>("Name")),
                new XElement(
                    "NickName",
                    fields.Field<string>("NickName"))),
            new XElement(
                "Details",
                new XElement(
                    "reference",
                    fields.Field<string>("reference"))))));

string filepath = @"C:\Users\admin\Desktop\test.xml";
xmlDoc.Save(filepath);

Answer 1

您可以使用xml方法为每个row创建ForEach

DataTable dtTest = new DataTable();
dtTest.Columns.Add("Name");
dtTest.Columns.Add("NickName");
dtTest.Columns.Add("Code");
dtTest.Columns.Add("reference");

dtTest.Rows.Add("Yash", "POPs", "Vapi", "None1");
dtTest.Rows.Add("shilpa", "shilpa", "valsad", "None2");
dtTest.Rows.Add("Dinesh", "dinu", "pune", "None3");
dtTest.Rows.Add("rahul", "mady", "pardi", "None4");

dtTest.AsEnumerable().ToList().ForEach(x => CreateXml(x));

public void CreateXml(DataRow row)
{
  XDocument xmlDoc = new XDocument(
      new XElement(
          "File",
           new XElement(
               "company_details",
               new XElement(
                   "company",
               new XElement(
                   "Name",
                    row.Field<string>("Name")),
               new XElement(
                   "NickName",
                    row.Field<string>("NickName"))),
               new XElement(
                    "Details",
               new XElement(
                    "reference",
                     row.Field<string>("reference"))))));
  xmlDoc.Save(Server.MapPath(row.Field<string>("Name")) + ".xml");
}

使用Linq C＃为DataTable的每个记录创建XML文件

1 个答案: