问:
根据用户的一些选择,我将从数据库中获取一组数据表。
我有三个数据元素:Teachers
,Subjects
,ClassRooms
我想要一个具有以下结构的xml文件:
<timetable importtype="database" options="idprefix:id">
<teachers options="" columns="id,name,short">
</teachers>
<subjects options="" columns="id,name,short">
</subjects>
<classrooms options="" columns="id,name,short">
</classrooms>
</timetable>
现在如何使用以前的结构生成xml文件并从数据表中填充xml文件数据?
例如:
<teachers options="" columns="id,name,short">
<teacher id="a" name="jhon" short="jo"/>
<teacher id="b" name="philips" short="Ph"/>
<teacher id="c" name="sara" short="Sa"/>
</teachers>
来自teacher datatable
。
请尽可能详细回答示例。
private static void ConvertDataTableToXML(string schemaFile, DataTable dtTeacher, DataTable dtSubject, DataTable dtClassRoom)
{
XDocument doc = new XDocument();
//Read the teachers element from xml schema file
XElement teachers = XDocument.Load(schemaFile).Descendants("teachers").SingleOrDefault();
XElement subjects = XDocument.Load(schemaFile).Descendants("subjects").SingleOrDefault();
XElement classes = XDocument.Load(schemaFile).Descendants("classrooms").SingleOrDefault();
if (teachers != null)
{
foreach (DataRow row in dtTeacher.Rows)
{
XElement teacher = new XElement("teacher");
teacher.SetAttributeValue("id", row["id"]);
teacher.SetAttributeValue("name", row["name"]);
teacher.SetAttributeValue("short", row["name"].ToString().Substring(0, 2));
teachers.Add(teacher);
}
}
if (subjects != null)
{
foreach (DataRow row in dtSubject.Rows)
{
XElement subject = new XElement("subject");
subject.SetAttributeValue("id", row["num"]);
subject.SetAttributeValue("name", row["name"]);
subject.SetAttributeValue("short", row["name"].ToString().Substring(0, 2));
subjects.Add(subject);
}
}
if (classes != null)
{
foreach (DataRow row in dtClassRoom.Rows)
{
XElement cls = new XElement("classroom");
cls.SetAttributeValue("id", row["id"]);
cls.SetAttributeValue("name", row["name"]);
cls.SetAttributeValue("short", row["name"].ToString().Substring(0, 2));
classes.Add(cls);
}
}
XElement xml = new XElement("timetable",
new XAttribute("importtype", "database"),
new XAttribute("options", "idprefix:id"),teachers,subjects,classes
);
doc.Add(xml);
string dirPath = System.Web.HttpContext.Current.Server.MapPath("~/import");
string targetFileName = Path.Combine(dirPath, "import.xml");
int counter = 1;
while (System.IO.File.Exists(targetFileName))
{
counter++;
targetFileName = Path.Combine(dirPath,
"import" + counter.ToString() + ".xml");
}
doc.Save(targetFileName);
}
答案 0 :(得分:1)
您可以尝试使用Linq-XML读取和更新XML文档。
看看样本:
//Read the teachers element from xml schema file
XElement teachers = XDocument.Load(schemaFile).Descendants("teachers").SingleOrDefault();
if (teachers != null)
{
DataTable dt = new DataTable();
dt.Columns.Add("id");
dt.Columns.Add("name");
dt.Rows.Add("1", "john");
dt.Rows.Add("2", "philips");
dt.Rows.Add("3", "sara");
XDocument doc = new XDocument();
foreach (DataRow row in dt.Rows)
{
XElement teacher = new XElement("teacher");
teacher.SetAttributeValue("id", row["id"]);
teacher.SetAttributeValue("name", row["name"]);
teacher.SetAttributeValue("short", row["name"].ToString().Substring(0,2));
teachers.Add(teacher);
}
doc.Add(teachers);
doc.Save(newFilename);
}