Xpath 查询 在线等~~~急~~~~~~~~~~~

Question

<?xml version='1.0' encoding="UTF-8" standalone="yes"?>
<Enrolment Register="李济明" Date="2001-09-01">
<Students>
<Student StudentNo="0208123" Sex="男">
<Name>王义平</Name>
<BirthDate>1984-08-20</BirthDate>
<CourseTaken>EE-122</CourseTaken>
<CourseTaken>EE-201</CourseTaken>
</Student>
<Student StudentNo="0104421" Sex="男">
<Name>周远行</Name>
<BirthDate>1984-07-16</BirthDate>
<CourseTaken>CS-110</CourseTaken>
<CourseTaken>EE-201</CourseTaken>
</Student>
<Student StudentNo="0309119" Sex="女">
<Name>李 维</Name>
<BirthDate>1984-08-01</BirthDate>
<CourseTaken>CS-110</CourseTaken>
<CourseTaken>CS-201</CourseTaken>
</Student>
</Students>
<Courses>
<Course CourseNo="CS-110" Credit="3">
<LectureHour>60</LectureHour>
<Semester>秋</Semester>
</Course>
<Course CourseNo="CS-201" Credit="4">
<LectureHour>80</LectureHour>
<Semester>春</Semester>
</Course>
<Course CourseNo="EE-122" Credit="5">
<LectureHour>106</LectureHour>
<Semester>秋</Semester>
</Course>
<Course CourseNo="EE-201" Credit="2">
<LectureHour>45</LectureHour>
<Semester>春</Semester>
</Course>
</Courses>
</Enrolment>列出选修CS-110课程的学生姓名、学号和出生日期 
我只能用三条语句分别列出：
/Enrolment/Students/Student[CourseTaken="CS-110"]/Name
/Enrolment/Students/Student[CourseTaken="CS-110"]/@StudentNo
/Enrolment/Students/Student[CourseTaken="CS-110"]/BirthDate请问如何用一条语句一起输出，谢谢

Answer 1

http://www.webdn.com/web_file/3sword/xml/0602016042/
不多说了,你看了那个该能试出来.

Answer 2

直接这样:/Enrolment/Students/Student[CourseTaken="CS-110"]然后遍历取出的结点,得到相关信息,不行吗?

Answer 3

for example:            XmlDocument doc = new XmlDocument();
            doc.Load("../../student.xml");
            XmlNodeList nodelist = doc.SelectNodes(@"/Enrolment/Students/Student[CourseTaken='CS-110']");
            foreach (XmlNode node in nodelist)
            {
                Console.WriteLine("StudentNo: " + node.ChildNodes[0].InnerText + "\tStudentName: " + node.Attributes["StudentNo"].Value + "\tBirthDate: " + node.ChildNodes[1].InnerText);
            }

Answer 4

to：liujia_0421(SnowLover) ( ) 
直接这样:/Enrolment/Students/Student[CourseTaken="CS-110"]这样只显示元素，不能显示属性的

Answer 5

TO:这样只显示元素，不能显示属性的你可以通过得到的结点,得到属性值啊..node.Attributes["StudentNo"].Value 不就可以得到吗?

Answer 6

try..
 
XmlDocument doc = new XmlDocument();
            doc.Load("../../student.xml");
            XmlNodeList nodelist = doc.SelectNodes(@"/Enrolment/Students/Student[CourseTaken='CS-110']");
            foreach (XmlNode node in nodelist)
            {
                Console.WriteLine("StudentNo: " + node.ChildNodes[0].InnerText + "\tStudentName: " + node.Attributes["StudentNo"].Value + "\tBirthDate: " + node.ChildNodes[1].InnerText);
            }输出:StudentNo: 周远行 StudentName: 0104421 BirthDate: 1984-07-16
StudentNo: 李 维 StudentName: 0309119 BirthDate: 1984-08-01

Answer 7

写反了..            XmlDocument doc = new XmlDocument();
            doc.Load("../../student.xml");
            XmlNodeList nodelist = doc.SelectNodes(@"/Enrolment/Students/Student[CourseTaken='CS-110']");
            foreach (XmlNode node in nodelist)
            {
               Debug.WriteLine("StudentName: " + node.ChildNodes[0].InnerText + "\tStudentNO: " + node.Attributes["StudentNo"].Value + "\tBirthDate: " + node.ChildNodes[1].InnerText);
            }输出:StudentName: 周远行 StudentNO: 0104421    BirthDate: 1984-07-16
StudentName: 李 维          StudentNO: 0309119    BirthDate: 1984-08-01

Answer 8

我要的是
Xpath查询路径表达式~~~