使用xmlstarlet将xml递归循环到csv
我想将复杂的xml转换为csv使用xmlstarlet将xml递归循环到csv,xml,csv,xmlstarlet,Xml,Csv,Xmlstarlet,我想将复杂的xml转换为csv <tests> <test> <name>AAA</name> <language>BBB</language> <Project> <name>XXX</name> <id>123</id> </Proj
<tests>
<test>
<name>AAA</name>
<language>BBB</language>
<Project>
<name>XXX</name>
<id>123</id>
</Project>
<fac>
<name>XXX</name>
<idt>
<number>99</number>
<idt>
<pers>YYY</pers>
</fac>
<fac>
<name>BBB</name>
<idt>
<number>70</number>
<idt>
<pers>MMM</pers>
</fac>
<fac>
<name>XXX</name>
<idt>
<number>40</number>
<idt>
<pers>XXX</pers>
</fac>
<date>2018</date>
</test>
<test>
<name>BBB</name>
<language>CCC</language>
<Project>
<name>AAA</name>
<id>12</id>
</Project>
<fac>
<name>YXX</name>
<idt>
<number>10</number>
<idt>
<pers>LLL</pers>
</fac>
<fac>
<name>BB</name>
<idt>
<number>7</number>
<idt>
<pers>MM</pers>
</fac>
<fac>
<name>XX</name>
<idt>
<number>40</number>
<idt>
<pers>XXX</pers>
</fac>
<date>2018</date>
</test>
<tests>
一切正常,但我只能获取第一个节点中包含的数据。我想要的是这样的东西:
第一次
name;language;name;id;data contained in the first node"fac";date
name(same as first line);language(same as first line); etc..; data contained in the second "fac" node;date (same as first line)
etc... as much as there are face nodes
然后是第二个节点
我不知道是否可以使用xmlstarlet实现这一点
提前感谢您的帮助
RFlow尝试以下xml linq:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.Xml.Linq;
using System.IO;
namespace ConsoleApplication1
{
class Program
{
const string XML_FILENAME = @"c:\temp\test.xml";
const string CSV_FILENAME = @"c:\temp\test.csv";
static void Main(string[] args)
{
string[] headers = {
"test name",
"language",
"project name",
"project id",
"fac name",
"idt number",
"pers"
};
StreamWriter writer = new StreamWriter(CSV_FILENAME);
writer.WriteLine(string.Join(",", headers));
XDocument doc = XDocument.Load(XML_FILENAME);
foreach (XElement test in doc.Descendants("test"))
{
string testName = (string)test.Element("name");
string language = (string)test.Element("language");
XElement project = test.Element("Project");
string projectName = (string)project.Element("name");
string projectId = (string)project.Element("id");
foreach(XElement fac in test.Elements("fac"))
{
string facName = (string)fac.Element("name");
string number = (string)fac.Descendants("number").FirstOrDefault();
string pers = (string)fac.Element("pers");
string csvLine = string.Join(",", new string[] {
testName,
language,
projectName,
projectId,
facName,
number,
pers
});
writer.WriteLine(csvLine);
}
}
writer.Flush();
writer.Close();
}
}
}
如果您希望为每个
fac
输入一个条目,则应匹配该条目。然后,您可以转到祖先测试
,以获取所需的其他数据
例如
xmlstarlet sel -T -t -m "//fac" -v "concat(ancestor::test/name,';',ancestor::test/language,';',ancestor::test/Project/name,';',ancestor::test/Project/id,';',name,';',idt/number,';',pers,';',ancestor::test/date)" -n test.xml
输出
AAA;BBB;XXX;123;XXX;99;YYY;2018
AAA;BBB;XXX;123;BBB;70;MMM;2018
AAA;BBB;XXX;123;XXX;40;XXX;2018
BBB;CCC;AAA;12;YXX;10;LLL;2018
BBB;CCC;AAA;12;BB;7;MM;2018
BBB;CCC;AAA;12;XX;40;XXX;2018
下面是为便于阅读而细分的concat()
concat(
ancestor::test/name,';',
ancestor::test/language,';',
ancestor::test/Project/name,';',
ancestor::test/Project/id,';',
name,';',
idt/number,';',
pers,';',
ancestor::test/date
)
您的示例输入不是有效的XML,看起来您的所有
都已转换为
?修复后,您的xmlstarlet
调用对我来说很好。谢谢,我的xml不是这样的,但在这个示例中我忘记了它。即使我改变了它,我也会得到第一个吗?我得到两行输出:AAA;BBB;XXX;123;XXX;99;YYY;2018年
之后是BBB;CCC;AAA;12;YXX;10;LLL;2018年
。我现在不知道如果我很清楚,我想要的是第一个节点:AAA;BBB;XXX;123;XXX;99;YYY;2018年
thenAAA;BBB;XXX;123;BBB;70;嗯,;2018年
thenAAA;BBB;XXX;123;XXX;40;XXX;2018
第二个节点也一样。@DanielHaley感谢您的帮助。现在一切都好了:)谢谢,我也要试试。
concat(
ancestor::test/name,';',
ancestor::test/language,';',
ancestor::test/Project/name,';',
ancestor::test/Project/id,';',
name,';',
idt/number,';',
pers,';',
ancestor::test/date
)