初学:返回xml字符串的处理问题?
返回如下xml字符串:
<?xml version="1.0" encoding="utf-8" ?>
<DeliverMsg xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns="igw.zsdx.com">
<returncode>0 </returncode>
<msg/>
<mos>
<MO>
<aaa>3150 </aaa>
<bbb>1182760 </bbb>
<ccc>07608888888 </ccc>
</MO>
<MO>
<aaa>3151 </aaa>
<bbb>1182760 </bbb>
<ccc>07608888888 </ccc>
</MO>
<MO>
<aaa>123 </aaa>
<bbb>118276039 </bbb>
<ccc>07608888888 </ccc>
</MO>
</mos>
</DeliverMsg>
请教,如何获取各MO下aaa等属性的值?
XmlDocument doc = new XmlDocument();
doc.Load("xml");
doc.SelectNodes("/DeliverMsg/mos/Mo");
下面代码是在winform开发环境下.
C# code
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XML code:
<upd:Update xmlns:lar="http://schemas.microsoft.com/msus/2002/12/LogicalApplicabilityRules" xmlns:cmd="http://schemas.microsoft.com/msus/2002/12/Up ......
package com.javabean.xml;
import java.net.MalformedURLException;
import java.net.URL;
import org.dom4j.io.SAXReader;
import org.dom4j.Attribute;
import org.dom4j.Document;
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<?xml version="1.0" encoding="utf-8" ?>
<configuration>
<id="1">
<user="test1" pass="12345" />
& ......
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public Document read(String fileName) throws MalformedURLException, DocumentException {
SAXReader reader = new SAXReader();
Document document = read ......