【发布时间】：2014-01-21 12:41:16
【问题描述】：
我正在使用 Xpath 和 Java。
XML 有大量的OBJECT_TYPES，每个对象类型都有属性和参数。
并且每个属性和参数都有元素。
如何从我的 XML 文件中执行以下操作。
我想知道如何使用 XPATH 字符串表达式选择所有属性元素，具体取决于 OBJECT_TYPE 字符串的名称。对象类型字符串名称取决于用户从列表中选择的名称。
我该怎么做？
应该是这样的：
String expression = "/getObjType()/prop/*";
但是getObjectType 是一种方法，所以我不能在字符串表达式中使用它。
XML 看起来像这样：
<type>
<OBJECT_TYPE>SiteData</OBJECT_TYPE>
<prop>
<DESCRIPTION>Site parameters</DESCRIPTION>
<PARENT>NULL</PARENT>
<VIRTUAL>0</VIRTUAL>
<VISIBLE>1</VISIBLE>
<PICTURE>NULL</PICTURE>
<HELP>10008</HELP>
<MIN_NO>1</MIN_NO>
<MAX_NO>1</MAX_NO>
<NAME_FORMAT>NULL</NAME_FORMAT>
</prop>
<param>
<PARAMETER>blabla</PARAMETER>
<DATA_TYPE>INTEGER</DATA_TYPE>
<DESCRIPTION>blaba</DESCRIPTION>
<MIN_NO>1</MIN_NO>
<MAX_NO>1</MAX_NO>
<ORDER1>1</ORDER1>
<NESTED>0</NESTED>
<DEFAULT1>NULL</DEFAULT1>
<FORMAT>0:16382</FORMAT>
</param>
<OBJECT_TYPE>Data</OBJECT_TYPE>
<prop>
<DESCRIPTION>Site parameters</DESCRIPTION>
<PARENT>NULL</PARENT>
<VIRTUAL>0</VIRTUAL>
<VISIBLE>1</VISIBLE>
<PICTURE>NULL</PICTURE>
<HELP>10008</HELP>
<MIN_NO>1</MIN_NO>
<MAX_NO>1</MAX_NO>
<NAME_FORMAT>NULL</NAME_FORMAT>
</prop>
<param>
<PARAMETER>gmgm</PARAMETER>
<DATA_TYPE>INTEGER</DATA_TYPE>
<DESCRIPTION>babla</DESCRIPTION>
<MIN_NO>1</MIN_NO>
<MAX_NO>1</MAX_NO>
<ORDER1>1</ORDER1>
<NESTED>0</NESTED>
<DEFAULT1>NULL</DEFAULT1>
<FORMAT>0:16382</FORMAT>
</param>
</type>
因此，根据 Object_type 的名称，我想获得这些属性，并且我列出了 122 个对象类型，因此我必须使用一个变量来选择用户选择的对象类型。
public class PropXMLParsing {
static PropXMLParsing instance = null;
private List<String> list = new ArrayList<String>();
ObjType obj = new ObjType();
public static PropXMLParsing getInstance() {
if (instance == null) {
instance = new PropXMLParsing();
try {
instance.ParserForObjectTypes();
} catch (SAXException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
} catch (ParserConfigurationException e) {
e.printStackTrace();
}
}
return instance;
}
public void ParserForObjectTypes() throws SAXException, IOException,
ParserConfigurationException {
try {
FileInputStream file = new FileInputStream(new File(
"xmlFiles/CoreDatamodel.xml"));
DocumentBuilderFactory builderFactory = DocumentBuilderFactory
.newInstance();
builderFactory.setNamespaceAware(true);
DocumentBuilder builder = builderFactory.newDocumentBuilder();
Document xmlDocument = builder.parse(file);
XPath xp = XPathFactory.newInstance().newXPath();
final Map<String, Object> vars = new HashMap<String, Object>();
xp.setXPathVariableResolver(new XPathVariableResolver() {
public Object resolveVariable(QName name) {
return vars.get(name.getLocalPart());
}
});
XPathExpression expr = xp
.compile("/type/OBJECT_TYPE[. = $type]/following-sibling::prop[1]");
vars.put("type", obj.getObjectType());
NodeList objectProps = (NodeList) expr.evaluate(xmlDocument,
XPathConstants.NODESET);
System.out.println(objectProps);
for (int i = 0; i < objectProps.getLength(); i++) {
System.out.println(objectProps.item(i).getFirstChild()
.getNodeValue());
list.add(objectProps.item(i).getFirstChild().getNodeValue());
}
} catch (FileNotFoundException e) {
e.printStackTrace();
} catch (SAXException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
} catch (ParserConfigurationException e) {
e.printStackTrace();
} catch (XPathExpressionException e) {
e.printStackTrace();
}
}
public String convertListToString() {
StringBuilder sb = new StringBuilder();
if (list.size() > 0) {
sb.append(list.get(0));
for (int i = 1; i < list.size(); i++) {
sb.append(list.get(i));
}
}
return sb.toString();
}
}
第二个解决方案我尝试过既不工作也不在控制台中打印出任何东西。
public void ParserForObjectTypes() throws SAXException, IOException,
ParserConfigurationException {
try {
FileInputStream file = new FileInputStream(new File(
"xmlFiles/CoreDatamodel.xml"));
DocumentBuilderFactory builderFactory = DocumentBuilderFactory
.newInstance();
DocumentBuilder builder = builderFactory.newDocumentBuilder();
Document xmlDocument = builder.parse(file);
XPath xPath = XPathFactory.newInstance().newXPath();
NodeList nodeList = (NodeList) xPath.compile(
"//OBJECT_TYPE[text() = '" + obj.getObjectType()
+ "']/following-sibling::prop[1]/*").evaluate(
xmlDocument, XPathConstants.NODESET);
for (int i = 0; i < nodeList.getLength(); i++) {
System.out.println(nodeList.item(i).getNodeName() + " = "
+ nodeList.item(i).getTextContent());
}
} catch (FileNotFoundException e) {
e.printStackTrace();
} catch (SAXException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
} catch (ParserConfigurationException e) {
e.printStackTrace();
} catch (XPathExpressionException e) {
e.printStackTrace();
}
}
【问题讨论】：
-
请添加您尝试查询的 XML 示例。
-
好的，我已经添加了。我认为不需要。