2016-12-02 122 views
0

由于XML片段:Xpath - 为什么只返回1个值?

<AddedExtras> 
    <AddedExtra Code="1234|ABCD" Quantity="1" Supplier="BDA"/> 
    <AddedExtra Code="5678|EFGH" Quantity="1" Supplier="BDA"/> 
    <AddedExtra Code="9111|ZXYW" Quantity="1" Supplier="BDA"/> 
</AddedExtras> 

以下XPath表达式:

//*["AddedExtra"]/@Code 

时,通过检查运行结果为:

Attribute='Code=1234|ABCD' 
Attribute='Code=5678|EFGH' 
Attribute='Code=9111|ZXYW' 

那么,为什么做下面的代码只回报第一行?

private String allCodes = "//*["AddedExtra"]/@Code"; 

从系统中获取我的XML并将其解析到一个文件:

public Document parseResponse(String response){ 
    DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance(); 
    factory.setNamespaceAware(true); 
    DocumentBuilder builder; 
    Document doc = null; 
    //Create a document reader and an XPath object 

    try { 
     builder = factory.newDocumentBuilder(); 
     doc = builder.parse(new InputSource((new StringReader(response)))); 

    } catch (ParserConfigurationException | org.xml.sax.SAXException | IOException e) { 
     e.printStackTrace(); 
    } 
    return doc; 
} 

获取新的文档:

public Document getParsedResponse(String response) { 
    return parseResponse(response); 
} 

返回从文档中的XPath值:

public String getAllCodeOptions(String response){ 
    Document doc = getParsedResponse(response); 
    return getNodeValueFromNodeList(doc, allCodes); 
} 

读取XML节点的新方法:

public String getNodeValueFromNodeList(Document doc, String expression){ 
    NodeList nodeList = null; 
    String nodes = null; 
    try { 
     nodeList = (NodeList) xpath.compile(expression).evaluate(doc, XPathConstants.NODESET); 
    } catch (XPathExpressionException e) { 
     e.printStackTrace(); 
    } 
    for(int i=0; i < nodeList.getLength(); i++){ 
     Node node = nodeList.item(i); 
     nodes = node.getNodeValue(); 
    } 
    return nodes; 
} 

回报:

Attribute='Code=1234|ABCD' 
+0

见http://stackoverflow.com/questions/2811001/how-to-read-xml-using-xpath-in-java – pringi

+0

请注意,你的断言是错误的。非空字符串的有效布尔值为true,因此'* [“AddedExtra”]'与'*'匹配相同的元素。我不会试图回答你的问题,因为这个问题似乎在不断变化。 –

回答

1

您将需要使用正确的评估方法,它将返回类型作为参数。类似下面,

NodeSet result = (NodeSet)e.evaluate(e, doc, XPathConstants.NODESET); 
for(int index = 0; index < result.getLength(); index ++) 
{  
    Node node = result.item(index); 
    String name = node.getNodeValue(); 
}