我的XML看起来像这样 -
<collected_objects>
<object flag="complete" id="objId" version="1">
<variable_value variable_id="varId">ValueGoesHere</variable_value>
<reference item_ref="2"/>
</object>
<object comment="objComment" flag="complete" id="objId" version="1">
<reference item_ref="1"/>
</object>
</collected_objects>
我正在使用以下代码处理它 -
Document dom = parser.getDocument();
NodeList collected_objects = dom.getElementsByTagName("object");
System.out.println("Number of collected objects are " + collected_objects.getLength());
for (int i = 0; i < collected_objects.getLength(); i++) {
Node aNode = collected_objects.item(i);
//get children of "objects"
NodeList refNodes = aNode.getChildNodes();
System.out.println("# of chidren are " + refNodes.getLength());
//print attributes of "objects"
NamedNodeMap attributes = aNode.getAttributes();
for (int a = 0; a < attributes.getLength(); a++) {
Node theAttribute = attributes.item(a);
System.out.println(theAttribute.getNodeName() + "=" + theAttribute.getNodeValue());
}
}
输出as-
Number of collected objects are 2
# of chidren are 5
flag=complete
id=objId
version=1
# of chidren are 3
comment=objComment
flag=complete
id=objId
version=1
我的问题是为什么&#34;#chidren是&#34;分别是5和3?我不应该分别期待2和1吗?
因为第一个对象有&#34; variable_value
&#34;和&#34; reference
&#34;第二个对象只有&#34; reference
&#34;
基本上,我的目的是处理&#34;对象&#34;。
的孩子答案 0 :(得分:16)
确保&lt; object&gt;之间没有空格节点儿童。空格被视为子节点并以此形式返回。
测试是否
childNode.getNodeType() == Node.ELEMENT_NODE
应该足够了。
答案 1 :(得分:8)
那是因为每个子节点之间有2个TEXT_NODE
(#text
)。
以下内容包括文本节点及其对应的值。
<object flag="complete" id="objId" version="1">
<TEXT_NODE />
<variable_value variable_id="varId">ValueGoesHere</variable_value>
<reference item_ref="2"/>
<TEXT_NODE />
</object>
可以通过修改代码来验证:
DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
Document dom = dBuilder.parse(new ByteArrayInputStream(S.getBytes()));
NodeList collected_objects = dom.getElementsByTagName("object");
System.out.println("Number of collected objects are "
+ collected_objects.getLength());
for (int i = 0; i < collected_objects.getLength(); i++) {
Node aNode = collected_objects.item(i);
// get children of "objects"
NodeList refNodes = aNode.getChildNodes();
System.out.println("# of chidren are " + refNodes.getLength());
//
for (int x = 0; x < refNodes.getLength(); x++) {
Node n = refNodes.item(x);
System.out.println(n.getNodeType() + " = " + n.getNodeName() + "/" + n.getNodeValue());
}
// print attributes of "objects"
NamedNodeMap attributes = aNode.getAttributes();
for (int a = 0; a < attributes.getLength(); a++) {
Node theAttribute = attributes.item(a);
System.out.println(theAttribute.getNodeName() + "="
+ theAttribute.getNodeValue());
}
}
输出:
Number of collected objects are 2
# of chidren are 5
3 = #text/
1 = variable_value/null
3 = #text/
1 = reference/null
3 = #text/
flag=complete
id=objId
version=1
# of chidren are 3
3 = #text/
1 = reference/null
3 = #text/
comment=objComment
flag=complete
id=objId
version=1
其中,3 = TEXT_NODE
和1 = ELEMENT_NODE
。
答案 2 :(得分:2)
您只计算ELEMENT节点类型。如果您只对子元素感兴趣,可以更改代码以包含以下检查
if (aNode.getNodeType() == Node.ELEMENT_NODE)
{
...
}