从使用 XmlParser 解析的文件确定结束节点
我有一个方法，需要搜索使用 XmlParser 按名称解析元素的 xml 文件，并且仅当该元素是结束节点时才返回它。例如：
class xmlTest extends GroovyTestCase {
def void test(){
def xmlBody = """
<rootElement>
<elementWithOneChild>
<endElement>Here is the end</endElement>
</elementWithOneChild>
<elementWithManyChildren>
<one>1</one>
<two>1</two>
<three>1</three>
</elementWithManyChildren>
</rootElement>"""
def parsedBody = new XmlParser().parseText(xmlBody)
def search1 = parsedBody.depthFirst().grep({it.name() == "elementWithOneChild"})
println search1[0].children().size()
def search2 = parsedBody.depthFirst().grep({it.name() == "endElement"})
println search2[0].children().size()
def search3 = parsedBody.depthFirst().grep({it.name() == "elementWithManyChildren"})
println search3[0].children().size()
}
}
我尝试使用 Node.children().size() 有效，除了元素包含一个子元素的 1 对 1 情况。在这种情况下，search1.children().size()和search2.children().size()都返回1。虽然，elementWithManyChildren的大小是3。我正在寻找某种方法来告诉结束节点除了具有一个子元素的元素之外。
我发现可行的一种方法是：
try{
search1[0].children().iterator().next().name()
}catch(e){
//If the next node does not have a name, it is an end node
}
但该解决方案似乎是一个糟糕的解决方案。
I have a method which needs to search an xml file which was parsed using XmlParser for an element by name and return it only if that element is an end node. For example:
class xmlTest extends GroovyTestCase {
def void test(){
def xmlBody = """
<rootElement>
<elementWithOneChild>
<endElement>Here is the end</endElement>
</elementWithOneChild>
<elementWithManyChildren>
<one>1</one>
<two>1</two>
<three>1</three>
</elementWithManyChildren>
</rootElement>"""
def parsedBody = new XmlParser().parseText(xmlBody)
def search1 = parsedBody.depthFirst().grep({it.name() == "elementWithOneChild"})
println search1[0].children().size()
def search2 = parsedBody.depthFirst().grep({it.name() == "endElement"})
println search2[0].children().size()
def search3 = parsedBody.depthFirst().grep({it.name() == "elementWithManyChildren"})
println search3[0].children().size()
}
}
My attempt to use Node.children().size() works except for the 1 to 1 case where an element contains one child element. In this case, search1.children().size() and search2.children().size() both return 1. Although, the size for elementWithManyChildren is 3. I am looking for some way to be able to tell an end node apart from an element with one child.
One way I have found to work is:
try{
search1[0].children().iterator().next().name()
}catch(e){
//If the next node does not have a name, it is an end node
}
But that solution just seems like a poor one.
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评论（1）
可能会浪费几个周期，但这应该可以让您找到终端节点
might waste a couple of cycles but this should allow you to find terminal nodes