xml中的节点位置 - Java [英] Node position in xml - Java

问题描述

我想准备一个函数，它会给我节点在xml中的位置.
例如:

I would like to prepare function which will give me position of node in xml.
For example:

int NodePosition = doc.getDocumentElement().getChildNodes().item(t).getNodeName().Position()

将来我想做一些这样的(当然这是例子):

and in future I would like make some like this (of course this is example):

System.out.println(Node.row[NodePosition].tostring()); 

有没有可能制作这样的东西?或者，也许您知道一些可以实现类似功能的函数.

Is it possible to make something like this? Or maybe you know some function which make something like this.

例如，当我深入时，获得位置时会出现问题.因为 m = 7

For example when I go deeper there is a problem to get position. Because m = 7

doc.getDocumentElement().getChildNodes().item(t).getChildNodes().item(m)

当我尝试制作时(但我必须保存位置):

When I try make this (but I have to save position):

System.out.println(doc.getDocumentElement().getChildNodes().item(7).getNodeName());

我收到错误

java.lang.NullPointerException

推荐答案

对我来说，您似乎在重新发明轮子 - 您正在尝试构建自己的 XPath 版本 - 请参阅:http://www.ibm.com/developerworks/library/x-javaxpathapi/index.html

To me it looks like you are reinventing the wheel - you are trying to build your own version of XPath - See: http://www.ibm.com/developerworks/library/x-javaxpathapi/index.html

如何选择Xpath 节点集中按索引指定的节点

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