我需要通过 XPath 获取一些 XML 节点,我该怎么做?
我尝试使用这个库https://github.com/djcsdy/haxe-xpath ,但是出了点问题https://github.com/djcsdy/haxe-xpath/issues/26
对于我的任务来说,xml-fast 并不是一个好的解决方案,因为它看起来比 XPath“稍差一些”,我认为:
js(xpath):
xml_doc.get('//project/classpaths/class[@path="' + src_path + '"]')
haxe(xml-fast):
(new Fast(xml_doc))).node.project.node.classpaths.nodes.class.filter(function (x:Fast) return x.has.path ? x.att.path == src_path : false)
谢谢
最佳答案
结果发现两个库,都需要一些修复。
1. "Haxe XPath"
仅当将目录“haxe-xpath/src/xpath”克隆到源中时才可以使用它(haxelib 存储库不包含此库)。
示例(删除第一个找到的元素):
package;
import xpath.XPathHx;
using Lambda;
class Main {
public static function main () {
var xml = Xml.parse("<a> <b> <c>qwe</c> <c>asd</c> </b> </a>");
trace(xml.toString());
var xpExpr = new XPathHx("//a/b/c"); // create new XPath expression object
var result = xpExpr.selectNodes(xml).array()[0]; // get first element from array of founded xml-nodes
result.parent.removeChild(result); // remove selected node from xml-tree
trace(xml.toString());
}
}
2. "xmlTools"
可以使用haxelib安装:
haxelib install xmlTools
haxelib install composure
Some fix was needed for this library (in my task)和一个nuance for xpath .
示例(删除第一个找到的元素):
package;
import xmlTools.XPath;
using Lambda;
class Main {
public static function main () {
var xml = Xml.parse("<a> <b> <c>qwe</c> <c>asd</c> </b> </a>");
trace(xml.toString());
var xpath = new XPath(); // create new XPath expression object
var result = xpath.resolve(xml, "*/a/b/c").array()[0]; // get first element from array of founded xml-nodes
result.parent.removeChild(result); // remove selected node from xml-tree
trace(xml.toString());
}
}
关于xml - Haxe 与 XPath 有友谊吗?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/37528220/