在Groovy中，泛型XML可以像简单的XML那样简单地解析吗?

Given a nice, simple XML structure, XmlSlurper() can allow me to read values from it very easily.

对于一个简单的XML结构，XmlSlurper()可以很容易地从它读取值。

def xml = "<html><head><title>groovy</title></head></html>"

def html = new XmlSlurper().parseText(xml)

println html.head.title

Is there a way to make this simple tree navigation possible for generic (type-based, etc) XML. Ideally, in the snippet of code below, I'd like to walk the values by their name attribute, but instead, I have to do all this searching:

是否有一种方法可以使这个简单的树导航成为通用的(基于类型的)XML。理想情况下，在下面的代码片段中，我想按它们的name属性遍历值，但是我必须执行所有这些搜索:

def genxml = """
<doc>
    <lst name = "head">
        <str name = "title">groovy</str>
        <str name = "keywords">java xml</str>
    </lst>
</doc>"""

def doc = new XmlSlurper().parseText(genxml)
println doc.lst.find { it.@name == "head" }.str.find { it.@name == "title" }

Is there a way to walk this just as:

有没有一种方法可以做到:

println doc.head.title

1 个解决方案

#1

head and title are attributes.

标题和标题是属性。

there are some really subtle differences between slurper and parser: http://www.ibm.com/developerworks/java/library/j-pg05199/

slurper和解析器之间有一些非常细微的区别:http://www.ibm.com/developerworks/java/library/j-pg05199/

you can do this:

你可以这样做:

println "${doc.lst.str[0]} ${doc.lst.str[0].@name}"
println doc.lst.str.each { 
    println "${it} ${it.@name}"
    }

but look at the output:

但是看看输出:

groovy title
groovy title
java xml keywords
groovyjava xml

#1