我有一个groovy系统批处理脚本,需要读入XML文档,更改值,然后保存它。
我已经弄清楚如何读取值并编写它。我无法弄清楚我的生活如何移除'成员'及其价值观。我需要能够删除所有成员并用自定义文件名替换它我无法弄明白。
XML看起来像这样:
<types>
<members>*</members>
<members>Account</members>
<members>Activity</members>
<members>Contact</members>
<members>Task</members>
<members>User</members>
<members>ContentVersion</members>
<name>CustomObject</name>
</types>
我会搜索名称&#34; CustomObject&#34;并使用此字符串删除所有兄弟姐妹的成员:
def replace = "MyCustomFile"
所以XML会这样:
<types>
<members>MyCustomFile</members>
<name>CustomObject</name>
</types>
我尝试了以下在网上找到的代码
println "Testing Slurper"
def root = new XmlSlurper().parse(new File(theFile))
root.types.each { types ->
println "names: ${types.name}"
types.members.each {
println "members: " + it.text()
}
}
println "Testing replace"
def book = "Booking__c"
def cleanUpNode(node) {
println node
def childs = node.children()
def remove = []
childs.each {
if (it instanceof Node) {
if (!it.children()) {
remove.add it
} else {
cleanUpNode it
if (!it.children()) {
remove.add it
}
}
}
}
remove.each { node.remove(it) }
}
cleanUpNode root.types.name
实际上并没有删除任何内容,而是输出:
<tag0:types>
<tag0:members>*</tag0:members>
<tag0:members>Account</tag0:members>
<tag0:members>Activity</tag0:members>
<tag0:members>Contact</tag0:members>
<tag0:members>Task</tag0:members>
<tag0:members>User</tag0:members>
<tag0:members>ContentVersion</tag0:members>
<tag0:name>CustomObject</tag0:name>
</tag0:types>
我仍然试图掌握这一点,所以任何帮助都会很棒
答案 0 :(得分:1)
您可以这样做:
List newMembersToAdd = ['myCustomFile', 'anotherCustomFile']
Node xml = new XmlParser(false, false, false).parse("myXml.xml")
xml.members?.each {
xml.remove it
}
newMembersToAdd.each { String newMember ->
new Node(xml, 'members', newMember)
}
new File("myNewXml.xml").withWriter { writer ->
def printer = new XmlNodePrinter( new PrintWriter(writer) )
printer.preserveWhitespace = true
printer.print( xml )
}