我使用XmlSlurper处理XML。它工作正常,直到我更新它。 appendNode不反映大小。
如何在结构更新后使用XmlSlurper?
XML定义:
def CAR_RECORDS = '''
<records>
<car name='HSV Maloo' make='Holden' year='2006'>
<country>Australia</country>
<record type='speed'>Production Pickup Truck with speed of 271kph</record>
</car>
<car name='P50' make='Peel' year='1962'>
<country>Isle of Man</country>
<record type='size'>Smallest Street-Legal Car at 99cm wide and 59 kg in weight</record>
</car>
<car name='Royale' make='Bugatti' year='1931'>
<country>France</country>
<record type='price'>Most Valuable Car at $15 million</record>
</car>
</records>
'''
代码失败:
def records = new XmlSlurper().parseText(CAR_RECORDS)
records.appendNode( { car(make:'BMW') } )
assert 4 == records.car.size() //fails!!! size == 3
打印包括宝马汽车在内的XML
def xmlOut = new groovy.xml.StreamingMarkupBuilder()
def temp = xmlOut.bind{
mkp.yield records
}
println temp
答案 0 :(得分:5)
您可以使用XmlSlurper再次阅读新创建的结构。
...
records = new XmlSlurper().parseText(temp as String)
assert 4 == records.car.size()
答案 1 :(得分:0)
试试这个
records = records.appendNode( { car(make:'BMW') } )
更新:实际上这段代码没有编译(请参阅注释),所以我稍后会发布自己的解决方法结果。
在某些项目中我使用了模板字符串,如:
static final String template = '<row><someList></someList></row>'
你可以像这样添加节点(lang Groovy):
def row = new XmlSlurper().parseText(template);
row.someList.appendNode(){
ListElement(){
field1('A')
field2('B')
field3('B')
}
}