我将JSON作为输出: -
def desiredJson = '{"count": 4, "max": "12", "min": 0, "details": [{"goBus": {"first": 12800, "second": 11900, "third": 12800},"goAir": {"first": 12800, "second": 11900, "third": 12800}, "gotTrain": {"first": 12800, "second": 11900},"sell": true, "darn": 2,"rate": [{ "busRate": 11900, "flag": false, "percent": 0}],}],}'
我想删除" count"键及其值,删除
"goBus": {
"first": 12800,
"second": 11900,
"third": 12800
},
删除"详细信息"的方括号节点
我已尝试将以下代码删除并替换为null: -
def slurper = new JsonSlurper();
def json = slurper.parse(file)
def newjson = JsonOutput.toJson(json).toString()
String j = "max"
newjson = newjson.replaceAll(""+ j +"", "")
log.info newjson
作为输出,不会删除最大值。或者我们是否可以通过其他方式从JSON中删除所有内容。
有人可以帮我吗?
我也试过这个: -
def json = new JsonSlurper().parseText(desiredJson)
def njson = json.details.goBus
def pjson = njson.remove()
log.info JsonOutput.toJson(pjson)
它返回false。
答案 0 :(得分:2)
通常没有理由使用字符串替换 - 它有很大的可能会弄乱一些东西。您可以在将地图写回JSON之前修改地图。 E.g:
import groovy.json.*
def jsonStr = '{"a": 1, "b": [{"c": 3, "d": 4}]}}'
def json = new JsonSlurper().parseText(jsonStr)
// XXX: first "de-array" `b`
json.b = json.b.first()
// next remove `c` from it
json.b.remove('c')
println JsonOutput.toJson(json)
// => {"a":1,"b":{"d":4}}
答案 1 :(得分:2)
这是具有所需输出的工作解决方案
此处的工作代码Working example
import groovy.json.*
def jsonStr = '''{
"count": 4,
"max": "12",
"min": 0,
"details": [{
"goBus": {
"first": 12800,
"second": 11900,
"third": 12800
},
"goAir": {
"first": 12800,
"second": 11900,
"third": 12800
},
"gotTrain": {
"first": 12800,
"second": 11900,
"third": 12800,
"fourth": 13000
},
"sell": true,
"darn": 2,
"rate": [{
"busRate": 11900,
"flag": false,
"percent": 0
}]
}]
}'''
def json = new JsonSlurper().parseText(jsonStr)
json.details[0].remove('goBus')
println JsonOutput.toJson(json)