Groovy将两个json与未知节点名称和值进行比较
发布时间:2020-12-14 16:25:18 所属栏目:大数据 来源:网络整理
导读:我有一个休息API来测试,我必须比较两个json响应.您可以在下面找到该文件的结构.要比较的两个文件应包含相同的元素,但顺序可能不同.不幸的是,名称,类型(简单,数组)和键数(root,nodeXYZ)也是未知的. {"root": [{ "node1": "value1","node2": "value1","node3":
我有一个休息API来测试,我必须比较两个json响应.您可以在下面找到该文件的结构.要比较的两个文件应包含相同的元素,但顺序可能不同.不幸的是,名称,类型(简单,数组)和键数(root,nodeXYZ)也是未知的.
{"root": [{ "node1": "value1","node2": "value1","node3": [ { "node311": "value311","node312": "value312" },{ "node321": "value321","node322": "value322" } ],"node4": [ { "node411": "value411","node412": "value413","node413": [ { "node4131": "value4131","node4132": "value4131" }],"node414": [] } { "node421": "value421","node422": "value422","node423": [ { "node4231": "value4231","node4232": "value4231" }],"node424": [] }] "node5": [ {"node51": "value51"},{"node52": "value52"},] }]} 我在中找到了一些有用的信息 take root get root children names check if child has children and get their names do it to the lowest leve child 与所有名称比较应该很容易(我猜) 解决方法
只需比较啜饮的地图:
def map1 = new JsonSlurper().parseText(document1) def map2 = new JsonSlurper().parseText(document2) assert map1 == map2 (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |