【发布时间】:2021-11-06 02:12:04
【问题描述】:
请有人帮我解决这种情况?
我有这个假 JSON...
[
{
"user": {
"type": "PF",
"code": 12345,
"Name": "Darth Vader",
"currency": "BRL",
"status": "SINGLE",
"localization": "NABOO",
"createDate": 1627990848665,
"olderAdress": [
{
"localization": "DEATH STAR",
"createDate": 1627990848775
},
{
"localization": "TATOOINE",
"createDate": 1627990555888
},
]
}
}
]
我的想法是,我需要提取“olderAdress”并创建新寄存器,但我也需要保留原始寄存器。 示例:这是我希望的结果。
[
{
"_id": ObjectId("5a3456e000102030405000000"),
"user": {
"Name": "Darth Vader",
"code": 12345,
"createDate":1627990848665,
"currency": "BRL",
"localization": "NABOO",
"status": "SINGLE",
"type": "PF"
}
},
{
"_id": ObjectId("5a789e000102030405000000"),
"user": {
"Name": "Darth Vader",
"code": 12345,
"createDate": 1627990848775,
"currency": "BRL",
"localization": "DEATH STAR",
"status": "SINGLE",
"type": "PF"
}
},
{
"_id": ObjectId("5a991e000102030405000000"),
"user": {
"Name": "Darth Vader",
"code": 12345,
"createDate": 1627990555888,
"currency": "BRL",
"localization": "TATOOINE",
"status": "SINGLE",
"type": "PF"
}
}
]
我在此链接 (Test to extract values) 中尝试相同的想法进行测试,但不幸的是我不能。有人可以帮帮我吗?
【问题讨论】:
标签: json mongodb spring-data-jpa aggregation-framework unwind