我的 couchbase 存盤桶中有多個檔案，其結構如下：
{
"objectType": "someObjectType",
"id": "1",
"latest": {
"id": "1",
"version": "biyr4jqi4nny"
},
"history": {
"bi11b295bjng": {
"id": "1",
"version": "bi11b295bjng"
}
"bi1189wx1h6v": {
"id": "1",
"version": "bi1189wx1h6v"
}
}
}
如上面的片段所示，歷史是物件的物件。我想要做的是選擇objectType、id、latest 和history，但history 應該只包括查詢中指定的物件，而不是所有的history（可能很大）。
我的查詢如下所示：
SELECT
bucket.id,
bucket.objectType,
bucket.latest,
bucket.history.bi11b295bjng
FROM bucket WHERE objectType = 'someObjectType'
這產生了以下回應：
[
{
"objectType": "someObjectType",
"id": 1,
"latest": {
"id": "9mobile_NG_001-ROW",
"version": "biyr4jqi4nny"
},
"biyr4jqi4nny": {
"id": "1",
"version": "biyr4jqi4nny"
}
}
]
查詢的物件從父物件中解開。所需的輸出應如下所示：
[
{
"objectType": "someObjectType",
"id": 1,
"latest": {
"id": "9mobile_NG_001-ROW",
"version": "biyr4jqi4nny"
},
"history": {
"biyr4jqi4nny": {
"id": "1",
"version": "biyr4jqi4nny"
}
}
}
]
如何在不丟失父物件的情況下獲取 history.{version} ？
uj5u.com熱心網友回復：
構造物件和別名作為歷史
SELECT
b.id,
b.objectType,
b.latest,
{ b.history.bi11b295bjng } AS history
FROM bucket AS b
WHERE b.objectType = "someObjectType";
如果需要同一欄位的多個物件
SELECT
b.id,
b.objectType,
b.latest,
{ b.history.bi11b295bjng, b.history.bi1189wx1h6v } AS history
FROM bucket AS b
WHERE b.objectType = "someObjectType";
如果您有很多想要洗掉其中之一
SELECT
b.id,
b.objectType,
b.latest,
OBJECT_REMOVE(b.history,"bi11b295bjng") AS history
FROM bucket AS b
WHERE b.objectType = "someObjectType";
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