arangodb AQL 查询以根据条件删除列表及其子项
arangodb AQL query to erase list and it's child depending of condition
我有一个包含日志列表的帐户集合
如果日志的“loggedAt”字段超过 5 天,我需要清除帐户“日志”列表中的每个日志,而不留下孤立日志或损坏“日志”列表的引用。
如何使用 AQL 或 arangosh 命令实现此目的?
像这样
FOR a IN account
FOR l IN log
FILTER l._id IN a.logs
FILTER DATE_DIFF(l.loggedAt, DATE_NOW(), 'days', false) > 5
UPDATE a WITH { logs: REMOVE_VALUE( a.logs, l ) } IN account
REMOVE { _key: l._key } IN log
账户文件示例
_id:account/1085466856
_rev:_bIYHFNu---
_key:1085466856
{
"_class": "com.app.Account",
"logs": [
"log/1085468455",
"log/1085468456",
"log/1085468457"
]
}
Account.class
@Data
@Document("account")
public class Account implements Serializable {
private static final long serialVersionUID = -1615384653229054932L;
@Id
private String id;
@Ref(lazy = false)
private Collection<Log> logs;
}
日志文档示例
_id:log/1085468455
_rev:_bHhi2zu-_1
_key:1085468455
{
"_class": "com.app.Log",
"loggedAt": "2020-08-24T17:44:43.600Z"
}
Log.class
@Data
@Document("log")
public class Log implements Serializable {
private static final long serialVersionUID = -1969696942832711654L;
@Id
private String id;
private Date loggedAt;
}
您可以在 2 个 AQL 查询中执行此操作:
- 先删除日志:
LET fiveDaysAgo = DATE_SUBTRACT(DATE_NOW(), 5, 'days')
FOR rec IN log
FILTER DATE_DIFF(rec.loggedAt, fiveDaysAgo, 'days') >= 0
REMOVE rec IN log
- 接下来清理帐户:
FOR acc IN account
LET validLogs = (
FOR accLog IN acc.logs
FOR rec IN log
FILTER accLog == rec._id
RETURN rec._id
)
FILTER LENGTH(acc.logs) != LENGTH(validLogs)
UPDATE acc WITH {logs: validLogs} IN account
如果需要,可以将上述查询合并为一个查询:
LET fiveDaysAgo = DATE_SUBTRACT(DATE_NOW(), 5, 'days')
FOR acc IN account
LET invalidLogs = (
FOR accLog IN acc.logs
FOR rec IN log
FILTER accLog == rec._id
FILTER DATE_DIFF(rec.loggedAt, fiveDaysAgo, 'days') >= 0
REMOVE rec IN log
RETURN OLD._id
)
FILTER LENGTH(invalidLogs) > 0
UPDATE acc WITH {logs: OUTERSECTION(acc.logs, invalidLogs)} IN account
但是,如果您在 log
集合中有僵尸记录(与 account
记录没有任何关联的记录),上面的查询不会影响它们。
我有一个包含日志列表的帐户集合
如果日志的“loggedAt”字段超过 5 天,我需要清除帐户“日志”列表中的每个日志,而不留下孤立日志或损坏“日志”列表的引用。
如何使用 AQL 或 arangosh 命令实现此目的?
像这样
FOR a IN account
FOR l IN log
FILTER l._id IN a.logs
FILTER DATE_DIFF(l.loggedAt, DATE_NOW(), 'days', false) > 5
UPDATE a WITH { logs: REMOVE_VALUE( a.logs, l ) } IN account
REMOVE { _key: l._key } IN log
账户文件示例
_id:account/1085466856
_rev:_bIYHFNu---
_key:1085466856
{
"_class": "com.app.Account",
"logs": [
"log/1085468455",
"log/1085468456",
"log/1085468457"
]
}
Account.class
@Data
@Document("account")
public class Account implements Serializable {
private static final long serialVersionUID = -1615384653229054932L;
@Id
private String id;
@Ref(lazy = false)
private Collection<Log> logs;
}
日志文档示例
_id:log/1085468455
_rev:_bHhi2zu-_1
_key:1085468455
{
"_class": "com.app.Log",
"loggedAt": "2020-08-24T17:44:43.600Z"
}
Log.class
@Data
@Document("log")
public class Log implements Serializable {
private static final long serialVersionUID = -1969696942832711654L;
@Id
private String id;
private Date loggedAt;
}
您可以在 2 个 AQL 查询中执行此操作:
- 先删除日志:
LET fiveDaysAgo = DATE_SUBTRACT(DATE_NOW(), 5, 'days')
FOR rec IN log
FILTER DATE_DIFF(rec.loggedAt, fiveDaysAgo, 'days') >= 0
REMOVE rec IN log
- 接下来清理帐户:
FOR acc IN account
LET validLogs = (
FOR accLog IN acc.logs
FOR rec IN log
FILTER accLog == rec._id
RETURN rec._id
)
FILTER LENGTH(acc.logs) != LENGTH(validLogs)
UPDATE acc WITH {logs: validLogs} IN account
如果需要,可以将上述查询合并为一个查询:
LET fiveDaysAgo = DATE_SUBTRACT(DATE_NOW(), 5, 'days')
FOR acc IN account
LET invalidLogs = (
FOR accLog IN acc.logs
FOR rec IN log
FILTER accLog == rec._id
FILTER DATE_DIFF(rec.loggedAt, fiveDaysAgo, 'days') >= 0
REMOVE rec IN log
RETURN OLD._id
)
FILTER LENGTH(invalidLogs) > 0
UPDATE acc WITH {logs: OUTERSECTION(acc.logs, invalidLogs)} IN account
但是,如果您在 log
集合中有僵尸记录(与 account
记录没有任何关联的记录),上面的查询不会影响它们。