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如何在MySQL中两个日期之间找到一组行?

[英]How do I find a group of rows exclusively BETWEEN two dates in MySQL?

我必须找到一组具有共享密钥的结果,这些结果已发布到我的本地数据存储区的API中,且不得早于过去90天,但不得迟于过去7天。

这将是一个有效的组,我需要为其捕获1234:

account_id,company_id,posted_date
1234,A,2018-02-28
1234,B,2018-03-13
1234,C.2018-04-23
1234,D,2018-05-15

这将是一个无效的组。 如果单个日期不在查询的上限或下限之内,则应从最终结果中排除帐户ID:

account_id,company_id,posted_date
5678,Z,2018-02-01
5678,Y,2018-03-13
5678,X.2018-04-23
5678,W,2018-05-21

这是使用子查询的查询的初稿:

SELECT DISTINCT account_id, company_id FROM local_data_store.result_api
WHERE account_id NOT IN (
    SELECT account_id FROM local_data_store.result_api
    GROUP BY account_id
    HAVING posted_date > DATE_SUB(NOW(), INTERVAL 7 DAY)
)
AND account_did IN (
    SELECT account_did FROM local_data_store.result_api
    GROUP BY account_did
    HAVING posted_date > DATE_SUB(NOW(), INTERVAL 90 DAY)
)
GROUP BY account_id, company_id
LIMIT 100000;

这是我现在正在处理的查询,没有子查询(我尝试了联接,但它们实际上不起作用):

SELECT DISTINCT account_id, company_id, 
COUNT(ra1.posted_date > DATE_SUB(NOW(), INTERVAL 90 DAY)) AS day90, 
COUNT(ra1.posted_date > DATE_SUB(NOW(), INTERVAL 7 DAY)) as day7
FROM local_data_store.result_api ra1
GROUP BY posted_date, account_id;

但是它运行了很长时间,以至于数据库连接超时。 这仅在包含375,000行的数据库表上。

这是我想出的解决方案。 希望它可以帮助别人。

SELECT DISTINCT account_did, cb_company_id, COUNT(did) as `# of jobs`
FROM cb_local_data_store.job_search_result_api 
WHERE account_did NOT IN (
SELECT account_did FROM cb_local_data_store.job_search_result_api
WHERE posted_date < DATE_SUB(NOW(), INTERVAL 60 DAY)
OR posted_date > DATE_SUB(NOW(), INTERVAL 4 DAY) )
GROUP BY account_did, cb_company_id

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