MySql获取一列等于另一列的另一表中一列相等的行

Question

我有两张桌子：

tbl_post 和 tbl_public

tbl_post：

id | post_id   |  author  |  date (unix time)
--------------------------------
 0 |  123456   | User 1   |  1489857236
 1 |  123457   | User 2   |  1489857237
 2 |  123458   | User 3   |  1489857238

tbl_public：

    id | post_id | public | date (unix time)
    -----------------------------------------
 --> 0 |  123456 |  0     |  1489857236
     1 |  123457 |  1     |
     2 |  123458 |  0     |
 --> 3 |  123456 |  1     |  1489857239

我想获取所有行，其中 user = "User 1" 并且表 tbl_public 中最新（按日期排序）公共列大于 0 按日期排序的 DESC

你能帮我查询命令吗？ :)

Answer 1

请简要说明您的需求，尽管我尝试了从您的问题中理解的内容。让我知道它是否对你有用。

SQL：

SELECT
    tp.post_id,
    tp.author,
    tpb.public,
    tpb.date
FROM
    tbl_post tp
    LEFT JOIN tbl_public tpb ON (tp.post_id = tpb.post_id)
WHERE
    tp.author = 'User 1'
ORDER BY
    tpb.date DESC;

Answer 2

您可以使用INNER JOIN 来执行此操作，语法如下：

SELECT column_name(s)
FROM table1
INNER JOIN table2
ON table1.column_name = table2.column_name;

Answer 3

SELECT
  *
FROM
  tbl_post 'a'
  INNER JOIN tbl_public 'b' ON a.post_id = b.post_id
WHERE
  author = 'User 1'
  AND public > 0