Question

我有一个这样组装的查询（使用Django后端）：

sql = 'SELECT * FROM `table` WHERE'
if condition1:
    sql += ' `col1` = "foo" OR'

if condition2:
    sql += ' `col2` = "bar" OR'
...

if conditionN:
    sql += ' `colN` = "foobar" OR'

sql = sql[:-2] //to remove the extra 'OR'

sql += 'ORDER BY `col1` LIMIT x OFFSET y'

print(sql) //'SELECT * FROM `table` WHERE `col1` = "foo" OR `col2` = "bar" ... OR `colN` == "foobar" ORDER BY `col1` LIMIT x OFFSET y'

我需要一个查询来获取COUNT(*)中所有行的table，但仍然需要从数据库中获取SELECT的条件和{{1} }

我试图做这样的事情：

LIMIT

如果您还不能确定的话，那么我对上面查询中发生的事情完全感到困惑，这是行不通的（显然）。

更新

该表如下所示：

SELECT t1.*, IFNULL(COUNT(t2.*), 0) AS childCount 
FROM `table` AS t1 LEFT JOIN `table` AS t2
WHERE `col1` = "foo" OR `col2` = 'bar' OR `colN` = 'foobar' 
ORDER BY `col1` desc LIMIT 10;

我需要的示例查询：

+--------+----------+
|  col1  |   col2   | 
+--------+----------+
|   a    |   foo    |      
|   b    |   foo    |      
|   c    |   bar    |      
|   d    |   bar    |     
|   e    |   foo    | 
+--------+----------+

它需要返回此值：

'SELECT *, COUNT(*) as childCount FROM `table` WHERE `col2` = "foo" LIMIT 1 OFFSET 0'

建议/说明将不胜感激：）

Answer 1

根据您更新的问题

可以

select col1,col2,count from (
select col2,count(col2) as count from table1 group by col2 order by col1)a
order by col1
limit 1
offset 0

Answer 2

IIUC

SELECT a.col1, a.col2, b.childCount
FROM table a
JOIN (
  SELECT c.col2, COUNT(1)
  FROM table c
  GROUP BY 1
  ORDER BY 2 DESC) b
ON a.col2 = b.col2
LIMIT 1
OFFSET 0

如何在同一查询中选择并获得COUNT？

更新

2 个答案: