Question

我有一张桌子

farmid chickens cows horses
10     104      50   2
12     62       110  7
17     74       12   154

我想查询表，只显示大于100的值。输出应该如下所示

farmid chickens cows horses
10     104      
12              110 
17                   154

有没有人对如何使这项工作有任何建议？

我试过这个：

SELECT farmid, chickens, cows, horses
FROM farms
WHERE chickens >=100 OR cows>=100 OR horses>=100;

然而，

返回

farmid chickens cows horses
10     104      50   2
12     62       110  7
17     74       12   154

Answer 1

如果你想让价值低于100的鸡，牛，马细胞为空或空，你应该使用case ... when。

例如......

select 
  (case when chickens > 100 then chickens else null end case) as chickens,
  (case when cows > 100 then cows else null end case) as cows,
  (case when horses > 100 then horses else null end case) as horses
from table_name
where
  chickens > 100 OR cows > 100 OR horses > 100

Answer 2

您需要WHERE子句。

语法：

SELECT column1, column2, ...
FROM table_name
WHERE condition;

您可以使用OR和＆gt; WHERE子句中的符号以获得所需的结果。

SELECT farmid,
    CASE WHEN chickens > 100 THEN chickens ELSE NULL
    END as chickens,
    CASE WHEN cows > 100 THEN cows ELSE NULL
    END as cows,
    CASE WHEN horses > 100 THEN horses ELSE NULL
    END as horses,
from tableName 
WHERE chickens > 100 OR cows > 100 OR horses > 100

这将选择farmid，并且取决于值>＆gt;的情况。 100，它将选择该列值。即使多个列具有值> gt，它也应该工作。 100。

Answer 3

这将返回所有表格行，如果您只想要chickens，cows或horses中的至少一个>＆gt; = 100的行，请将最后一行包含为好。

SELECT 
    farmid,
    CASE
        WHEN chickens >= 100 THEN chickens
        ELSE NULL
    END AS chickens,
    CASE
        WHEN cows >= 100 THEN cows
        ELSE NULL
    END AS cows,
    CASE
        WHEN horses >= 100 THEN horses
        ELSE NULL
    END AS horses
FROM
    <enter table name here>
WHERE chickens >= 100 OR cows >= 100 OR horses >= 100

仅显示大于100 mysql的值

3 个答案: