好， group by 以及 min() 请记住：

select tool_id,
       (case when count(*) < 3 then 0
             else min(available)
        end) as min_available_for_all_three_days
from tools t
where date >= '20180501' and date < '20180504'
group by tool_id;

这假设如果缺少一行，则该工具不可用。

select IF(
    (select count(*) 
     from tools 
     where (date >= '20180501' and date <= '20180503')
         and tool_id = 1 and available > 0
    ) = 3,
    (
        (select count(*) 
         from tools 
         where (date >= '20180501' and date <= '20180503')
             and tool_id = 2 and available > 0
        ) = 3,
        "Yes all the tools are available",
        "Tool 2 not available"
    ),

    "Tool 1 not available"
);

select tool_id from
(
select tool_id,date,sum(available) as tot_avl_on_day from tools
group by tool_id,date
)t
where tot_avl_on_day=0;

这将检查任意天数，以及是否所有天数都有可用的工具。

它将可用天数与日期范围中的天数匹配。您可能需要将日期值转换为日期类型。

SELECT
    `Tool_ID`,
    SUM(IF(`Available` > 0,1,0)) as `daysAvailable`
FROM `Tools`
WHERE `Sate` BETWEEN '2018-05-01' AND '2018-05-03'
    AND `daysAvailable` = DATEDIFF('2018-05-03','2018-05-01');