mysql一些语句

<!-- 报警量排行按创建时间每月来排行 -->
<select >
SELECT b.`name` as areaName,
COUNT(c.area_code) as companySum
FROM sys_area b
LEFT JOIN amp_ent_info c on b.`CODE` =c.area_code
GROUP BY b.`code` HAVING b.`code` LIKE '3307__'
</select>

 

<!-- 报警等级近30天的数据排行 -->
<select >
SELECT
truncate(t2.levelHigh/t1.levelAll*100,2) as levelHighPercent,
truncate(t3.levelMid/t1.levelAll*100,2) as levelMidPercent,
truncate(t4.levelLow/t1.levelAll*100,2) as levelLowPercent
from
(SELECT count(*) as levelAll from amp_alarm_log where DATE_SUB(CURDATE(), INTERVAL 30 DAY) &lt;= create_date) t1,
(SELECT count(*) as levelHigh from amp_alarm_log a where a.level=3 and DATE_SUB(CURDATE(), INTERVAL 30 DAY) &lt;= create_date) t2,
(SELECT count(*) as levelMid from amp_alarm_log a where a.level=2 and DATE_SUB(CURDATE(), INTERVAL 30 DAY) &lt;= create_date) t3,
(SELECT count(*) as levelLow from amp_alarm_log a where a.level=1 and DATE_SUB(CURDATE(), INTERVAL 30 DAY) &lt;= create_date) t4
</select>