能夠使用以下方法mysql
SELECT DISTINCT pg.part_grp_id, pg.part_grp_name, pg.equip_category_id FROM cost_part_grp pg, cost_part_kit pk, cost_part_event pe WHERE pe.mdl_ver_id IN ( SELECT s.mdl_ver_id FROM cost_fleet_model s WHERE s.fleet_id = 1002 ) AND pe.part_kit_id = pk.part_kit_id AND pk.part_grp_id = pg.part_grp_id ORDER BY CASE pg.equip_category_id WHEN 2 THEN 1 WHEN 1 THEN 2 WHEN 4 THEN 3 ELSE 4 END
原文連接:https://blog.csdn.net/skytalemcc/article/details/5883728sql
UPDATE t_bss_employees SET mobilephone = REPLACE (mobilephone, "2129", "0000") WHERE id IN ( SELECT a.id FROM ( SELECT id FROM t_bss_employees WHERE mobilephone IN ( '18121299262', '18121299247', '18121299206', '18121299209' ) ) AS a )
將mobilephone字段中的2129字符串替換爲0000
使用當前表爲條件更新當前表。須要在條件處再添加一個()構建一個虛擬表ui
select a.style, ROUND( b.num / a.sum * 100, 2 ) as styleRate from (SELECT count(qspc.id) sum ,qq.style style from t_qc_security_plan_comment qspc inner join t_qc_question qq on qq.id = qspc.question_id LEFT JOIN t_qc_address qa ON qa.id = qspc.address_id LEFT JOIN t_system_organ so on qa.hospital = so.id WHERE qq.scr_level is not NULL and qq.type = 3 and so.tenant_code = 'zzyy' GROUP BY style) a left join (SELECT count(qspc.id) num ,qq.style style from t_qc_security_plan_comment qspc inner join t_qc_question qq on qq.id = qspc.question_id LEFT JOIN t_qc_address qa ON qa.id = qspc.address_id LEFT JOIN t_system_organ so on qa.hospital = so.id WHERE qq.scr_level is not NULL and qq.type = 3 and so.tenant_code = 'zzyy' and qspc.is_bad =0 GROUP BY style) b on a.style = b.style
計算邏輯是分別求總數和平均數。而後兩數相除再用round求精度.net
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