我在附加图像中得到这个结果,我需要122和&的总和。 455来自“trans_inventory”专栏。
trans_id trans_items trans_user trans_date trans_comment trans_inventory location_id
8236 1488 1 2016-05-02 02:14:35 Manual Edit of Quantity 10.0000000000 1
8237 1488 1 2016-05-02 02:15:05 RECV 58 122.0000000000 1
8238 1488 1 2016_05-02 02:15:29 RECV 59 455.0000000000 1
8239 1488 1 2016-05-02 02:17:32 NEWP- 506 -55.0000000000 1
8240 1488 1 2016-05-02 02:18:00 NEWP- 507 -121.0000000000 1
我怎样才能得到它?
抱歉我的英语不好。也许,我无法解释清楚。答案 0 :(得分:0)
尝试以下查询:
SELECT SUM(trans_inventory) FROM (YOUR_QUERY_OF_RESULT_HERE)x
WHERE trans_comment LIKE "RECV %"
......或当您使用桌子时:
SELECT SUM(trans_inventory) FROM table_name
WHERE trans_comment LIKE "RECV %"
答案 1 :(得分:0)
使用SUM函数:
SELECT SUM("trans_inventory") FROM table_name WHERE trans_id = 8237 OR trans_id = 8238;