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DB提琴
; }7 c- i) T' C5 I3 t1 t7 I6 G' FCREATE TABLE logistics ( id int primary key, campaign VARCHAR quantity_offered VARCHAR quantity_ordered VARCHAR quantity_delivered VARCHAR quantity_recorded VARCHAR quantity_completed VARCHAR(255));INSERT INTO logistics(id,campaign,quantity_offered,quantity_ordered,quantity_delivered,quantity_recorded,quantity_completed)VALUES ("1","C001","500","450","465","462","465"),("2","C002","700","570",NULL,NULL,NULL),("3","C003","600","610","605","602",NULL),("4","C004","300",NULL,NULL,NULL,NULL),("5","C005","400","425",NULL,NULL,NULL),("6","C006","900","870","868",NULL,NULL),("7","C007","350","360","372","375","390"),("8","C008","250","290",NULL,NULL,NULL);在上表中我campaigns不同于它们对应的不同quantities。
0 J x/ n- g; d( y8 E5 m7 [! q的quantities不同的填写方法columns。- o1 P% N/ ?0 {: s
现在,我想根据以下层次结构获得每个 最新可用数量campaign:/ S6 f v- J6 @, R8 ]
quantity_completed > quantity_recorded > quantity_delivered > quantity_ordered > quantity_offered结果如下:8 J } z5 j! U* A3 i
Campaign QuantityC001 46555 C 570005C003 C 30000C5005 42555C 8700C 3900C008 2900002我需要执行什么查询?5 |& }3 _2 W( {
* h2 O/ i, k+ s6 Z 解决方案: |
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