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我有这样的查询(在函数中):% }+ v# n9 N& `8 j. ^
UPDATE some_table SET column_1 = param_1, column_2 = param_2, column_3 = param_3, column_4 = param_4, column_5 = param_5WHERE id = some_id;param_x我的函数参数在哪里?有没有办法不更新这些参数的列?NULL?例如-; ~( i$ j& t) v8 Y
如果param_4和param_5是NULL,然后只更新前三列,离开旧值column_4和column_5。
; w4 {! w) x8 L0 c4 t我现在做的是:- D( {- @8 T9 c* R0 }! e1 R
SELECT * INTO temp_row FROM some_table WHERE id = some_id;UPDATE some_table SET column_1 = COALESCE(param_1,temp_row.column_1), column_2 = COALESCE(param_2,temp_row.column_2), column_3 = COALESCE(param_3,temp_row.column_3), column_4 = COALESCE(param_4,temp_row.column_4), column_5 = COALESCE(param_5,temp_row.column_5)WHERE id = some_id;有没有更好的办法?& x. b2 D0 Q: @' h V* s
) x4 I% S6 G( R* l+ D 解决方案: 2 T7 j/ o5 z6 A" P7 e4 J
删除SELECT句子,不需要使用,只需要使用当前值:% s; {0 ^& V+ R1 R4 V
UPDATE some_table SET column_1 = COALESCE(param_1,column_1), column_2 = COALESCE(param_2,column_2), column_3 = COALESCE(param_3,column_3), column_4 = COALESCE(param_4,column_4), column_5 = COALESCE(param_5,column_5)WHERE id = some_id; |
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