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我有这样的查询(在函数中):
; r5 z2 s+ v# y Q# p3 _ q BUPDATE 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-3 L# G0 M2 m. x: [; {
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。
$ C3 ]: ]( B1 J5 v, `& S我现在做的是:! y& k9 N3 I( V- _
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?& b x: v; [4 O8 t" I! K& ]
- I7 }" A" S* v/ H% w 解决方案: & `! R; ?8 b! I" ~
删除SELECT句子,不需要使用,只需要使用当前值:
# C( o' Q5 M4 Y1 w' R* G. {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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