Mysql sql更新中出错
我将2012年参与索赔的运营商的工资提高了11%, 试一试 但是它说 “错误代码:1241。操作数应包含1列” 有人知道我如何修复这些错误吗Mysql sql更新中出错,mysql,sql,sql-update,Mysql,Sql,Sql Update,我将2012年参与索赔的运营商的工资提高了11%, 试一试 但是它说 “错误代码:1241。操作数应包含1列” 有人知道我如何修复这些错误吗 sqlfiddele这里:我认为您的查询应该是 update operator o SET o.sueldo = o.sueldo + (o.sueldo * 0.11) where exists (select 1 from claim where r.cod_operador r = o.cod_operador and r.fecha Between
sqlfiddele这里:我认为您的查询应该是
update operator o
SET o.sueldo = o.sueldo + (o.sueldo * 0.11)
where exists (select 1 from claim where r.cod_operador r = o.cod_operador
and r.fecha Between '2012-01-01 'and '2012-12-31');
我想你的问题应该是
update operator o
SET o.sueldo = o.sueldo + (o.sueldo * 0.11)
where exists (select 1 from claim where r.cod_operador r = o.cod_operador
and r.fecha Between '2012-01-01 'and '2012-12-31');
你对这份声明做了什么 这样,它至少适合您的桌子:
update operador o
set o.sueldo = o.sueldo + (o.sueldo * 0.11)
where exists (select 1 from reclamo r where r.cod_operador = o.cod_operador
and r.fecha Between '2012-01-01 'and '2012-12-31');
但我不知道它最终是否会做正确的事情…你对这句话做了什么 这样,它至少适合您的桌子:
update operador o
set o.sueldo = o.sueldo + (o.sueldo * 0.11)
where exists (select 1 from reclamo r where r.cod_operador = o.cod_operador
and r.fecha Between '2012-01-01 'and '2012-12-31');
但我不知道它最终是否会做正确的事情……o.sueldo+(o.sueldo*0.11)
可能会表示为o.sueldo*1.11
o.sueldo+(o.sueldo*0.11)
可能会表示为o.sueldo*1.11
。