x=0.623
k_p ze_01 ze_02 ze_03 ......
1 0.072 0.115 0.216 ......
2 0.207 0.297 0.711 ......
3 0.676 0.872 1.237 ......
4 0.989 1.239 1.695 ......
5 1.344 1.646 2.187 ......
... ... ... ...求与X值最相近的值所在列,最后返回列名!!!各位大侠帮忙!!!
k_p ze_01 ze_02 ze_03 ......
1 0.072 0.115 0.216 ......
2 0.207 0.297 0.711 ......
3 0.676 0.872 1.237 ......
4 0.989 1.239 1.695 ......
5 1.344 1.646 2.187 ......
... ... ... ...求与X值最相近的值所在列,最后返回列名!!!各位大侠帮忙!!!
解决方案 »
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- 请问如下sql查询语句怎么写
(select min(ze_01-0.623) a,'ze_01' col_name from t
union all
select min(ze_02-0.623) a,'ze_02' col_name from t
union all
select min(ze_03-0.623) a,'ze_03' col_name from t
)
select col_name from s1
where a =(select min(a) from s1)
一个sql会死人的,用存储过程
with s1 as
(select abs(ze_01-0.623) a,'ze_01' col_name from t
union all
select abs(ze_02-0.623) a,'ze_02' col_name from t
union all
select abs(ze_03-0.623) a,'ze_03' col_name from t
)
select col_name from s1
where a in (select min(a) from s1);