public void send(Map<String, List<Integer>> mapOf){
mapOf.forEach((k ,v) -> {
v.stream().forEach(o ->{
myList.add(o)
message.push(o);
});
});
}
给出结果
message.push(o), pushes data to the database. <br/>
但是我需要
之类的结果select ite_code,specification,x_unit ,
Cast (( Row_number() OVER( partition BY ite_code, specification, x_unit ORDER BY ite_code, specification, x_unit ) ) AS NUMERIC(18, 4)) partition_no from (
select ite_code ='110001010001', specification= 'Sample 1',x_unit= 'NOS' union all
select ite_code ='110001010001', specification= 'Sample 1',x_unit= 'NOS' union all
select ite_code ='110001010001', specification= 'Sample 2',x_unit= 'NOS' union all
select ite_code ='110001010001', specification= 'Sample 2',x_unit= 'NOS' ) a
我的代码有什么问题?
答案 0 :(得分:4)
也许您打算在此处使用等级函数:
SELECT
ite_code,
specification,
x_unit,
DENSE_RANK() OVER (ORDER BY ite_code, specification, x_unit) AS partition_no
FROM yourTable
ORDER BY
ite_code,
specification,
x_unit;
我对查询所做的主要更改是将ROW_NUMBER
交换为DENSE_RANK
,并删除了分区。