排除最少 6 位数字并替换 Hive 中的尾随数字

发布于 2025-01-12 09:45:05 字数 767 浏览 0 评论 0原文

有人可以帮我在配置单元中编写以下逻辑吗？我在列中有一个值，其中的数字尾随 0。我需要将所有这些 0 替换为 9，同时将 0 替换为 9 ，我还需要考虑在 9 之前至少应该有 6 位数字，否则需要排除一些 0，以便 9 之前至少可以有 6 位数字。 PFB 一些场景。

1234506600000000000

这里我们可以看到尾随 0 之前的位数是 8 (12345066)，所以我只需要把 0 保留 9 个，输出就会像这样。 1234506699999999999。

1234500000000000000

这里我在尾随 0 之前只有 5 位数字，所以我需要考虑第 6 个位置的 0 也是一个数字，并且需要在用 9 替换 0 时排除它，因此输出将为 1234509999999999999。

1000000000000000000

这里我在尾随 0 之前只有 1 位数字，所以我需要排除 5 个额外的 0，并且需要将剩余的 0 替换为 9，所以最终输出将类似于 1000009999999999999。

 Input                        Output

1234506600000000000        1234506699999999999
1234500000000000000        1234509999999999999
1000000000000000000        1000009999999999999

原文

Can someone help me to write below logic in hive.
I have one value in column in which I have digits with trailing 0's. I need to Replace all these 0's by 9,while replacing the 0's by 9 ,I need to also consider that before 9 minimun 6 digits should be there, else need to exclude some 0's so that before 9 can have at-least 6 digits.PFB some Scenarios.

1234506600000000000

Here we can see the number of digits before trailing 0's is 8 (12345066) so i Just need to remaining 0's by 9 and the output will be like. 1234506699999999999.

1234500000000000000

Here I have only 5 digits before trailing 0's so I need to consider 6th Position's 0 also a digit and need to exclude this while replacing the 0's by 9 so the output will be 1234509999999999999.

1000000000000000000

Here I have only 1 digit before trailing 0's ,so I need to exclude 5 extra 0's and need to replace remaining 0's by 9, so final output will be like 1000009999999999999.

 Input                        Output

1234506600000000000        1234506699999999999
1234500000000000000        1234509999999999999
1000000000000000000        1000009999999999999

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平生欢 2025-01-19 09:45:05

如果您想修改另一个问题中的 leftjoin 技术，我们可以调整正则表达式以匹配至少 6 位数字（包括 0）

with mytable as (
select '1234560000000' as input union all
select '123450000000' union all
select '12340000000' union all
select '1230000000'   
)

select lpad(concat(splitted[0], translate(splitted[1],'0','9')),13,0)
from
(
select split(regexp_replace(input,'(\\d{6,}?)(0+)
如果您想走我建议的替换/填充/替换路线，你会在 rtrim'd 后检查数字的长度，如果它小于 6，则用零将其填充到 6。如果字符串长度超过 6 个字符，rpad 的大多数实现都会将其截断为 6 个字符 - 如果不这样做，那么在 rtrim 之后调用 rpad 会很好而且很简单。如果 hive 的 rpad 执行子字符串操作，那么创建自己的 rpad 函数可能值得，该函数可以单独留下比 N 长的字符串
,'$1|$2'),'\\|') splitted
  from mytable
)s

如果您想走我建议的替换/填充/替换路线，你会在 rtrim'd 后检查数字的长度，如果它小于 6，则用零将其填充到 6。如果字符串长度超过 6 个字符，rpad 的大多数实现都会将其截断为 6 个字符 - 如果不这样做，那么在 rtrim 之后调用 rpad 会很好而且很简单。如果 hive 的 rpad 执行子字符串操作，那么创建自己的 rpad 函数可能值得，该函数可以单独留下比 N 长的字符串

If you want to modify leftjoin's technique from the other question we can tweak the Regex to match at least 6 digits including 0s

with mytable as (
select '1234560000000' as input union all
select '123450000000' union all
select '12340000000' union all
select '1230000000'   
)

select lpad(concat(splitted[0], translate(splitted[1],'0','9')),13,0)
from
(
select split(regexp_replace(input,'(\\d{6,}?)(0+)
If you want to go the replace/pad/replace route I proposed, you'd check the length of the number after it's rtrim'd and if it's less than 6, rpad it out to 6 with zeroes. Most implementations of rpad would chop the string off at 6 chars if it were longer than 6 - if they didn't it would be nice and simple to just call rpad after rtrim. It might be worth making your own rpad function that leaves strings longer than N alone, if hive's rpad performs a substring op
,'$1|$2'),'\\|') splitted
  from mytable
)s

If you want to go the replace/pad/replace route I proposed, you'd check the length of the number after it's rtrim'd and if it's less than 6, rpad it out to 6 with zeroes. Most implementations of rpad would chop the string off at 6 chars if it were longer than 6 - if they didn't it would be nice and simple to just call rpad after rtrim. It might be worth making your own rpad function that leaves strings longer than N alone, if hive's rpad performs a substring op

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