如何使用 python 从字符串中删除 .ds 模式字符串

Question

我有这个字符串

a="""SELECT
    transform_abc.ds AS "ds",
    SUM(transform_abc.dollars) AS "dollars",
    transform_abc.unit AS "unit"
FROM fct_table_abc transform_abc
WHERE
    (
        transform_abc.is_charged > 0
        OR transform_abc.account_status = 0
    )
    AND transform_abc.ds = '2020-02-20'
GROUP BY
    transform_abc.ds,
    transform_abc.unit"""

我需要从这个字符串中删除 SELECT 和 GROUP BY 之后但不是 WHERE 之后的带有 ds 的列。

需要输出：

a="""SELECT
    SUM(transform_abc.dollars) AS "dollars",
    transform_abc.unit AS "unit"
FROM fct_table_abc transform_abc
WHERE
    (
        transform_abc.is_charged > 0
        OR transform_abc.account_status = 0
    )
    AND transform_abc.ds = '2020-02-20'
GROUP BY
    transform_abc.unit"""

Tranform_abc 只是一个 table 名称，它可以是任何 table 名称。所以我们不能在正则表达式中使用它。不确定如何解决这个问题

Answer 1

请在下面找到解决方案。这可能在语法上不正确，没有编译它但应该符合逻辑。

#!/usr/bin/python
import re

a="""SELECT 
transform_abc.ds AS "ds",
    SUM(transform_abc.dollars) AS "dollars",
    transform_abc.unit AS "unit"
FROM fct_table_abc transform_abc
WHERE
    (
        transform_abc.is_charged > 0
        OR transform_abc.account_status = 0
    )
    AND transform_abc.ds = '2020-02-20'
GROUP BY
    transform_abc.ds,
    transform_abc.unit"""


a = re.sub("SELECT[\s]+([a-zA-Z0-9.])+_([a-zA-Z0-9.])+.ds", "SELECT ", a)

a = re.sub("SELECT[\s]+[a-zA-Z]+[\s]+(\")ds(\",)","SELECT ",a)

a = res.sub("GROUP BY[\s]+([a-zA-Z0-9.])+_([a-zA-Z0-9.])+.ds,","GROUP BY ",a)

a 将包含最终的预期结果

谢谢 KM

Answer 2

这里有一个方法：

import re

a="""SELECT
    transform_abc.ds AS "ds",
    SUM(transform_abc.dollars) AS "dollars",
    transform_abc.unit AS "unit"
FROM fct_table_abc transform_abc
WHERE
    (
        transform_abc.is_charged > 0
        OR transform_abc.account_status = 0
    )
    AND transform_abc.ds = '2020-02-20'
GROUP BY
    transform_abc.ds,
    transform_abc.unit"""

res = re.sub(r'((?:SELECT|GROUP BY)\s+(?:(?!WHERE)[\s\S])*?)\s+[\w.]+\.ds.+', r'', a)
print res

输出：

SELECT

    SUM(transform_abc.dollars) AS "dollars",
    transform_abc.unit AS "unit"
FROM fct_table_abc transform_abc
WHERE
    (
        transform_abc.is_charged > 0
        OR transform_abc.account_status = 0
    )
    AND transform_abc.ds = '2020-02-20'
GROUP BY

    transform_abc.unit

Demo & explanation

如果 SQL 在一行，使用：

((?:SELECT|GROUP BY)\s+(?:(?!WHERE)[\s\S])*?)\s*[\w.]+\.ds[^,]*,

Demo & explanation

如何使用 python 从字符串中删除 .ds 模式字符串

How to remove .ds patterned string from a string using python

python

regex

regex-lookarounds

regexp-replace

a 将包含最终的预期结果

谢谢 KM