【问题标题】：How to remove .ds patterned string from a string using python如何使用python从字符串中删除.ds模式字符串
【发布时间】：2020-08-15 14:20:57
【问题描述】：

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"""

我需要在 SELECT 和 GROUP BY 之后从这个字符串中删除带有 ds 的列，而不是在 WHERE 之后。

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 只是一个表名，它可以是任何表名。所以我们不能在正则表达式中使用它。不知道如何解决这个问题

【问题讨论】：

标签： python regex regex-lookarounds regexp-replace

【解决方案1】：

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'\1', 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 在一行上，请使用：

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

Demo & explanation

【讨论】：

可以格式化也可以不格式化。例如从 fct_table_abc transform_abc WHERE (transform_abc.is_charged > 0 OR transform_abc.account_status = 0) 和 transform_abc.ds = ' 2020-02-20' GROUP BY transform_abc.ds,ransform_abc.unit
@analyst：你的意思是单行吗？喜欢this

【解决方案2】：

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，但应该在逻辑上工作。

#!/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

感谢KM

【讨论】：