在python中使用正则表达式排除某些字符串

Question

我想对下面的代码应用正则表达式，这样我就可以删除出现在逗号和单词'AS'之间的任何字符串。

Select customer_name, customer_type, COUNT(*) AS volumenFROM tablenGROUP BY customer_name, customer_typenORDER BY volume DESCnLIMIT 10

预期输出:

Select customer_name, customer_type, volumenFROM tablenGROUP BY customer_name, customer_typenORDER BY volume DESCnLIMIT 10

我尝试了下面的命令，但是没有得到想要的输出

result = re.sub(r",s*COUNT(*)s*ASs*w+", "", text)

Answer 1

您可以使用捕获组并在替换中使用该组。

(,s*)[^,]*sASbs*

• (,s*)捕获组1，匹配逗号和可选的空白字符
• [^,]*匹配除逗号以外的任何字符
• sASbs*匹配一个空白字符，然后是AS，后跟可选的空格

Regex demo | Python demo

import re

pattern = r"(,s*)[^,]*sASbs*"
s = ("Select customer_name, customer_type, COUNT(*) AS volume\nFROM table\nGROUP BY customer_name, customer_type\nORDER BY volume DESC\nLIMIT 10n")

print(re.sub(pattern, r"1", s))

输出

Select customer_name, customer_type, volumenFROM tablenGROUP BY customer_name, customer_typenORDER BY volume DESCnLIMIT 10

Answer 2

我将使用:

text = "Select customer_name, customer_type, COUNT(*) AS volumenFROM tablenGROUP BY customer_name, customer_typenORDER BY volume DESCnLIMIT 10"
result = re.sub(r',s*S+s+ASbs*', ', ', text)
print(result)

这个打印:

Select customer_name, customer_type, volume
FROM table
GROUP BY customer_name, customer_type
ORDER BY volume DESC
LIMIT 10

这里使用的正则表达式模式表示匹配:

• ,a逗号
• s*可选空白
• S+a非空白项
• s+一个或多个空白字符
• ASliteral "AS">
• b字边界
• s*更多可选空格