正则表达式从括号中获取 id

Question

I was fetching data from Microsoft regarding their plans, here's the webpage for reference of data

https://docs.microsoft.com/en-us/azure/active-directory/enterprise-users/licensing-service-plan-reference#feedback

I'm working on table data to fetch the respective products with their guid initially for the first column it was easy but for the last, they were just using break tags in it. Here's my code for it.

import requests
from requests.api import head
from bs4 import BeautifulSoup
import pandas as pd
import json
import re

url = "https://docs.microsoft.com/en-us/azure/active-directory/enterprise-users/licensing-service-plan-reference#feedback"

payload = {}
headers = {}

response = requests.request("GET", url, headers=headers, data=payload)
soup = BeautifulSoup(response.content, 'lxml')
table = soup.find( "table" )
df = pd.read_html(str(table))[0]
df = df.drop(labels=['Service plans included'],
  axis='columns')
json_dict = json.loads(df.to_json(orient='records'))
regex = r"([A-Z ]+ \(.*?\))"
microsoft_processed_data = []
for item in json_dict:
    plan_data = item["Service plans included (friendly names)"]
    matches = re.findall(regex, plan_data)
    dict = {}
    for match in matches:
        dict_key = match.split("(", )[1]
        dict_key = dict_key.replace(")", "")
        dict_value = match.split(" (")[0]
        print(dict_key + " : " + dict_value)
        dict[dict_key] = dict_value
    item["Service plans included (friendly names)"] = dict
    microsoft_processed_data.append(item)

with open('data.json', 'w') as f:
    json.dump(microsoft_processed_data, f, indent = 4)

It worked till they started using brackets in their plan names as well and my regex failed which was working in a group.

If we consider this sample row out of all

EXCHANGE ONLINE (PLAN 1) (9aaf7827-d63c-4b61-89c3-182f06f82e5c)

then as per my regex, it was picking up text starting from the beginning to the end of closed brackets.

so, my regex my picked up data till > EXCHANGE ONLINE (PLAN 1)

But I'm looking to get the data till the guid id of it and then separate the name for dictionary.

Here's my sample Expected Dictionary

{
    "EXCHANGE ONLINE (PLAN 1)" : "9aaf7827-d63c-4b61-89c3-182f06f82e5c"
}

Answer 1

Try: regex = r"([AZ ]+ \\(.*?\\))\\s*\\((.*?)\\)" , It will give tuple of the values you are looking for.

Let's focus on \\s*\\((.*?)\\) part only

• \\s* will match any number of white space characters
• Then \\((.*?)\\) will take any thing between the parenthesis.

re.findall(regex, text)
[('EXCHANGE ONLINE (PLAN 1)', '9aaf7827-d63c-4b61-89c3-182f06f82e5c')]

You can directly pass it to the dict if you want to create a dictionary out of it:

dict(re.findall(regex, text))
{'EXCHANGE ONLINE (PLAN 1)': '9aaf7827-d63c-4b61-89c3-182f06f82e5c'}

Answer 2

([AZ,0-9,\\-,\\(,\\),\\s\\+\\s*\\(.*?\\))

如果有人在 Microsoft Office 365 许可证的未来查看数据集，然后在您的爬虫中使用它来获取每种许可证类型的所有子许可证，则此正则表达式非常完美。