使用正则表达式从 URL 中提取文件名——需要排除一些字符

Question

I have a resource formatted like the following below:

{"url": "http://res1.icourses.cn/share/process17//mp4/2017/3/17/6332c641-28b5-43a0-894c-972bd804f4e1_SD.mp4", "name": "1-课程导学"}, 
{"url": "http://res2.icourses.cn/share/process17//mp4/2017/3/17/a21902b6-8680-4bdf-8f47-4f99d1354475_SD.mp4", "name": "2-计算机网络的定义与分类"}

I want to extract the file names 6332c641-28b5-43a0-894c-972bd804f4e1_SD.mp4 and a21902b6-8680-4bdf-8f47-4f99d1354475_SD.mp4 from the url.

How would I write a regular expression to match the string at this location?

Answer 1

Based on the strings you provided, you could iterate over the dictionaries, get value for "url" and use the following regex

([^\\/]*)$

Explanation:

() - defines capturing group
[^\/] - Match a single character not present after the ^
\/ - matches the character / literally (case sensitive)
* - Quantifier — Matches between zero and unlimited times, as many times as possible, giving back as needed (greedy)
$ - asserts position at the end of the string, or before the line terminator right at the end of the string (if any)

For example:

for record in records:
    print(re.search("([^\/]*)$", record['url']).group(1))

In this case, we are exploiting the fact that the filename occurs at the end of the string. Using the $ anchor makes the only valid match one that terminates the string.

If you wanted to do this to a dictionary cast as a string, you could by changing the ending condition. Like so ([^\\/]*?)\\", . Now ", terminates the match (note the \\ to escape the " . See https://regex101.com/r/k9VwC6/25

Finally, if we weren't so lucky that the capturing group was at the end of the string (meaning we couldn't use $ ) we could use a negative look behind. You can read up on those here .

Answer 2

You can use re.findall :

import re
s = [{"url": "http://res1.icourses.cn/share/process17//mp4/2017/3/17/6332c641-28b5-43a0-894c-972bd804f4e1_SD.mp4", "name": "1-课程导学"}, {"url": "http://res2.icourses.cn/share/process17//mp4/2017/3/17/a21902b6-8680-4bdf-8f47-4f99d1354475_SD.mp4", "name": "2-计算机网络的定义与分类"}]
filenames = [re.findall('(?<=/)[\w\-\_]+\.mp4', i['url'])[0] for i in s]

Output:

['6332c641-28b5-43a0-894c-972bd804f4e1_SD.mp4', 'a21902b6-8680-4bdf-8f47-4f99d1354475_SD.mp4']

Answer 3

You can use a short regex [^/]*$

Code:

import re
s = [{"url": "http://res1.icourses.cn/share/process17//mp4/2017/3/17/6332c641-28b5-43a0-894c-972bd804f4e1_SD.mp4", "name": "1-课程导学"}, {"url": "http://res2.icourses.cn/share/process17//mp4/2017/3/17/a21902b6-8680-4bdf-8f47-4f99d1354475_SD.mp4", "name": "2-计算机网络的定义与分类"}]
filenames = [re.findall('[^/]*$', i['url'])[0] for i in s]
print(filenames)`

Output:

['6332c641-28b5-43a0-894c-972bd804f4e1_SD.mp4', 'a21902b6-8680-4bdf-8f47-4f99d1354475_SD.mp4']

Check the regex - https://regex101.com/r/k9VwC6/30