Python：如何只打印以特定字符开头的特定字符串？

Question

我有一个包含 url 列表的文本文件，如下所示：

https://www.ebay.com/itm/Egyptian-Comfort-1800-Count-4-Piece-Bed-Sheet-Set-Deep-Pocket-Bed-Sheets/142436469971?epid=1760442729&hash=item2129e00cd3%3Ag%3A7gIAAOSw3YBdRVJd&_trkparms=%2526rpp_cid%253D601435485fceeb223c6f4511&var=442541824291

这里我只想在读取文本文件时打印epid=1760442729。

我试过：

result = []`
with open('deals.txt', 'r') as f:
for line in f:
    if line.startswith('?epid='):
        break
    result.append(line)
print(result[0].split('epid='))

但是我没有得到预期的结果。

任何帮助或建议都会对我有所帮助。提前致谢

Answer 1

import re

s = """https://www.ebay.com/itm/Egyptian-Comfort-1800-Count-4-Piece-Bed-Sheet-Set-Deep-Pocket-Bed-Sheets/142436469971?epid=1760442729&hash=item2129e00cd3%3Ag%3A7gIAAOSw3YBdRVJd&_trkparms=%2526rpp_cid%253D601435485fceeb223c6f4511&var=442541824291
https://www.ebay.com/itm/Egyptian-Comfort-1800-Count-4-Piece-Bed-Sheet-Set-Deep-Pocket-Bed-Sheets/142436469971?epid=172442729&hash=item2129e00cd3%3Ag%3A7gIAAOSw3YBdRVJd&_trkparms=%2526rpp_cid%253D601435485fceeb223c6f4511&var=442541824291"""

for i in re.findall(r'epid=(\d+)&', s, re.MULTILINE):
    print(f'epid = {i}')

epid = 1760442729
epid = 172442729

Answer 2

如果总是使用相同的结构，我会用子串来做

result = []
with open('deals.txt', 'r') as f:
for line in f:
    a= line.find('epid=')
    b= line.find('&hash=')
    print(line[a:b])
    result.append(line)

Answer 3

使用设计用于解析 URL 的库。

示例：

from urllib.parse import urlparse, parse_qs

URL='https://www.ebay.com/itm/Egyptian-Comfort-1800-Count-4-Piece-Bed-Sheet-Set-Deep-Pocket-Bed-Sheets/142436469971?epid=1760442729&hash=item2129e00cd3%3Ag%3A7gIAAOSw3YBdRVJd&_trkparms=%2526rpp_cid%253D601435485fceeb223c6f4511&var=442541824291'

url_component = urlparse(URL)
query_component = parse_qs(url_component.query)
epid_data = query_component['epid'][0]
print(f'epid = {epid_data}')

Python：如何只打印以特定字符开头的特定字符串？

Python: How i Can print only a particular string starting with a specific character?

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