一文详解Python的re.match使用方法和技巧

import re
result = re.match(pattern, string, flags=0)

text = "2023-05-15生日"
match = re.match(r"(\d{4})-(\d{2})-(\d{2})", text)
if match:
    print(match.group(0))  # 输出：2023-05-15
    print(match.group(1))  # 输出：2023（年份）

text = "Hello 2023 World"
print(re.match(r"\d{4}", text))   # None（开头无数字）
print(re.search(r"\d{4}", text))  # <re.Match object; span=(6, 10), match='2023'>

pattern = r"(\d{4})-(\d{2})-(\d{2})"
text = "2025-08-15"
match = re.match(pattern, text)
print(match.group(1))  # 2025（年）
print(match.group(2))  # 08（月）

pattern = r"(?P<year>\d{4})-(?P<month>\d{2})-(?P<day>\d{2})"
match = re.match(pattern, "2025-08-15")
print(match.group("month"))  # 08

text = "Python3"
match = re.match(r"\d+", text)  # 实际匹配到"3"，但文本开头无数字
if not match:
    raise ValueError("匹配失败")  # 正确处理方式

text = "2023-05-15"
new_text = re.sub(r"(\d{4})-(\d{2})-(\d{2})", r"\2/\3/\1", text)
print(new_text)  # 05/15/2023

log_line = '192.168.1.10 - - [10/Mar/2024:12:34:56 +0000] "GET /index.html HTTP/1.1" 200 1024'
pattern = r"(\d+\.\d+\.\d+\.\d+).*?\[(.*?)\]"
match = re.search(pattern, log_line)
if match:
    print("IP:", match.group(1))      # 192.168.1.10
    print("Timestamp:", match.group(2)) # 10/Mar/2024:12:34:56 +0000

def validate_password(password):
    pattern = r"^(?=.*[a-z])(?=.*[A-Z])(?=.*\d).{8,}$"
    return bool(re.match(pattern, password))

print(validate_password("Pass123"))  # True
print(validate_password("weak"))     # False

text = "价格：100元"
new_text = re.sub(r"(\d+)", lambda m: str(int(m.group(1))*1.1), text)
print(new_text)  # 价格：110.0元

text = "apple, banana; cherry|date"
parts = re.split(r"[,;|]\s*", text)
print(parts)  # ['apple', 'banana', 'cherry', 'date']