Python使用re模块的findall方法获取所有匹配结果的操作指南

import re

result = re.findall(pattern, string, flags=0)

text = "我的电话是13812345678和15987654321，邮箱是abc@xyz.com"

# 匹配所有数字（连续的一串）
phone_numbers = re.findall(r'\d{11}', text)
print(phone_numbers)  # ['13812345678', '15987654321']

text = """
用户信息如下：
姓名：张三
电话：13912345678
邮箱：zhangsan@example.com
备用电话：15898765432
备用邮箱：lisi@company.org
"""

# 同时提取手机号和邮箱
pattern = r'(1[3-9]\d{9})|(?:\w+@\w+\.\w+)'
matches = re.findall(pattern, text)

print(matches)
# [('13912345678', ''), ('zhangsan@example.com', ''), ('15898765432', ''), ('lisi@company.org', '')]

pattern = r'(?P<phone>1[3-9]\d{9})|(?P<email>\w+@\w+\.\w+)'
results = []

for match in re.finditer(pattern, text):
    if match.group('phone'):
        results.append(f"电话: {match.group('phone')}")
    elif match.group('email'):
        results.append(f"邮箱: {match.group('email')}")

print(results)
# ['电话: 13912345678', '邮箱: zhangsan@example.com', '电话: 15898765432', '邮箱: lisi@company.org']

text = "今天天气不错，气温25度，明天预计28度，后天可能30度"

# 使用 search 只能拿到第一个
first_temp = re.search(r'\d+', text)
print(first_temp.group())  # '25'

# 而 findall 一次性拿走全部
all_temps = re.findall(r'\d+', text)
print(all_temps)  # ['25', '28', '30']

2024-04-05 14:23:12 [ERROR] Failed to connect to DB: Connection refused (code: 503)
2024-04-05 14:23:15 [WARN] Disk usage exceeds threshold (code: 400)
2024-04-05 14:23:18 [ERROR] Timeout occurred (code: 504)

log_data = """
2024-04-05 14:23:12 [ERROR] Failed to connect to DB: Connection refused (code: 503)
2024-04-05 14:23:15 [WARN] Disk usage exceeds threshold (code: 400)
2024-04-05 14:23:18 [ERROR] Timeout occurred (code: 504)
"""

error_codes = re.findall(r'code:\s*(\d+)', log_data)

print("提取到的错误码:", error_codes)
# ['503', '400', '504']

titles = [
    "2024年度技术峰会 - 构建智能未来",
    "2023年开发者大会 - 创新驱动增长",
    "2022全球云论坛 - 云端协作新范式"
]

# 匹配年份和后面的标题部分
pattern = r'(\d{4})\s*年?[\-\–]?\s*(.+?)$'

results = []
for title in titles:
    match = re.findall(pattern, title)
    if match:
        year, topic = match[0]
        results.append({"year": year, "topic": topic.strip()})

print(results)
# [
#   {'year': '2024', 'topic': '构建智能未来'},
#   {'year': '2023', 'topic': '创新驱动增长'},
#   {'year': '2022', 'topic': '云端协作新范式'}
# ]

multi_line_text = """
https://example.com/page1
http://test.org/api
ftp://files.net/data/
"""

# 默认模式下，. 无法跨越换行
urls = re.findall(r'^https?://[^\s]+', multi_line_text, flags=re.MULTILINE)
print(urls)
# ['https://example.com/page1', 'http://test.org/api']

13812345678
+8613812345678
(138)12345678
138-1234-5678

raw_phones = [
    "13812345678",
    "+8613812345678",
    "(138)12345678",
    "138-1234-5678"
]

# 提取所有数字
digits = []
for phone in raw_phones:
    nums = re.findall(r'\d+', phone)
    digits.extend(nums)

# 统一格式化
formatted = []
for num in digits:
    if len(num) == 11:
        formatted.append(f"+86-{num[:3]}-{num[3:7]}-{num[7:]}")
    else:
        print(f"无效号码: {num}")

print(formatted)
# ['+86-138-1234-5678', '+86-138-1234-5678', '+86-138-1234-5678', '+86-138-1234-5678']

text = "abc123def456"
result = re.findall(r'\d+', text)
print(result)  # ['123', '456']

for match in re.finditer(r'\d+', text):
    print(f"数字: {match.group()}, 位置: {match.start()}-{match.end()}")
# 数字: 123, 位置: 3-6
# 数字: 456, 位置: 9-12

# 错误做法
re.findall(r'.', "Hello!")  # 匹配所有字符，包括感叹号！
# 但你想匹配的是句号？
re.findall(r'\.', "Hello!")  # 正确：只匹配句号

re.findall(r'(a)(b)', "ab")  # [('a', 'b')]