我要评论正则写复杂了很麻烦，难写难调试，只需要两个函数，就能用简单正则组合构建复杂正则：
比如输入一个字符串规则，可以使用 {name} 引用前面定义的规则：
# rules definition
rules = r'''
protocol = http|https
login_name = [^:@\r\n\t ]+
login_pass = [^@\r\n\t ]+
login = {login_name}(:{login_pass})?
host = [^:/@\r\n\t ]+
port = \d+
optional_port = (?:[:]{port})?
path = /[^\r\n\t ]*
url = {protocol}://({login}[@])?{host}{optional_port}{path}?
'''
然后调用 regex_build 函数，将上面的规则转换成一个字典并输出：
# expand patterns in a dictionary
m = regex_build(rules, capture = true)
# list generated patterns
for k, v in m.items():
print(k, '=', v)
结果：
protocol = (?p<protocol>http|https)
login_name = (?p<login_name>[^:@\r\n\t ]+)
login_pass = (?p<login_pass>[^@\r\n\t ]+)
login = (?p<login>(?p<login_name>[^:@\r\n\t ]+)(:(?p<login_pass>[^@\r\n\t ]+))?)
host = (?p<host>[^:/@\r\n\t ]+)
port = (?p<port>\d+)
optional_port = (?p<optional_port>(?:[:](?p<port>\d+))?)
path = (?p<path>/[^\r\n\t ]*)
url = (?p<url>(?p<protocol>http|https)://((?p<login>(?p<login_name>[^:@\r\n\t ]+)(:(?p<login_pass>[^@\r\n\t ]+))?)[@])?(?p<host>[^:/@\r\n\t ]+)(?p<optional_port>(?:[:](?p<port>\d+))?)(?p<path>/[^\r\n\t ]*)?)
用手写直接写是很难写出这么复杂的正则的，写出来也很难调试，而组合方式构建正则的话，可以将小的简单正则提前测试好，要用的时候再组装起来，就不容易出错，上面就是组装替换后的结果。
下面用里面的 url 这个规则来匹配一下：
# 使用规则 "url" 进行匹配
pattern = m['url']
s = re.match(pattern, 'https://name:pass@www.baidu.com:8080/haha')
# 打印完整匹配结果
print('matched: "%s"'%s.group(0))
print()
# 打印分组匹配结果
for name in ('url', 'login_name', 'login_pass', 'host', 'port', 'path'):
print('subgroup:', name, '=', s.group(name))
输出：
match text with pattern "url"
matched: "https://name:pass@www.baidu.com:8080/haha"
subgroup: url = https://name:pass@www.baidu.com:8080/haha
subgroup: login_name = name
subgroup: login_pass = pass
subgroup: host = www.baidu.com
subgroup: port = 8080
subgroup: path = /haha
可以取完整结果，也可以按照规则名字，取得里面具体某个部件得匹配结果。
这下可以方便的写复杂正则表达式了。
再 python 的正则表达式里 {xxx} 是用来表示长度的，里面都是数字，如果里面是变量名的话不会和原有规则冲突，因此这个写法是安全的。
实现代码：
import re
# 将 pattern 里形如 {name} 的文本，用 macros 里的预定义规则替换
def regex_expand(macros, pattern, guarded = true):
output = []
pos = 0
size = len(pattern)
while pos < size:
ch = pattern[pos]
if ch == '\\':
output.append(pattern[pos:pos + 2])
pos += 2
continue
elif ch != '{':
output.append(ch)
pos += 1
continue
p2 = pattern.find('}', pos)
if p2 < 0:
output.append(ch)
pos += 1
continue
p3 = p2 + 1
name = pattern[pos + 1:p2].strip('\r\n\t ')
if name == '':
output.append(pattern[pos:p3])
pos = p3
continue
elif name[0].isdigit():
output.append(pattern[pos:p3])
pos = p3
continue
elif ('<' in name) or ('>' in name):
raise valueerror('invalid pattern name "%s"'%name)
if name not in macros:
raise valueerror('{%s} is undefined'%name)
if guarded:
output.append('(?:' + macros[name] + ')')
else:
output.append(macros[name])
pos = p3
return ''.join(output)
# 给定规则文本，构建规则字典
def regex_build(code, macros = none, capture = true):
defined = {}
if macros is not none:
for k, v in macros.items():
defined[k] = v
line_num = 0
for line in code.split('\n'):
line_num += 1
line = line.strip('\r\n\t ')
if (not line) or line.startswith('#'):
continue
pos = line.find('=')
if pos < 0:
raise valueerror('%d: not a valid rule'%line_num)
head = line[:pos].strip('\r\n\t ')
body = line[pos + 1:].strip('\r\n\t ')
if (not head):
raise valueerror('%d: empty rule name'%line_num)
elif head[0].isdigit():
raise valueerror('%d: invalid rule name "%s"'%(line_num, head))
elif ('<' in head) or ('>' in head):
raise valueerror('%d: invalid rule name "%s"'%(line_num, head))
try:
pattern = regex_expand(defined, body, guarded = not capture)
except valueerror as e:
raise valueerror('%d: %s'%(line_num, str(e)))
try:
re.compile(pattern)
except re.error:
raise valueerror('%d: invalid pattern "%s"'%(line_num, pattern))
if not capture:
defined[head] = pattern
else:
defined[head] = '(?p<%s>%s)'%(head, pattern)
return defined
# 定义一套组合规则
rules = r'''
protocol = http|https
login_name = [^:@\r\n\t ]+
login_pass = [^@\r\n\t ]+
login = {login_name}(:{login_pass})?
host = [^:/@\r\n\t ]+
port = \d+
optional_port = (?:[:]{port})?
path = /[^\r\n\t ]*
url = {protocol}://({login}[@])?{host}{optional_port}{path}?
'''
# 将上面的规则展开成字典
m = regex_build(rules, capture = true)
# 输出字典内容
for k, v in m.items():
print(k, '=', v)
print()
# 用最终规则 "url" 匹配文本
pattern = m['url']
s = re.match(pattern, 'https://name:pass@www.baidu.com:8080/haha')
# 打印完整匹配
print('matched: "%s"'%s.group(0))
print()
# 按名字打印分组匹配
for name in ('url', 'login_name', 'login_pass', 'host', 'port', 'path'):
print('subgroup:', name, '=', s.group(name))
完事，主要逻辑 84 行代码。
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