49 lines
1.3 KiB
Python
49 lines
1.3 KiB
Python
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import re
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if __name__ == '__main__':
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# 如果只是简单的文字匹配或者查找，下面三个方法足以解决问题：
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url = "http://www.baidu.com"
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url.startswith("http")
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url.endswith(".com")
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url.find("baidu")
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# 但如果是更加复杂的匹配，就要用到re库的正则了
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text1 = '11/27/2012'
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text2 = 'Nov 27, 2012'
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if re.match(r'\d+/\d+/\d+', text1):
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print("yes")
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else:
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print("no")
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if re.match(r'\d+/\d+/\d+', text2):
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print("yes")
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else:
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print("no")
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# match可以被一次性消费，但是如果想要多次匹配，就要先把正则编译
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datepat = re.compile(r'\d+/\d+/\d+')
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if datepat.match(text1):
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print("yes")
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else:
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print("no")
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if datepat.match(text2):
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print("yes")
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else:
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print("no")
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# 这里要注意的是，match方法是从头匹配，如果要匹配的内容在一堆垃圾里面，请使用findall
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# 我们还会使用捕获组，这样可以把每个组单独提取出来
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datepat = re.compile(r'(\d+)/(\d+)/(\d+)')
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m = datepat.match(text1)
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print(m.group(0))
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print(m.group(1))
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print(m.group(2))
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print(m.group(3))
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# match只能匹配开头，它不管结尾，如果想要精确匹配需要加休止符$
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datepat = re.compile(r'(\d+)/(\d+)/(\d+)$')
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