Python regex匹配API路径

Question

我需要一个正则表达式匹配api路径与以下规则

1. 路径以“/”开头，但可以有多个“/”

2. “/”后面必须跟一个以小写开头的单词，但以后可以是大写

   x = ["/word/worD/sdfsfsd","/fsdfsdf","/","/{sfsdf",'/','/_','/{','/{"type":"tnt"}',"/allGear{Exe","/Grear"]
   
   import re
   
   pattern = re.compile("(/[a-z][a-zA-Z]+)+")
   
   for y in x:
        print (pattern.match(y))
   

因此在本例中，只有前两个元素必须生成匹配。你知道吗

Answer 1

使用以下方法：

import re
x = ["/word/worD/sdfsfsd","/fsdfsdf","/","/{sfsdf",'/','/_','/{','/{"type":"tnt"}',"/allGear{Exe","/Grear"]
result = [p for p in x if re.match(r'^(\/[a-z][a-zA-z]+)+$', p)]

print(result)

输出：

['/word/worD/sdfsfsd', '/fsdfsdf']

Answer 2

我想这应该适合你。但是，正如在评论中所说的，我认为根据你的描述/allGear{Exe也应该包括在内。我给出的代码也返回了这个结果

 x = ["/word/worD/sdfsfsd","/fsdfsdf","/","/{sfsdf",'/','/_','/{','/{"type":"tnt"}',"/allGear{Exe","/Grear"]


import re

for i in x:
    pattern = re.search("""\/[a-z][a-zA-Z]+""", i, re.S)
#If you don't want the /allGear, change the regex to """\/[a-z][a-zA-Z]+$""";
    if(pattern is not None):
        print i