[英]Python regex library can't match even though it should
My script: 我的剧本:
#!/usr/bin/env python
import os
import re
def grep(filepath, regex):
regObj = re.compile(regex)
res = []
with open(filepath) as f:
for line in f:
if regObj.match(line):
res.append(line)
return res
print(grep('/opt/conf/streaming.cfg', 'Port='))
Supposed to loop through the lines in the file given and match the regex provided, if exists, append to res
and eventually return res
. 假定遍历给定文件中的各行并匹配提供的regex(如果存在),追加到
res
并最终返回res
。
The content of /opt/conf/streaming.cfg
contains a line: /opt/conf/streaming.cfg
的内容包含一行:
SocketAcceptPort=8003
Still prints []
仍然打印
[]
How come? 怎么会?
Checking the docs for re.match
gives us this first sentence: 检查文档是否存在
re.match
,这是我们的第一句话:
If zero or more characters at the beginning of string match
如果在字符串开头匹配零个或多个字符
Notice the part about "beginning of string"? 注意“字符串开始”部分吗? You need to use a different re function to match anywhere in the line.
您需要使用其他功能来匹配行中的任何地方。 For example, further down in the docs for
match
is this note: 例如,以下文档中的
match
为:
If you want to locate a match anywhere in string, use search() instead
如果要在字符串中的任意位置找到匹配项,请使用search()代替
If you're looking for a list of ports, could you not use a string comparison instead: 如果要查找端口列表,则可以不使用字符串比较:
#!/usr/bin/env python
import os
import re
def grep(filepath, substring):
res = []
with open(filepath) as f:
for line in f:
if substring in line:
res.append(line.rsplit("\n")[0])
return res
print(grep('/opt/conf/streaming.cfg', 'Port='))
giving the result: 给出结果:
['SocketAcceptPort=8003']
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