[英]Python regex search with greedy operator
I am trying to use regex to find an expression, and here is what i am trying to do 我正在尝试使用正则表达式来查找表达式,这就是我正在尝试做的
import re
h1= open("test.txt")
for ln in h1:
ln = ln.rstrip()
if re.search("^From:.+@", ln):
print(ln)
As per the logic it should stop the search at '@', however in my case that doesn't happen i get the following output: From: ray@media.berkeley.edu 按照逻辑,它应该在“ @”处停止搜索,但是在我没有发生的情况下,我得到以下输出:从:ray@media.berkeley.edu
I also tried to use a qualifier '?' 我也尝试使用限定词'?' but it doesn't seem to work.
但它似乎不起作用。 Can someone suggest why it isn't working or how can i make it work specifically using regex.
有人可以建议为什么它不起作用,或者我如何使用正则表达式使它特别起作用。
您正在打印整行,而可以执行以下操作:
print(ln.split('@')[0])
Regex solution: 正则表达式解决方案:
# regex = r"(.+?)@"
import re
h1= open("test.txt")
regex = r"(.+?)@"
for ln in h1:
ln = ln.rstrip()
matches = re.search(regex, ln)
if matches:
for groupNum in range(0, len(matches.groups())):
groupNum = groupNum + 1
print ("Group {groupNum} found at {start}-{end}: {group}".format(groupNum = groupNum, start = matches.start(groupNum), end = matches.end(groupNum), group = matches.group(groupNum)))
Other way by splitting and using startswith: 通过拆分和使用startswith的其他方法:
h1= open("test.txt")
for ln in h1:
if ln.startswith("From: "):
print(ln.split("@")[0])
# Output From: ray
If Regex
is the only thing which you are looking for, Then you can try something like this. 如果正则
Regex
是您唯一需要的东西,那么您可以尝试这样的事情。
Problems: 问题:
1. You should not print whole line.
1.您不应打印整行。
2. You should use
^From:[^@]+
to prevent matching of@
2.您应使用
^From:[^@]+
来防止匹配@
Try this code snippet here 在此处尝试此代码段
import re
h1= open("test.txt")
for ln in h1:
ln = ln.rstrip()
result = re.search("^From:[^@]+", ln)
if result is not None:
print(result.group(0))
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