My problem is that I want to check the browserstring with pure regex.
Mozilla/5.0 (Linux; U; Android 3.0; en-us; Xoom Build/HRI39) AppleWebKit/534.13 (KHTML, like Gecko) Version/4.0 Safari/534.13
1. 适当
Mozilla/5.0 (Linux; U; Android 2.2.1; en-us; Nexus One Build/FRG83) AppleWebKit/533.1 (KHTML, like Gecko) Version/4.0 Mobile Safari/533.1
不适用
My tried solution is: /?((?<=Android)(?:[^])*?(?=Mobile))/i
but it matches exactly wrong.
如果扼杀含有一个字句,请先检查。
如果你想保证,在某个地方,扼杀含有“roid”,你可以这样做:
^(?=.*Android).*
你们也可以把它们结合起来,确保其中某些地方包含“roid”和“Mobile”:
^(?=.*Android)(?=.*Mobile).*
如果你想确保,在扼杀中某个词是NOT,则使用前面的负面看法:
^(?=.*Android)(?!.*Mobile).*
This would require the word "Android to be in the string and the word "Mobile" is not allowed in the string. The .* part matches then the complete string/row when the assertions at the beginning are true.
See it here on Regexr
With some implementations of regular expressions, you can use a negative lookbehind assertion.
Per the docs, a negative lookbehind written as (?<!...) matches only if the current position in the string is not preceded by a match for ...
这里有一份互动文字,显示如何用你的抽样插图来进行负面研究:
>>> s = "Mozilla/5.0 (Linux; U; Android 3.0; en-us; Xoom Build/HRI39) AppleWebKit/534.13 (KHTML, like Gecko) Version/4.0 Safari/534.13"
>>> bool(re.search(r Android.*(?<! Mobile) Safari , s))
True
>>> t = "Mozilla/5.0 (Linux; U; Android 2.2.1; en-us; Nexus One Build/FRG83) AppleWebKit/533.1 (KHTML, like Gecko) Version/4.0 Mobile Safari/533.1"
>>> bool(re.search(r Android.*(?<! Mobile) Safari , t))
False
I d just break it up
if ((m/Android/i) && (m/Safari/i) && !(m/Mobile Safari/i))
既然这样的话,根据斯拉夫,你可以把这一点结合起来。
if ((m/Android/i) && (m/(?<!Mobile )Safari/i))
or even
if (m/Android.*(?<!Mobile )Safari/i)
FYI see Lookahead/lookbehind
Update Tested these fine with Perl5 regex flavour (arguably the most popular flavour):
perl -ne print "$. yes
" if m/Android.*(?<!Mobile )Safari/i
表演:
1 yes
附属履行机构
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