[英]PHP Get version from url
How can I download the versions from the link?如何从链接下载版本? I tried to do it with regEx but it didn't work as I wanted.
我试图用 regEx 来做,但它没有按我的意愿工作。
I wish it would work like this (Where I gave NULL, there is nothing to choose from)我希望它能像这样工作(我给了 NULL,没有什么可以选择的)
https://ajax.googleapis.com/ajax/libs/jquery/1.11.2/jquery.min.js > 1.11.2
https://ajax.googleapis.com/ajax/libs/jquery/v1.11.2/jquery.min.js > 1.11.2
https://ajax.googleapis.com/ajax/libs/jquery/version-1.11.2/jquery.min.js > 1.11.2
https://ajax.googleapis.com/ajax/libs/jquery/1-11-2/jquery.min.js > NULL
https://ajax.googleapis.com/ajax/libs/jquery/1112/jquery.min.js > NULL
https://www.google.com/recaptcha/api.js?render=6LdBsjgVaoTJ8rC-Npzz16bnAE > NULL
My regEx: /([0-9]+\.?)+/
我的正则表达式:/(
/([0-9]+\.?)+/
.?)+/
Thank you in advance for your help:)预先感谢您的帮助:)
You can use您可以使用
\/(?:v(?:ersion)?-?)?\K\d+(?:\.\d+)+
See the regex demo .请参阅正则表达式演示。 NOTE: if you want to make sure after the version number there is a
/
or end of string, add a (?![^\/])
lookahead at the end of the pattern, \/(?:v(?:ersion)?-?)?\K\d+(?:\.\d+)+(?![^\/])
(see this regex demo ).注意:如果要确保版本号后面有
/
或字符串结尾,请在模式末尾添加(?![^\/])
前瞻, \/(?:v(?:ersion)?-?)?\K\d+(?:\.\d+)+(?![^\/])
(参见这个正则表达式演示)。
Details :详情:
\/
- a /
char \/
- 一个/
字符(?:v(?:ersion)?-?)?
- an optional sequence of a v
and then an optional ersion
and then an optional -
char - \K
- omit the matched text v
的可选序列,然后是一个可选的ersion
,然后是一个可选的-
char - \K
- 省略匹配的文本\d+(?:\.\d+)+
- match and consumer one or more digits and then one or more sequences of a dot and one or more digits. \d+(?:\.\d+)+
- 匹配并使用一个或多个数字,然后是一个或多个点序列和一个或多个数字。
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