我已经证明了这一点:
执行生产者(3)
生产者(0)
1名助理摄影师(12名)
我愿利用一个校车,对插头的第一部分进行校对,并删除“(南)”部分(括号前面的空间和括号中的括号/数)。 在使用I号区之后,我想把我的遗体等同于:“执行制作人”、“制作人”、“最助理摄像机”
如果你知道学习考试的任何资源也是巨大的。
除最后括号及其数字内容外,你只得选择所有特性:
(.+) (d+)
The first two parenthesis capture the content (here, all content, declared by the point). Then, you want two parenthesis (be careful to the slash), meaning we do not want these parenthesis to capture the "d+" expression, which is a number.
我的首选网站之一:。
Maybe s/([sw]+w)s*(d+)/1/?
I don t know Ruby, so you d have to translate it to its own regexp syntax.
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