找出顺序有元音的单词的数量 [英] find the number of words that have vowels in order

问题描述

谁能告诉我如何从包含a，e，i，o，u的特定目录中查找所有单词.

太过分了在这里，"a"之后是"a"，然后是"i"，然后是"o"，然后是"u".元音永不重复.

我有猫/home/user1/tmp/* | grep [aeiou]

但这并没有显示我的预期答案.

请帮忙.

在此先感谢您.

Hi,

Can any one please tell me how to find all words from a particular directory which contains a,e,i,o,u in order.

Ex abstancious. Here ''a'' followed by ''a'' followed by ''i'' followed by ''o'' followed by ''u''. Vowels never repeated.

I have cat /home/user1/tmp/* | grep [aeiou]

But it''s not displaying my expected answer.

Please help.

Thanks in advance.

推荐答案

您的正则表达式只是在寻找任何单个元音.您想要更多类似"^ [^ aeiou] * a [^ aeiou] * e [^ aeiou] * i [^ aeiou] * o [^ aeiou] * u [^ aeiou] *

Your regular expression is just looking for any single vowel. You want something more like "^[^aeiou]*a[^aeiou]*e[^aeiou]*i[^aeiou]*o[^aeiou]*u[^aeiou]*

"我想(假设它必须全部包含).这个搜索零个或多个辅音，a，零个或多个辅音，等等.

" I think (assuming it must have all of them). This one searches for zero or more consonants, a, zero or more consonants, e, etc.

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