Bash显示charcaters如果不是在字符串 [英] Bash show charcaters if not in string

问题描述

一切正常，但我不明白如何做一件事：

我向用户显示了带有猜测字母的单词（例如世界是hello world，用户猜到了'l' ll * *** l *）

我存储用户已经尝试过的字母var guess

我这样做与以下：

  echo$ {word // [^ [：space：] $ guess] / * }
  

我现在要做的事情是回声字母表，用户已经尝试了，所以在这种情况下显示没有L的完整字母表。

我已经尝试以和我刚才显示的相同的方式做，

如果您需要更多信息，请告诉我。

谢谢，

解决方案

您不会显示您尝试的内容，但参数扩展可以正常工作。

  $ alphabet = abcdefghijklmnopqrstuvwxyz 
 $ word =hello world
 $ guesses = aetl 
 $ echo$ {word / / [^ [：space：] $ guesses] / *}
 * el * *** l * 
 $ echo$ {alphabet // [$ guesses] / *}
 * bcd * fghijk * mnopqrs * uvwxyz 
  

I am trying out bash, and I am trying to make a simple hangman game now.

Everything is working but I don't understand how to do one thing:

I am showing the user the word with guessed letters (so for example is the world is hello world, and the user guessed the 'l' I show them **ll* ***l* )

I store the letters that the user already tried in var guess

I do that with the following:

echo "${word//[^[:space:]$guess]/*}"

The thing I want to do now is echo the alphabet, but leave out the letters that the user already tried, so in this case show the full alphabet without the L.

I already tried to do it the same way as I shown just yet, but it won't quite work.

If you need any more info please let me know.

Thanks, Tim

解决方案

You don't show what you tried, but parameter expansion works fine.

$ alphabet=abcdefghijklmnopqrstuvwxyz
$ word="hello world"
$ guesses=aetl
$ echo "${word//[^[:space:]$guesses]/*}"
*ell* ***l*
$ echo "${alphabet//[$guesses]/*}"
*bcd*fghijk*mnopqrs*uvwxyz

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