Bash shell中测试，如果在一个字符串中的所有字符都在另一个字符串 [英] Bash shell test if all characters in one string are in another string

问题描述

我有两个字符串，我想比较平等的字符，字符串必须包含确切的字符，但是可以mychars有多余的字符。

I have two strings which I want to compare for equal chars, the strings must contain the exact chars but mychars can have extra chars.

mychars="abcdefg"
testone="abcdefgh"        # false h is not in mychars
testtwo="abcddabc"        # true all char in testtwo are in mychars

function test() {
    if each char in $1 is in $2  # PSEUDO CODE
    then
      return 1
    else
      return 0
    fi
}

if test $testone $mychars; then
   echo "All in the string" ;
else ;  echo "Not all in the string" ; fi

# should echo "Not all in the string" because the h is not in the string mychars

if test $testtwo $mychars; then
   echo "All in the string" ;
else ;  echo "Not all in the string" ; fi

# should echo 'All in the string'

什么是做到这一点的最好方法是什么？我的猜测是遍历第一个参数的所有字符。

What is the best way to do this? My guess is to loop over all the chars in the first parameter.

推荐答案

您可以使用 TR 从 mychars 用一个符号，那么你就可以测试，如果生成的字符串是从符号有什么不同，PE：

You can use tr to replace any char from mychars with a symbol, then you can test if the resulting string is any different from the symbol, p.e.,:

tr -s "[$mychars]" "." <<< "ggaaabbbcdefg"

输出：

.

不过：<​​/ P>

But:

tr -s "[$mychars]" "." <<< "xxxggaaabbbcdefgxxx"

打印：

xxx.xxx

所以，你的函数可能是这样的：

So, your function could be like the following:

function test() {
    local dictionary="$1"
    local res=$(tr -s "[$dictionary]" "." <<< "$2")
    if [ "$res" == "." ]; then 
        return 1
    else
        return 0
    fi
}

---

更新：：由 @ mklement0 建议，全功能可以缩短（和由以下逻辑固定）：

---

Update: As suggested by @mklement0, the whole function could be shortened (and the logic fixed) by the following:

function test() {
    local dictionary="$1"
    [[ '.' == $(tr -s "[$dictionary]" "." <<< "$2") ]] 
}

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