POSIX sh에 문자열에 다른 문자열이 포함되어 있는지 어떻게 알 수 있습니까?

POSIX sh에 문자열에 다른 문자열이 포함되어 있는지 어떻게 알 수 있습니까? [복제]

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이 질문에는 이미 답변이 있습니다.

• Bash 22 답변에 문자열에 하위 문자열이 포함되어 있는지 확인하는 방법

다른 문자열 안에 문자열이 있으면 다양한 논리를 수행하는 Unix 쉘 스크립트를 작성하고 싶습니다. 예를 들어, 특정 폴더에있는 경우 분기합니다. 누군가 이것을 달성하는 방법을 말해 줄 수 있습니까? 가능하다면 나는 이것을 쉘 특정이 아닌 것으로 만들고 싶습니다 (즉, bash만이 아닙니다). 다른 방법이 없다면 그렇게 할 수 있습니다.

#!/usr/bin/env sh

if [ "$PWD" contains "String1" ]
then
    echo "String1 present"
elif [ "$PWD" contains "String2" ]
then
    echo "String2 present"
else
    echo "Else"
fi

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여기 또 다른 해결책이 있습니다. 이 사용 POSIX 하위 문자열 매개 변수 확장을 가 bash는, 대시, KSH에서 작동, 그래서 ...

test "${string#*$word}" != "$string" && echo "$word found in $string"

편집 : 좋은 생각입니다, C. Ross. 다음은 몇 가지 예가 포함 된 기능화 된 버전입니다.

# contains(string, substring)
#
# Returns 0 if the specified string contains the specified substring,
# otherwise returns 1.
contains() {
    string="$1"
    substring="$2"
    if test "${string#*$substring}" != "$string"
    then
        return 0    # $substring is in $string
    else
        return 1    # $substring is not in $string
    fi
}

contains "abcd" "e" || echo "abcd does not contain e"
contains "abcd" "ab" && echo "abcd contains ab"
contains "abcd" "bc" && echo "abcd contains bc"
contains "abcd" "cd" && echo "abcd contains cd"
contains "abcd" "abcd" && echo "abcd contains abcd"
contains "" "" && echo "empty string contains empty string"
contains "a" "" && echo "a contains empty string"
contains "" "a" || echo "empty string does not contain a"
contains "abcd efgh" "cd ef" && echo "abcd efgh contains cd ef"
contains "abcd efgh" " " && echo "abcd efgh contains a space"

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순수한 POSIX 쉘 :

#!/bin/sh
CURRENT_DIR=`pwd`

case "$CURRENT_DIR" in
  *String1*) echo "String1 present" ;;
  *String2*) echo "String2 present" ;;
  *)         echo "else" ;;
esac

Extended shells like ksh or bash have fancy matching mechanisms, but the old-style case is surprisingly powerful.

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Sadly, I am not aware of a way to do this in sh. However, using bash (starting in version 3.0.0, which is probably what you have), you can use the =~ operator like this:

#!/bin/bash
CURRENT_DIR=`pwd`

if [[ "$CURRENT_DIR" =~ "String1" ]]
then
 echo "String1 present"
elif [[ "$CURRENT_DIR" =~ "String2" ]]
then
 echo "String2 present"
else
 echo "Else"
fi

As an added bonus (and/or a warning, if your strings have any funny characters in them), =~ accepts regexes as the right operand if you leave out the quotes.

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#!/usr/bin/env sh

# Searches a subset string in a string:
# 1st arg:reference string
# 2nd arg:subset string to be matched

if echo "$1" | grep -q "$2"
then
    echo "$2 is in $1"
else 
    echo "$2 is not in $1"
fi

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Here is a link to various solutions of your issue.

This is my favorite as it makes the most human readable sense:

The Star Wildcard Method

if [[ "$string" == *"$substring"* ]]; then
    return 1
fi
return 0

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case $(pwd) in
  *path) echo "ends with path";;
  path*) echo "starts with path";;
  *path*) echo "contains path";;
  *) echo "this is the default";;
esac

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There's bash regexps. Or there's 'expr':

 if expr "$link" : '/.*' > /dev/null; then
    PRG="$link"
  else
    PRG=`dirname "$PRG"`/"$link"
  fi

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See the manpage for the 'test' program. If you're just testing for the existence of a directory you would normally do something like so:

if test -d "String1"; then
  echo "String1 present"
end

If you're actually trying to match a string you can use bash expansion rules & wildcards as well:

if test -d "String*"; then
  echo "A directory starting with 'String' is present"
end

If you need to do something more complex you'll need to use another program like expr.

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In special cases where you want to find whether a word is contained in a long text, you can iterate through the long text with a loop.

found=F
query_word=this
long_string="many many words in this text"
for w in $long_string; do
    if [ "$w" = "$query_word" ]; then
          found=T
          break
    fi
done

This is pure Bourne shell.

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test $(echo "stringcontain" "ingcon" |awk '{ print index($1, $2) }') -gt 0 && echo "String 1 contain string 2"

--> output: String 1 contain string 2

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If you want a ksh only method that is as fast as "test", you can do something like:

contains() # haystack needle
{
    haystack=${1/$2/}
    if [ ${#haystack} -ne ${#1} ] ; then
        return 1
    fi
    return 0
}

It works by deleting the needle in the haystack and then comparing the string length of old and new haystacks.

참고URL : https://stackoverflow.com/questions/2829613/how-do-you-tell-if-a-string-contains-another-string-in-posix-sh