如何判断一个字符串是否包含POSIX中的另一个字符串？

Question

我想编写一个Unix shell脚本,如果另一个字符串中有一个字符串,它将执行各种逻辑.例如,如果我在某个文件夹中,请分支.有人可以告诉我如何做到这一点？如果可能的话,我想让这不是特定于shell的(即不仅仅是bash),但如果没有别的方法可以做到这一点.

#!/usr/bin/env sh

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

Answer 1

这是另一个解决方案.这使用POSIX子串参数扩展,所以它适用于bash,dash,ksh ......

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

编辑:这是一个好主意,C.罗斯.这是一个带有一些例子的功能化版本:

# 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"

Answer 2

纯POSIX外壳:

#!/bin/sh
CURRENT_DIR=`pwd`

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

像ksh或bash这样的扩展shell具有花哨的匹配机制,但旧式的case功能非常强大.

Answer 3

可悲的是,我不知道有什么方法可以做到这一点.但是,使用bash(从版本3.0.0开始,这可能就是你所拥有的),你可以像这样使用=〜运算符:

#!/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

作为额外的奖励(和/或警告,如果你的字符串中有任何有趣的字符),=〜如果你省略引号,则接受正则表达式作为正确的操作数.

Answer 4

#!/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

Answer 5

以下是您问题的各种解决方案的链接.

这是我的最爱,因为它具有最人性化的感觉:

星形通配符方法

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

Answer 6

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

Answer 7

有bash regexps.或者有'expr':

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