我想写一个Unix shell脚本,如果在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,则将其分支。有人能告诉我怎么做到这一点吗?如果可能的话,我希望这不是特定于shell的(例如,不只是bash ),但如果没有其他方法,我可以凑合使用。
#!/usr/bin/env sh
if [ "$PWD" contains "String1" ]
then
echo "String1 present"
elif [ "$PWD" contains "String2" ]
then
echo "String2 present"
else
echo "Else"
fi
发布于 2012-01-11 07:01:05
这是另一个解决方案。这使用了POSIX substring parameter expansion,因此它可以在Bash、Dash、KornShell (ksh)、zsh (zsh)等中工作。
test "${string#*$word}" != "$string" && echo "$word found in $string"
一个带有一些示例的功能化版本:
# 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 $string
fi
}
contains "abcd" "e" || echo "abcd does not contain e"
contains "abcd" "ab" && echo "abcd contains ab"
contains "abcd" "bc" && echo "abcd contains bc"
contains "abcd" "cd" && echo "abcd contains cd"
contains "abcd" "abcd" && echo "abcd contains abcd"
contains "" "" && echo "empty string contains empty string"
contains "a" "" && echo "a contains empty string"
contains "" "a" || echo "empty string does not contain a"
contains "abcd efgh" "cd ef" && echo "abcd efgh contains cd ef"
contains "abcd efgh" " " && echo "abcd efgh contains a space"
发布于 2010-05-14 05:24:00
纯POSIX shell:
#!/bin/sh
CURRENT_DIR=`pwd`
case "$CURRENT_DIR" in
*String1*) echo "String1 present" ;;
*String2*) echo "String2 present" ;;
*) echo "else" ;;
esac
像ksh或bash这样的扩展case
具有奇特的匹配机制,但老式shell的功能却出人意料地强大。
发布于 2010-05-14 03:52:43
遗憾的是,我不知道在sh中有什么方法可以做到这一点。但是,使用bash (从3.0.0版开始,可能就是您拥有的版本),您可以像这样使用=~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
作为额外的好处(和/或警告,如果字符串中有任何有趣的字符),如果省略引号,=~将接受正则表达式作为正确的操作数。
https://stackoverflow.com/questions/2829613
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