如果您运行bash脚本，那么它将在其当前环境或其子环境上操作，而不会在父环境上操作。

如果目标是运行命令: goto.sh /home/test 然后在/home/test中交互式地工作，一种方法是在你的脚本中运行一个bash交互式子shell:

#!/bin/bash
cd $1
exec bash

这样，您将处于/home/test中，直到退出(退出或按Ctrl+C)这个shell。

启动脚本时，将创建一个只继承您的环境的新进程。当它结束时，它就结束了。您当前的环境保持不变。

相反，你可以这样开始你的脚本:

. myscript.sh

的。将评估当前环境中的脚本，因此它可能会被更改

使用pushd，当前目录被推送到目录堆栈上，并被更改为给定的目录，popd获取堆栈顶部的目录并更改为该目录。

pushd ../new/dir > /dev/null
# do something in ../new/dir
popd > /dev/null

在你的shell脚本中添加以下cd行:

exec $SHELL

这是我目前为bash做这件事的方式(在Debian上测试)。也许有更好的办法:

不要用exec bash这样做，例如:

#!/bin/bash
cd $1
exec bash

because while it appears to work, after you run it and your script finishes, yes you'll be in the correct directory, but you'll be in it in a subshell, which you can confirm by pressing Ctrl+D afterwards, and you'll see it exits the subshell, putting you back in your original directory. This is usually not a state you want a script user to be left in after the script they run returns, because it's non-obvious that they're in a subshell and now they basically have two shells open when they thought they only had one. They might continue using this subshell and not realize it, and it could have unintended consequences. If you really want the script to exit and leave open a subshell in the new directory, it's better if you change the PS1 variable so the script user has a visual indicator that they still have a subshell open. Here's an example I came up with. It is two files, an outer.sh which you call directly, and an inner.sh which is sourced inside the outer.sh script. The outer script sets two variables, then sources the inner script, and afterwards it echoes the two variables (the second one has just been modified by the inner script). Afterwards it makes a temp copy of the current user's ~/.bashrc file, adds an override for the PS1 variable in it, as well as a cleanup routine, and finally it runs exec bash --rcfile pointing at the .bashrc.tmp file to initialize bash with a modified environment, including the modified prompt and the cleanup routine. After outer.sh exits, you'll be left inside a subshell in the desired directory (in this case testdir/ which was entered into by the inner.sh script) with a visual indicator making it clear to you, and if you exit out of the subshell, the .bashrc.tmp file will be deleted by the cleanup routine, and you'll be back in the directory you started in. Maybe there's a smarter way to do it, but that's the best way I could figure out in about 40 minutes of experimenting:

文件1:outer.sh

#!/bin/bash

var1="hello"
var2="world"

source inner.sh

echo $var1
echo $var2

cp ~/.bashrc .bashrc.tmp

echo 'export PS1="(subshell) $PS1"' >> .bashrc.tmp

cat <<EOS >> .bashrc.tmp
cleanup() {
    echo "cleaning up..."
    rm .bashrc.tmp
}

trap 'cleanup' 0
EOS

exec bash --rcfile .bashrc.tmp

文件2:inner.sh

cd testdir
var2="bird"

然后运行:

$ mkdir testdir
$ chmod 755 outer.sh

$ ./outer.sh

它应该输出:

hello
bird

然后使用exec bash将您放到子shell中，但是带有 修改后的提示符使其更明显，如下所示:

(subshell) user@computername:~/testdir$

如果你用Ctrl-D退出子shell，它应该通过删除来清理 testdir/目录下的临时.bashrc.tmp文件 我想知道是否有比复制.bashrc文件更好的方法 就像这样，在subshell中适当地改变PS1的变量…

这种方法对我来说更容易。

假设在个人iMac上，你是管理员，当打开命令窗口时，在默认目录下/Users/jdoe，这将是进入的目录:/Users/jdoe/Desktop/Mongo/db.3.2.1/bin。

下面这些步骤可以帮助你完成工作:

vi mongobin，我进去了: cd /Users/jdoe/Desktop/Mongo/db.3.2.1/bin作为第一行。 Chmod 755 mongobin 源mongobin 松材线虫病

拖鞋!