Shell脚本在文件中的相对位置 [英] shell script relative position from the file

问题描述

我几乎不了解Linux中的Shell脚本或命令

I have almost no idea about shell scripts or commands in linux

我有一个名为projectx

projectX恰好位于users/hardik/desktop/projectx

我创建了一个shell脚本start.sh.这就是shell脚本的内容

I have created a shell script start.sh. This is the content of shell script

echo "Starting typescript build in new terminal.."

osascript -e 'tell application "Terminal" to do script "npm run build"' 

sleep 3

echo "Starting firebase functions...."

osascript -e 'tell application "Terminal" to do script "firebase emulators:start --only functions"'


echo "Process compelete.. Check if there were two terminals window open"

现在可以，但是在这里说

now this works but say here

osascript -e 'tell application "Terminal" to do script "npm run build"'

它在根目录中运行，因此出现以下错误

it runs that in the root and hence gives the following error

ENOENT:没有这样的文件或目录，请打开/Users/hardik/package.json

ENOENT: no such file or directory, open /Users/hardik/package.json

如何使其在相对于start.sh

更新:我尝试过

echo "Starting typescript build in new terminal.."
path=`pwd`
osascript -e 'tell application "Terminal" to do script "cd ${path} npm run watch:scss"' 
osascript -e 'tell application "Terminal" to do script "npm run watch"'

echo "Process compelete.. Check if there were two terminals window open"

但这不适用于错误cd: too many arguments

推荐答案

插入该行:

cd `dirname $0`

在start.sh的顶部. dirname命令生成文件包含的目录，$ 0是调用start.sh的路径，因此这是您想要的相对目录.请注意，这些引号用左引号引起，因此运行目录名的结果将传递给cd.

at the top of start.sh. The dirname command produces the directory that a file contains, and $0 is the path that invoked start.sh, so that's the relative directory that you wanted. Note those quotes a left-quotes, so the result of running dirname is passed to cd.

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