Bash获取命令的返回值并使用此值退出 [英] Bash get return value of a command and exit with this value

问题描述

我想执行一个命令，然后获取该命令的返回值.完成后，我想退出并给退出我之前执行的命令的返回值.我已经有这段代码，但是似乎没有用.

I want to execute a command then get the return value of this command. when it's done i want to exit and give exit the return value of the previous command, the one i just executed. I already have this piece of code but it doesn't seem to work.

exec gosu user /var/www/bin/phpunit -c app
typeset ret_code
ret_code=$?
if [ $ret_code == 0 ]; then
    exit 0
fi

exit 1

我该怎么办?谢谢

推荐答案

我认为您的问题是 typeset 本身创建的返回值为0.请尝试

I think your problem is that typeset itself creates a return value of 0. Try

gosu user /var/www/bin/phpunit -c app
ret_code=$?
return $ret_code

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