如何在 Bash 中抛出错误？
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How to throw an error in Bash?
提问by boop
How do I throw an error in bash to get in my catchclause (I'm not sure what this expression is actually called)
我如何在 bash 中抛出错误以进入我的catch子句（我不确定这个表达式实际上叫什么）
{
# ...
if [ "$status" -ne "200" ]
# throw error
fi
} || {
# on error / where I want to get if status != 200
}
I know I just could use a function but this case made me curious if it would be possible to do this
我知道我只能使用一个函数，但是这个案例让我很好奇是否可以这样做
采纳答案by Chris Maes
There are multiple ways to do something similar:
有多种方法可以做类似的事情：
use subshells(might not be the best solution if you want to set parameters etc...)
使用子外壳（如果你想设置参数等，可能不是最好的解决方案......）
(
if [[ "$status" -ne "200" ]]
then
exit 1
fi
) || (
# on error / where I want to get if status != 200
echo "error thrown"
)
use an intermediate errorvariable(you can catch multiple errors by setting different numbers. Also: less indentation depth)
使用中间错误变量（您可以通过设置不同的数字来捕获多个错误。另外：减少缩进深度）
if [[ "$status" -ne "200" ]]
then
error=1
fi
if [ $error != 0 ]
then
echo "error $error thrown"
fi
use immediately the exit value of your test(note that I changed -neto -eq)
立即使用测试的退出值（请注意，我更改-ne为-eq）
[[ "$status" -eq "200" ]] || echo "error thrown"
回答by toth
One way to do it is with just exit as shellter mentions, but you have to create a subshell for that work (which exists on the exit). To do this, replace the curly brackets with parentheses.
一种方法是像 shellter 提到的那样直接退出，但您必须为该工作创建一个子 shell（它存在于出口上）。为此，请用括号替换大括号。
The code will be
代码将是
(
# ...
if [ "$status" -ne "200" ]
exit 1
fi
) || {
# on error / where I want to get if status != 200
}
回答by John1024
The return code for the braces is the return code of the last statement executed in the braces. So, here is one method that uses the shell-builtin false:
大括号的返回码是大括号中执行的最后一条语句的返回码。所以，这是一种使用 shell-builtin 的方法false：
{
# ...
if [ "$status" -ne "200" ]; then
false # throws "error" (return code=1)
fi
} || {
echo "Caught error" # on error / where I want to get if status != 200
}
In this particular case, the ifstatement is superfluous. The test statement, [ "$status" -ne "200" ], throws the exit code that you need:
在这种特殊情况下，该if语句是多余的。测试语句 ，[ "$status" -ne "200" ]抛出您需要的退出代码：
{
# ...
! [ "$status" -ne "200" ]
} || {
echo "Caught error" # on error / where I want to get if status != 200
}
