Bash:有条件，基于命令的特定退出代码 [英] Bash: conditional based on specific exit code of command

问题描述

此问题是对>基于退出的Bash条件的后续操作命令代码.

我了解并已利用该问题的公认答案:

I understand, and have made use of the accepted answer to that question:

$ cat ./txt
foo
bar
baz
$ if ! grep -Fx bax txt &>/dev/null; then echo "not found"; fi
not found
$

但是我无法弄清楚从表面上看似简单的转折的语法:如何为命令的特定退出代码编写条件?

But I can't figure out the syntax to pull off a seemingly simple twist to the premise: how can I write a conditional for a specific exit code of a command?

这篇文章指出指出 grep 可能会失败，错误代码> 1；所以我想写上面的条件来打印未找到"，特别是在 grep 返回错误代码1的情况下.如何编写这样的语句?以下似乎不起作用(我显然在挣扎):

This post pointed out that grep might fail with error code > 1; so I wanted to write the above conditional to print "not found" specifically if grep returned error code 1. How can I write such a statement? The following don't seem to work (I'm obviously flailing):

$ # Sanity-check that grep really fails with error code 1:
$ grep -Fx bax txt &>/dev/null
$ echo $?
1
$
$
$ if grep -Fx bax txt &>/dev/null -eq 1; then echo "not found"; fi
$ if grep -Fx bax txt &>/dev/null == 1; then echo "not found"; fi
$ if grep -Fx bax txt &>/dev/null = 1; then echo "not found"; fi
$ if [ grep -Fx bax txt &>/dev/null -eq 1 ]; then echo "not found"; fi
$ if [ grep -Fx bax txt &>/dev/null == 1 ]; then echo "not found"; fi
$ if [ grep -Fx bax txt &>/dev/null = 1 ]; then echo "not found"; fi
$

---

注意:出于特定的原因，我特别想避免先运行该命令，然后再在条件语句中使用 $?，

推荐答案

如果您想对不同的错误代码做出不同的响应(与链接的问题中的成功/失败相反)，则需要运行以下命令并然后检查 $?以获得其特定的退出状态:

If you want to respond differently to different error codes (as opposed to just success/failure, as in the linked question), you need to run the command and then check $? to get its specific exit status:

grep -Fx bax txt &>/dev/null
if [ $? -eq 1 ]; then
    echo "not found"
fi

(在对链接问题的回答中，我描述了如何测试 $?是否为零，这是针对货物崇拜的编程；这不适用于检查特定的非零值.)

(In my answer to the linked question, I described testing whether $? is zero as cargo cult programming; this does not apply to checking it for specific nonzero values.)

请注意，对于每个运行的命令，都将重置 $?-包括测试上一个命令中的 $?的命令-因此，如果要执行此操作您必须立即将其存储在一个变量中，然后在该变量上运行测试:

Note that $? is re-set for every command that runs -- including one that tests $? from the previous command -- so if you want to do more than one thing with it you must immediately store it in a variable, then run your tests on that variable:

curl "$url"
curl_status=$?
if [ "$curl_status" -eq 6 ]; then
    echo "Couldn't resolve host name"
elif [ "$curl_status" -eq 7 ]; then
    echo "Couldn't connect to host"
elif [ "$curl_status" -ne 0 ]; then
    echo "Some sort of error occurred; curl status was $curl_status"
fi

请注意，使用！取消命令的成功/失败将有效地破坏特定的错误代码，因为它将所有非零代码转换为相同的结果:零.如果要执行 if的等效功能！命令... 并仍然可以访问特定的代码，我最喜欢的习惯用法是使用 ||| 和一个大括号组，例如:

Note that using ! to negate the success/failure of a command effectively destroys the specific error code, because it converts all nonzero codes to the same result: zero. If you want to do the equivalent of if ! command ... and still have access to the specific code, my favorite idiom is to use || and a brace group, like this:

curl "$url" || {
    curl_status=$?
    ...
    # Do error handling/reporting based on $curl_status
    ...
}

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