bash 使用“set -o nounset”测试是否在bash中设置了变量
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Test if a variable is set in bash when using "set -o nounset"
提问by vinodkone
The following code exits with a unbound variable error. How to fix this, while still using the set -onounset option?
以下代码以未绑定变量错误退出。如何解决这个问题，同时仍然使用set -onounset 选项？
#!/bin/bash
set -o nounset
if [ ! -z ${WHATEVER} ];
then echo "yo"
fi
echo "whatever"
采纳答案by Angelom
#!/bin/bash
set -o nounset
VALUE=${WHATEVER:-}
if [ ! -z ${VALUE} ];
then echo "yo"
fi
echo "whatever"
In this case, VALUEends up being an empty string if WHATEVERis not set. We're using the {parameter:-word}expansion, which you can look up in man bashunder "Parameter Expansion".
在这种情况下，VALUE如果WHATEVER未设置，则最终为空字符串。我们正在使用{parameter:-word}扩展，您可以man bash在“参数扩展”下查找。
回答by l0b0
You need to quote the variables if you want to get the result you expect:
如果你想得到你期望的结果，你需要引用变量：
check() {
if [ -n "${WHATEVER-}" ]
then
echo 'not empty'
elif [ "${WHATEVER+defined}" = defined ]
then
echo 'empty but defined'
else
echo 'unset'
fi
}
Test:
测试：
$ unset WHATEVER
$ check
unset
$ WHATEVER=
$ check
empty but defined
$ WHATEVER=' '
$ check
not empty
回答by NublaII
How about a oneliner?
单线呢？
[ -z "${VAR:-}" ] && echo "VAR is not set or is empty" || echo "VAR is set to $VAR"
-zchecks both for empty or unset variable
-z检查空变量或未设置变量
回答by Acumenus
Assumptions:
假设：
$ echo $SHELL
/bin/bash
$ /bin/bash --version | head -1
GNU bash, version 4.1.2(1)-release (x86_64-redhat-linux-gnu)
$ set -o nounset
If you want a non-interactive script to print an error and exit if a variable is null or not set:
如果您希望非交互式脚本打印错误并在变量为空或未设置时退出：
$ [[ "${HOME:?}" ]]
$ [[ "${IAMUNBOUND:?}" ]]
bash: IAMUNBOUND: parameter null or not set
$ IAMNULL=""
$ [[ "${IAMNULL:?}" ]]
bash: IAMNULL: parameter null or not set
If you don't want the script to exit:
如果您不希望脚本退出：
$ [[ "${HOME:-}" ]] || echo "Parameter null or not set."
$ [[ "${IAMUNBOUND:-}" ]] || echo "Parameter null or not set."
Parameter null or not set.
$ IAMNULL=""
$ [[ "${IAMUNNULL:-}" ]] || echo "Parameter null or not set."
Parameter null or not set.
You can even use [and ]instead of [[and ]]above, but the latter is preferable in Bash.
您甚至可以使用[and]代替[[and]]上面，但后者在 Bash 中更可取。
Note what the colon does above. From the docs:
注意上面冒号的作用。从文档：
Put another way, if the colon is included, the operator tests for both parameter's existence and that its value is not null; if the colon is omitted, the operator tests only for existence.
换句话说，如果包含冒号，则运算符会测试两个参数是否存在，并且其值不为空；如果省略冒号，则运算符仅测试是否存在。
There is apparently no need for -nor -z.
显然不需要-nor -z。
In summary, I may typically just use [[ "${VAR:?}" ]].Per the examples, this prints an error and exits if a variable is null or not set.
总之，我通常可能只使用[[ "${VAR:?}" ]]. 根据示例，如果变量为空或未设置，则会打印错误并退出。
回答by Ale? Friedl
You can use
您可以使用
if [[ ${WHATEVER:+$WHATEVER} ]]; then
but
但
if [[ "${WHATEVER:+isset}" == "isset" ]]; then
might be more readable.
可能更具可读性。
回答by Max Bileschi
While this isn't exactlythe use case asked for above, I've found that if you want to use nounset(or -u) the default behavior is the one you want: to exit nonzero with a descriptive message.
虽然这不是正是用例要求上面，我发现，如果你想使用nounset（或-u）的默认行为是你想要的：退出非零一个描述性的消息。
It took me long enough to realize this that I figured it was worth posting as a solution.
我花了很长时间才意识到这一点，我认为值得将其作为解决方案发布。
If all you want is to echo something else when exiting, or do some cleanup, you can use a trap.
如果您只想在退出时回显其他内容或进行一些清理，则可以使用陷阱。
The :-operator is probably what you want otherwise.
:-否则，操作员可能就是您想要的。
