在 Bash 中，如何测试变量是否以“-u”模式定义
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In Bash, how do I test if a variable is defined in "-u" mode
提问by Ramon
I just discovered set -uin bash and it helped me find several previously unseen bugs. But I also have a scenario where I need to test if a variable is defined before computing some default value. The best I have come up with for this is:
我刚刚set -u在 bash 中发现，它帮助我找到了几个以前看不见的错误。但我也有一个场景，我需要在计算一些默认值之前测试是否定义了一个变量。我为此想到的最好的方法是：
if [ "${variable-undefined}" == undefined ]; then
variable="$(...)"
fi
which works (as long as the variable doesn't have the string value undefined). I was wondering if there was a better way?
哪个有效（只要变量没有字符串值undefined）。我想知道是否有更好的方法？
采纳答案by Todd A. Jacobs
What Doesn't Work: Test for Zero-Length Strings
什么不起作用：测试零长度字符串
You can test for undefined strings in a few ways. Using the standard test conditional looks like this:
您可以通过几种方式测试未定义的字符串。使用标准测试条件如下所示：
# Test for zero-length string.
[ -z "$variable" ] || variable='foo'
This will not work with set -u, however.
但是，这不适用于set -u。
What Works: Conditional Assignment
什么有效：条件分配
Alternatively, you can use conditional assignment, which is a more Bash-like way to do this. For example:
或者，您可以使用条件赋值，这是一种更类似于 Bash 的方式来执行此操作。例如：
# Assign value if variable is unset or null.
: "${variable:=foo}"
Because of the way Bash handles expansion of this expression, you can safely use this with set -uwithout getting a "bash: variable: unbound variable" error.
由于 Bash 处理此表达式扩展的方式，您可以安全地使用它set -u而不会出现“bash：变量：未绑定变量”错误。
回答by Ramon
This is what I've found works best for me, taking inspiration from the other answers:
这是我发现最适合我的方法，从其他答案中汲取灵感：
if [ -z "${varname-}" ]; then
...
varname=$(...)
fi
回答by Felix Leipold
In bash 4.2 and newer there is an explicit way to check whether a variable is set, which is to use -v. The example from the question could then be implemented like this:
在 bash 4.2 和更新版本中，有一种明确的方法来检查是否设置了变量，即使用 -v。然后可以像这样实现问题中的示例：
if [[ ! -v variable ]]; then
variable="$(...)"
fi
See http://www.gnu.org/software/bash/manual/bashref.html#Bash-Conditional-Expressions
请参阅http://www.gnu.org/software/bash/manual/bashref.html#Bash-Conditional-Expressions
If you only want to set the variable, if it is not already set you are probably better of doing something along these lines:
如果您只想设置变量，如果它尚未设置，您可能最好按照以下方式做一些事情：
variable="${variable-$(...)}"
variable="${variable-$(...)}"
Note that this does not deal with a defined but empty variable.
请注意，这不处理已定义但为空的变量。
回答by kevinarpe
The answers above are not dynamic, e.g., how to test is variable with name "dummy" is defined? Try this:
上面的答案不是动态的，例如，如何测试定义了名称为“dummy”的变量？尝试这个：
is_var_defined()
{
if [ $# -ne 1 ]
then
echo "Expected exactly one argument: variable name as string, e.g., 'my_var'"
exit 1
fi
# Tricky. Since Bash option 'set -u' may be enabled, we cannot directly test if a variable
# is defined with this construct: [ ! -z "$var" ]. Instead, we must use default value
# substitution with this construct: [ ! -z "${var:-}" ]. Normally, a default value follows the
# operator ':-', but here we leave it blank for empty (null) string. Finally, we need to
# substitute the text from as 'var'. This is not allowed directly in Bash with this
# construct: [ ! -z "${:-}" ]. We need to use indirection with eval operator.
# Example: ="var"
# Expansion for eval operator: "[ ! -z ${:-} ]" -> "[ ! -z ${var:-} ]"
# Code execute: [ ! -z ${var:-} ]
eval "[ ! -z ${:-} ]"
return $? # Pedantic.
}
回答by efa
Unfortunatly [[ -v variable ]]is not supported in older versions of bash (at least not in version 4.1.5 I have on Debian Squeeze)
不幸的[[ -v variable ]]是，旧版本的 bash 不支持（至少在 Debian Squeeze 上的 4.1.5 版本中不支持）
You could instead use a sub shell as in this :
您可以改为使用子 shell，如下所示：
if (true $variable)&>/dev/null; then
variable="$(...)"
fi
回答by Luc M
In the beginning of your script, you could define your variables with an empty value
在脚本的开头，您可以使用空值定义变量
variable_undefined=""
Then
然后
if [ "${variable_undefined}" == "" ]; then
variable="$(...)"
fi
回答by George
if [ "${var+SET}" = "SET" ] ; then
echo "$var = ${var}"
fi
I don't know how far back ${var+value} is supported, but it works at least as far back as 4.1.2. Older versions didn't have ${var+value}, they only had ${var:+value}. The difference is that ${var:+value} will only evaluate to "value" if $var is set to a nonemptystring, while ${var+value} will also evaluate to "value" if $var is set to the empty string.
我不知道支持多远的 ${var+value}，但它至少可以追溯到 4.1.2。旧版本没有 ${var+value}，他们只有 ${var:+value}。不同之处在于，如果 $var 设置为非空字符串，则${var:+value} 只会计算为“value”，而如果 $var 设置为空字符串，则 ${var+value} 也会计算为“value”细绳。
Without [[ -v var ]] or ${var+value} I think you'd have to use another method. Probably a subshell test as was described in a previous answer:
如果没有 [[ -v var ]] 或 ${var+value} 我认为你必须使用另一种方法。可能是先前答案中描述的子外壳测试：
if ( set -u; echo "$var" ) &> /dev/null; then
echo "$var = ${var}
fi
If your shell process has "set -u" active already it'll be active in the subshell as well without the need for "set -u" again, but including it in the subshell command allows the solution to also work if the parent process hasn't got "set -u" enabled.
如果您的 shell 进程已经激活了“set -u”，它也会在子 shell 中处于活动状态，而无需再次使用“set -u”，但是将它包含在 subshell 命令中允许解决方案在父进程中也能工作没有启用“set -u”。
(You could also use another process like "printenv" or "env" to test for the presence of the variable, but then it'd only work if the variable is exported.)
（您也可以使用“printenv”或“env”等其他过程来测试变量是否存在，但只有在导出变量时它才有效。）
