当启用Bash选项集-u时，上面的答案不起作用。此外，它们不是动态的，例如，如何测试是否定义了名为“dummy”的变量？试试看：

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 $1 as 'var'.  This is not allowed directly in Bash with this
    # construct: [ ! -z "${$1:-}" ].  We need to use indirection with eval operator.
    # Example: $1="var"
    # Expansion for eval operator: "[ ! -z \${$1:-} ]" -> "[ ! -z \${var:-} ]"
    # Code  execute: [ ! -z ${var:-} ]
    eval "[ ! -z \${$1:-} ]"
    return $?  # Pedantic.
}

相关：在Bash中，如何测试变量是否以“-u”模式定义

以下是如何测试参数是否未设置、是否为空（“Null”）或是否设置了值：

+--------------------+----------------------+-----------------+-----------------+
|   Expression       |       parameter      |     parameter   |    parameter    |
|   in script:       |   Set and Not Null   |   Set But Null  |      Unset      |
+--------------------+----------------------+-----------------+-----------------+
| ${parameter:-word} | substitute parameter | substitute word | substitute word |
| ${parameter-word}  | substitute parameter | substitute null | substitute word |
| ${parameter:=word} | substitute parameter | assign word     | assign word     |
| ${parameter=word}  | substitute parameter | substitute null | assign word     |
| ${parameter:?word} | substitute parameter | error, exit     | error, exit     |
| ${parameter?word}  | substitute parameter | substitute null | error, exit     |
| ${parameter:+word} | substitute word      | substitute null | substitute null |
| ${parameter+word}  | substitute word      | substitute word | substitute null |
+--------------------+----------------------+-----------------+-----------------+

来源：POSIX：参数扩展：

在所有显示为“替换”的情况下，表达式将替换为显示的值。在所有显示为“assign”的情况下，参数都被指定该值，该值也会替换表达式。

要在操作中显示此内容，请执行以下操作：

+--------------------+----------------------+-----------------+-----------------+
|   Expression       |  When FOO="world"    |  When FOO=""    |    unset FOO    |
|   in script:       |  (Set and Not Null)  |  (Set But Null) |     (Unset)     |
+--------------------+----------------------+-----------------+-----------------+
| ${FOO:-hello}      | world                | hello           | hello           |
| ${FOO-hello}       | world                | ""              | hello           |
| ${FOO:=hello}      | world                | FOO=hello       | FOO=hello       |
| ${FOO=hello}       | world                | ""              | FOO=hello       |
| ${FOO:?hello}      | world                | error, exit     | error, exit     |
| ${FOO?hello}       | world                | ""              | error, exit     |
| ${FOO:+hello}      | hello                | ""              | ""              |
| ${FOO+hello}       | hello                | hello           | ""              |
+--------------------+----------------------+-----------------+-----------------+

if [ "$1" != "" ]; then
  echo \$1 is set
else
  echo \$1 is not set
fi

尽管对于参数，通常最好测试$#，我认为这是参数的数量。

if [ $# -gt 0 ]; then
  echo \$1 is set
else
  echo \$1 is not set
fi

我很惊讶没有人尝试编写一个shell脚本来以编程方式生成这个臭名昭著的难以摸索的表。既然我们在这里试图学习编码技术，为什么不用代码表达答案？：）这是我的看法（应该在任何POSIX shell中都适用）：

H="+-%s-+-%s----+-%s----+-%s--+\n"       # table divider printf format
R="| %-10s | %-10s | %-10s | %-10s |\n"  # table row printf format

S='V'     # S is a variable that is set-and-not-null
N=''      # N is a variable that is set-but-null (empty "")
unset U   # U is a variable that is unset

printf "$H" "----------" "-------" "-------" "---------";
printf "$R" "expression" "FOO='V'" "FOO='' " "unset FOO";
printf "$H" "----------" "-------" "-------" "---------";
printf "$R" "\${FOO:-x}" "${S:-x}" "${N:-x}" "${U:-x}  "; S='V';N='';unset U
printf "$R" "\${FOO-x} " "${S-x} " "${N-x} " "${U-x}   "; S='V';N='';unset U
printf "$R" "\${FOO:=x}" "${S:=x}" "${N:=x}" "${U:=x}  "; S='V';N='';unset U
printf "$R" "\${FOO=x} " "${S=x} " "${N=x} " "${U=x}   "; S='V';N='';unset U
#                                  "${N:?x}" "${U:?x}  "
printf "$R" "\${FOO:?x}" "${S:?x}" "<error>" "<error>  "; S='V';N='';unset U
#                                            "${U?x}   "
printf "$R" "\${FOO?x} " "${S?x} " "${N?x} " "<error>  "; S='V';N='';unset U
printf "$R" "\${FOO:+x}" "${S:+x}" "${N:+x}" "${U:+x}  "; S='V';N='';unset U
printf "$R" "\${FOO+x} " "${S+x} " "${N+x} " "${U+x}   "; S='V';N='';unset U
printf "$H" "----------" "-------" "-------" "---------";

以及运行脚本的输出：

+------------+------------+------------+------------+
| expression | FOO='V'    | FOO=''     | unset FOO  |
+------------+------------+------------+------------+
| ${FOO:-x}  | V          | x          | x          |
| ${FOO-x}   | V          |            | x          |
| ${FOO:=x}  | V          | x          | x          |
| ${FOO=x}   | V          |            | x          |
| ${FOO:?x}  | V          | <error>    | <error>    |
| ${FOO?x}   | V          |            | <error>    |
| ${FOO:+x}  | x          |            |            |
| ${FOO+x}   | x          | x          |            |
+------------+------------+------------+------------+

该脚本缺少了一些功能，比如在发生（或不发生）副作用分配时显示，但也许其他更有野心的人希望以此为出发点，并以此为出发。

Note

由于Bash标签，我给出了一个非常注重Bash的答案。

简短的回答

只要您只处理Bash中的命名变量，这个函数应该始终告诉您变量是否已设置，即使它是一个空数组。

variable-is-set() {
    declare -p "$1" &>/dev/null
}

为什么这样做

在Bash（至少可以追溯到3.0）中，如果var是一个声明的/set变量，那么declare-p var输出一个声明命令，将变量var设置为当前的类型和值，并返回状态代码0（成功）。如果var未声明，则declare-p var将向stderr输出错误消息并返回状态代码1。使用&>/dev/null，将常规stdout和stderr输出重定向到/dev/null，永不可见，并且不更改状态代码。因此，函数只返回状态代码。

为什么其他方法（有时）在Bash中失败

[-n“$var”]：仅检查${var[0]｝是否为非空。（在Bash中，$var与${var[0]｝相同。）[-n“${var+x}”]：仅检查是否设置了${var[0]｝。[“${#var[@]}”！=0]：仅检查是否至少设置了$var的一个索引。

当此方法在Bash中失败时

这只适用于命名变量（包括$_），而不适用于某些特殊变量（$！，$@，$#，$$，$*，$？，$-，$0，$1，$2，…，以及我可能忘记的任何变量）。由于这些都不是数组，POSIX样式[-n“${var+x}”]适用于所有这些特殊变量。但请注意不要将其包装在函数中，因为在调用函数时，许多特殊变量会更改值/存在性。

外壳兼容性说明

如果您的脚本有数组，并且试图使其与尽可能多的shell兼容，那么考虑使用typeset-p而不是declare-p。我读到ksh只支持前者，但还没有能够测试这一点。我知道Bash 3.0+和Zsh 5.5.1都支持typeset-p和declare-p，不同之处仅在于其中一个是另一个的替代。但除了这两个关键字之外，我还没有测试过它们的差异，也没有测试过其他外壳。

如果您需要脚本与POSIX sh兼容，那么就不能使用数组。如果没有数组，[-n“｛$var+x｝”]可以工作。

Bash中不同方法的比较代码

此函数取消设置变量var，对传递的代码求值，运行测试以确定var是否由evald代码设置，最后显示不同测试的结果状态代码。

我跳过了测试-vvar、[-vvar]和[[-vvar]]，因为它们产生了与POSIX标准[-n“${var+x}”]相同的结果，同时需要Bash 4.2+。我也跳过了typeset-p，因为它与我测试过的shell（Bash 3.0到5.0和Zsh 5.5.1）中的声明-p相同。

is-var-set-after() {
    # Set var by passed expression.
    unset var
    eval "$1"

    # Run the tests, in increasing order of accuracy.
    [ -n "$var" ] # (index 0 of) var is nonempty
    nonempty=$?
    [ -n "${var+x}" ] # (index 0 of) var is set, maybe empty
    plus=$?
    [ "${#var[@]}" != 0 ] # var has at least one index set, maybe empty
    count=$?
    declare -p var &>/dev/null # var has been declared (any type)
    declared=$?

    # Show test results.
    printf '%30s: %2s %2s %2s %2s\n' "$1" $nonempty $plus $count $declared
}

测试用例代码

请注意，如果变量尚未声明为关联数组，则由于Bash将非数字数组索引视为“0”，测试结果可能是意外的。此外，关联数组仅在Bash 4.0+中有效。

# Header.
printf '%30s: %2s %2s %2s %2s\n' "test" '-n' '+x' '#@' '-p'
# First 5 tests: Equivalent to setting 'var=foo' because index 0 of an
# indexed array is also the nonindexed value, and non-numerical
# indices in an array not declared as associative are the same as
# index 0.
is-var-set-after "var=foo"                        #  0  0  0  0
is-var-set-after "var=(foo)"                      #  0  0  0  0
is-var-set-after "var=([0]=foo)"                  #  0  0  0  0
is-var-set-after "var=([x]=foo)"                  #  0  0  0  0
is-var-set-after "var=([y]=bar [x]=foo)"          #  0  0  0  0
# '[ -n "$var" ]' fails when var is empty.
is-var-set-after "var=''"                         #  1  0  0  0
is-var-set-after "var=([0]='')"                   #  1  0  0  0
# Indices other than 0 are not detected by '[ -n "$var" ]' or by
# '[ -n "${var+x}" ]'.
is-var-set-after "var=([1]='')"                   #  1  1  0  0
is-var-set-after "var=([1]=foo)"                  #  1  1  0  0
is-var-set-after "declare -A var; var=([x]=foo)"  #  1  1  0  0
# Empty arrays are only detected by 'declare -p'.
is-var-set-after "var=()"                         #  1  1  1  0
is-var-set-after "declare -a var"                 #  1  1  1  0
is-var-set-after "declare -A var"                 #  1  1  1  0
# If 'var' is unset, then it even fails the 'declare -p var' test.
is-var-set-after "unset var"                      #  1  1  1  1

测试输出

标题行中的测试助记符对应于[-n“$var”]，[-n“$｛var+x｝”]，[“$｛#var[@]｝”！=0]，并分别声明-p var。

                         test: -n +x #@ -p
                      var=foo:  0  0  0  0
                    var=(foo):  0  0  0  0
                var=([0]=foo):  0  0  0  0
                var=([x]=foo):  0  0  0  0
        var=([y]=bar [x]=foo):  0  0  0  0
                       var='':  1  0  0  0
                 var=([0]=''):  1  0  0  0
                 var=([1]=''):  1  1  0  0
                var=([1]=foo):  1  1  0  0
declare -A var; var=([x]=foo):  1  1  0  0
                       var=():  1  1  1  0
               declare -a var:  1  1  1  0
               declare -A var:  1  1  1  0
                    unset var:  1  1  1  1

总结

声明-p var&>/dev/null对于测试Bash中的命名变量是（100%？）可靠的，至少从3.0开始。[-n“${var+x}”]在符合POSIX的情况下是可靠的，但不能处理数组。其他测试用于检查变量是否为非空，以及检查其他shell中声明的变量。但这些测试既不适合Bash也不适合POSIX脚本。

在现代版本的Bash（我认为是4.2或更高版本；我不确定）上，我会这样做：

if [ ! -v SOMEVARIABLE ] #note the lack of a $ sigil
then
    echo "Variable is unset"
elif [ -z "$SOMEVARIABLE" ]
then
    echo "Variable is set to an empty string"
else
    echo "Variable is set to some string"
fi