这是我每天使用的：

#
# Check if a variable is set
#   param1  name of the variable
#
function is_set() { [[ $(eval echo "\${${1}+x}") ]]; }

这在Linux和Solaris以及Bash 3.0下运行良好。

$ myvar="TEST"
$ is_set myvar ; echo $?
0

$ myvar=
$ is_set myvar ; echo $?
0

$ unset myvar
$ is_set myvar ; echo $?
1

要检查是否设置了变量，请执行以下操作：

var=""; [[ $var ]] && echo "set" || echo "not set"

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。

我喜欢辅助功能来隐藏Bash的粗糙细节。在这种情况下，这样做会增加更多（隐藏的）粗糙度：

# The first ! negates the result (can't use -n to achieve this)
# the second ! expands the content of varname (can't do ${$varname})
function IsDeclared_Tricky
{
  local varname="$1"
  ! [ -z ${!varname+x} ]
}

因为我在这个实现中首先遇到了bug（灵感来自Jens和Lionel的回答），所以我想出了一个不同的解决方案：

# Ask for the properties of the variable - fails if not declared
function IsDeclared()
{
  declare -p $1 &>/dev/null
}

我发现它更直接，更害羞，更容易理解/记住。测试用例表明它是等效的：

function main()
{
  declare -i xyz
  local foo
  local bar=
  local baz=''

  IsDeclared_Tricky xyz; echo "IsDeclared_Tricky xyz: $?"
  IsDeclared_Tricky foo; echo "IsDeclared_Tricky foo: $?"
  IsDeclared_Tricky bar; echo "IsDeclared_Tricky bar: $?"
  IsDeclared_Tricky baz; echo "IsDeclared_Tricky baz: $?"

  IsDeclared xyz; echo "IsDeclared xyz: $?"
  IsDeclared foo; echo "IsDeclared foo: $?"
  IsDeclared bar; echo "IsDeclared bar: $?"
  IsDeclared baz; echo "IsDeclared baz: $?"
}

main

测试用例还显示，局部var不声明var（除非后面跟着“=”）。很长一段时间以来，我以为我是这样声明变量的，只是为了发现我只是表达了我的意图。。。我想这是不可能的。

IsDeclared_Tricky xyz:1IsDeclared_Tricky foo:1IsDeclared_Tricky条：0我声明的Tricky baz:0IsDeclared xyz:1IsDeclared foo:1IsDeclared栏：0IsDeclared baz:0

奖金：用例

我主要使用此测试以某种“优雅”和安全的方式（几乎类似于接口…）向函数提供（和返回）参数：

# Auxiliary functions
function die()
{
  echo "Error: $1"; exit 1
}

function assertVariableDeclared()
{
  IsDeclared "$1" || die "variable not declared: $1"
}

function expectVariables()
{
  while (( $# > 0 )); do
    assertVariableDeclared $1; shift
  done
}

# Actual example
function exampleFunction()
{
  expectVariables inputStr outputStr
  outputStr="$inputStr, World!"
}

function bonus()
{
  local inputStr='Hello'
  local outputStr= # Remove this to trigger the error
  exampleFunction
  echo $outputStr
}

bonus

如果调用时声明了所有必需的变量：

你好，世界！

其他：

错误：未声明变量：outputStr

[[ $foo ]]

Or

(( ${#foo} ))

Or

let ${#foo}

Or

declare -p foo

略读所有答案后，这同样有效：

if [[ -z $SOME_VAR ]]; then read -p "Enter a value for SOME_VAR: " SOME_VAR; fi
echo "SOME_VAR=$SOME_VAR"

如果您不使用SOME_VAR代替我的$SOME_VVAR，它会将其设置为空值；$这是工作所必需的。