Question

如this answer中所述，检查变量是否在bash中设置的“正确”方式如下所示：

if [ -z ${var+x} ]; then
    echo "var is unset"
else
    echo "var is set to '$var'"
fi

我感兴趣的是如何将其提取到一个可以重用于不同变量的函数中。

迄今为止我能做的最好的事情是：

is_set() {
  local test_start='[ ! -z ${'
  local test_end='+x} ]'
  local tester=$test_start$1$test_end

  eval $tester
}

它似乎有效，但有没有更好的方法不诉诸eval？

Answer 1

在Bash中，您可以使用[[ -v var ]]。不需要功能或复杂的方案。

从联系手册：

   -v varname
          True if the shell variable varname is set (has been assigned a value).

前2个命令序列打印ok：

[[ -v PATH ]] && echo ok

var="" ; [[ -v var ]] && echo ok

unset var ; [[ -v var ]] && echo ok

Answer 2

使用[[ ... ]]，您只需执行此操作（不需要双引号）：

[[ $var ]] && echo "var is set"
[[ $var ]] || echo "var is not set or it holds an empty string"

如果你真的想要一个函数，那么我们就可以编写单行代码：

is_empty()     { ! [[ $1 ]]; } # check if not set or set to empty string
is_not_empty() {   [[ $1 ]]; } # check if set to a non-empty string

var="apple"; is_not_empty "$var" && echo "var is not empty" # shows "var is not empty"
var=""     ; is_empty "$var"     && echo "var is empty"     # shows "var is empty"
unset var  ; is_empty "$var"     && echo "var is empty"     # shows "var is empty"
var="apple"; is_empty "$var"     || echo "var is not empty" # shows "var is not empty"

最后，可以实施is_unset和is_set来将$1视为变量的名称：

is_unset() {   [[ -z "${!1+x}" ]]; }           # use indirection to inspect variable passed through $1 is unset
is_set()   { ! [[ -z "${!1+x}" ]]; }           # check if set, even to an empty string

unset var; is_unset var && echo "var is unset" # shows "var is unset"
var=""   ; is_set var   && echo "var is set"   # shows "var is set"
var="OK" ; is_set var   && echo "var is set"   # shows "var is set"

Bash函数用于检查是否设置了给定变量

2 个答案: