我如何知道是否在Bash中设置了变量?
例如,如何检查用户是否向函数提供了第一个参数?
function a {
# if $1 is set ?
}
我如何知道是否在Bash中设置了变量?
例如,如何检查用户是否向函数提供了第一个参数?
function a {
# if $1 is set ?
}
当前回答
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
其他回答
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
if [[ ${!xx[@]} ]] ; then echo xx is defined; fi
有许多方法可以做到这一点,以下是其中之一:
if [ -z "$1" ]
如果$1为空或未设置,则此操作成功。
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
[[ $foo ]]
Or
(( ${#foo} ))
Or
let ${#foo}
Or
declare -p foo