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
[[ $foo ]]
Or
(( ${#foo} ))
Or
let ${#foo}
Or
declare -p foo
其他回答
if [[ ${1:+isset} ]]
then echo "It was set and not null." >&2
else echo "It was not set or it was null." >&2
fi
if [[ ${1+isset} ]]
then echo "It was set but might be null." >&2
else echo "It was was not set." >&2
fi
如果未设置,则要退出
这对我很有用。如果没有设置参数,我希望脚本退出并显示错误消息。
#!/usr/bin/env bash
set -o errexit
# Get the value and empty validation check all in one
VER="${1:?You must pass a version of the format 0.0.0 as the only argument}"
运行时返回错误
peek@peek:~$ ./setver.sh
./setver.sh: line 13: 1: You must pass a version of the format 0.0.0 as the only argument
仅检查,不退出-空和未设置无效
如果您只想检查值set=VALID或unset/empty=INVALID,请尝试此选项。
TSET="good val"
TEMPTY=""
unset TUNSET
if [ "${TSET:-}" ]; then echo "VALID"; else echo "INVALID";fi
# VALID
if [ "${TEMPTY:-}" ]; then echo "VALID"; else echo "INVALID";fi
# INVALID
if [ "${TUNSET:-}" ]; then echo "VALID"; else echo "INVALID";fi
# INVALID
或者,即使是短期测试;-)
[ "${TSET:-}" ] && echo "VALID" || echo "INVALID"
[ "${TEMPTY:-}" ] && echo "VALID" || echo "INVALID"
[ "${TUNSET:-}" ] && echo "VALID" || echo "INVALID"
仅检查,不退出-仅空为无效
这就是问题的答案。如果您只想检查值set/empty=VALID或unset=INVALID,请使用此选项。
注意,“..-1}”中的“1”无关紧要,它可以是任何东西(比如x)
TSET="good val"
TEMPTY=""
unset TUNSET
if [ "${TSET+1}" ]; then echo "VALID"; else echo "INVALID";fi
# VALID
if [ "${TEMPTY+1}" ]; then echo "VALID"; else echo "INVALID";fi
# VALID
if [ "${TUNSET+1}" ]; then echo "VALID"; else echo "INVALID";fi
# INVALID
短期测试
[ "${TSET+1}" ] && echo "VALID" || echo "INVALID"
[ "${TEMPTY+1}" ] && echo "VALID" || echo "INVALID"
[ "${TUNSET+1}" ] && echo "VALID" || echo "INVALID"
我把这个答案献给了@mklement0(comments),他要求我准确回答这个问题。
参考:2.6.2参数扩展
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。
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
声明一个简单函数is_set,它使用Declare-p直接测试变量是否存在。
$ is_set() {
declare -p $1 >/dev/null 2>&1
}
$ is_set foo; echo $?
0
$ declare foo
$ is_set foo; echo $?
1