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 $1 is set ?
}
当前回答
我很惊讶没有人尝试编写一个shell脚本来以编程方式生成这个臭名昭著的难以摸索的表。既然我们在这里试图学习编码技术,为什么不用代码表达答案?:)这是我的看法(应该在任何POSIX shell中都适用):
H="+-%s-+-%s----+-%s----+-%s--+\n" # table divider printf format
R="| %-10s | %-10s | %-10s | %-10s |\n" # table row printf format
S='V' # S is a variable that is set-and-not-null
N='' # N is a variable that is set-but-null (empty "")
unset U # U is a variable that is unset
printf "$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。
其他回答
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 $? # Pedantic.
}
相关:在Bash中,如何测试变量是否以“-u”模式定义
要检查变量是否设置为非空值,请使用[-n“$x”],正如其他人已经指出的那样。
大多数情况下,最好将具有空值的变量与未设置的变量以相同的方式处理。但如果需要,您可以区分这两个:[-n“${x+set}”](如果设置了x,则“${x+set}”扩展为set,如果未设置x,则扩展为空字符串)。
要检查是否传递了参数,请测试$#,这是传递给函数(或不在函数中时传递给脚本)的参数数(请参见Paul的答案)。
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 [ $# -gt 0 ]; then
echo \$1 is set
else
echo \$1 is not set
fi
我总是使用这个,因为任何第一次看到代码的人都很容易理解:
if [ "$variable" = "" ]
then
echo "Variable X is empty"
fi
如果要检查是否为空;
if [ ! "$variable" = "" ]
then
echo "Variable X is not empty"
fi
就是这样。
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