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
以下是如何测试参数是否未设置、是否为空(“Null”)或是否设置了值:
+--------------------+----------------------+-----------------+-----------------+
| Expression | parameter | parameter | parameter |
| in script: | Set and Not Null | Set But Null | Unset |
+--------------------+----------------------+-----------------+-----------------+
| ${parameter:-word} | substitute parameter | substitute word | substitute word |
| ${parameter-word} | substitute parameter | substitute null | substitute word |
| ${parameter:=word} | substitute parameter | assign word | assign word |
| ${parameter=word} | substitute parameter | substitute null | assign word |
| ${parameter:?word} | substitute parameter | error, exit | error, exit |
| ${parameter?word} | substitute parameter | substitute null | error, exit |
| ${parameter:+word} | substitute word | substitute null | substitute null |
| ${parameter+word} | substitute word | substitute word | substitute null |
+--------------------+----------------------+-----------------+-----------------+
来源:POSIX:参数扩展:
在所有显示为“替换”的情况下,表达式将替换为显示的值。在所有显示为“assign”的情况下,参数都被指定该值,该值也会替换表达式。
要在操作中显示此内容,请执行以下操作:
+--------------------+----------------------+-----------------+-----------------+
| Expression | When FOO="world" | When FOO="" | unset FOO |
| in script: | (Set and Not Null) | (Set But Null) | (Unset) |
+--------------------+----------------------+-----------------+-----------------+
| ${FOO:-hello} | world | hello | hello |
| ${FOO-hello} | world | "" | hello |
| ${FOO:=hello} | world | FOO=hello | FOO=hello |
| ${FOO=hello} | world | "" | FOO=hello |
| ${FOO:?hello} | world | error, exit | error, exit |
| ${FOO?hello} | world | "" | error, exit |
| ${FOO:+hello} | hello | "" | "" |
| ${FOO+hello} | hello | hello | "" |
+--------------------+----------------------+-----------------+-----------------+
其他回答
要检查变量是否设置为非空值,请使用[-n“$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)。
case "$1" in
"") echo "blank";;
*) echo "set"
esac
虽然这里所述的大多数技术都是正确的,但Bash 4.2支持对变量的存在进行实际测试(man-Bash),而不是测试变量的值。
[[ -v foo ]]; echo $?
# 1
foo=bar
[[ -v foo ]]; echo $?
# 0
foo=""
[[ -v foo ]]; echo $?
# 0
值得注意的是,与许多其他方法(如使用[-z)不同,这种方法在set-u/set-o nounset模式下用于检查未设置的变量时不会导致错误。
要检查是否设置了变量,请执行以下操作:
var=""; [[ $var ]] && echo "set" || echo "not set"
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