[英]getopts not working - bash
I am writing a bash script which accept parameters. 我正在编写一个接受参数的bash脚本。 I am using getopts to achieve it.
我正在使用getopts来实现它。
#!/bin/bash
while getopts ":a" opt; do
case $opt in
a)
echo "-a was triggered!" >&2
;;
\?)
echo "Invalid option: -$OPTARG" >&2
;;
esac
done
but above code return's me this error. 但是上面的代码返回是我这个错误。
'etOpts_test.sh: line 4: syntax error near unexpected token `in
'etOpts_test.sh: line 4: ` case $opt in
I am using CentOs 5.5 我正在使用CentOs 5.5
At line 4 you probably want case "$opt" in
(quote $opt
). 在第4行,您可能希望在
case "$opt" in
(quote $opt
)。 Otherwise if it contains a metacharacter it could fail. 否则,如果它包含一个元字符,则可能会失败。
It should be a: , not :a to denote a flag requiring an argument, also question mark should not be quoted as it serves as wildcard symbol. 它应该是: ,而不是:a来表示需要参数的标志,也不应加问号,因为它用作通配符。 Overall code would be (also demonstrating a flag -h not taking arguments):
总体代码将是(还显示一个标志-h不带参数):
function usage {
echo "usage: ..."
}
a_arg=
while getopts a:h opt; do
case $opt in
a)
a_arg=$OPTARG
;;
h)
usage && exit 0
;;
?)
usage && exit 2
;;
esac
done
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.