在 Bash 函数中使用 getopts
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Using getopts inside a Bash function
提问by Magnus
I'd like to use getoptsinside a function that I have defined in my .bash_profile.
The idea is I'd like to pass in some flags to this function to alter its behavior.
我想getopts在我在 .bash_profile 中定义的函数中使用。这个想法是我想将一些标志传递给这个函数来改变它的行为。
Here's the code:
这是代码：
function t() {
echo $*
getopts "a:" OPTION
echo $OPTION
echo $OPTARG
}
When I invoke it like this:
当我像这样调用它时：
t -a bc
I get this output:
我得到这个输出：
-a bc
?
?
What's wrong? I'd like to get the value bcwithout manually shifting and parsing. How do I use getoptscorrectly inside a function?
怎么了？我想在bc不手动移动和解析的情况下获得该值。如何getopts在函数内部正确使用？
EDIT: corrected my code snippet to try $OPTARG, to no avail
编辑：更正了我的代码片段以尝试 $OPTARG，但无济于事
EDIT #2: OK turns out the code is fine, my shell was somehow messed up. Opening a new window solved it. The arg value was indeed in $OPTARG.
编辑 #2：好的，结果代码很好，我的外壳不知何故搞砸了。打开一个新窗口解决了它。arg 值确实在 $OPTARG 中。
回答by Adrian Frühwirth
As @Ansgar points out, the argument to your option is stored in ${OPTARG}, but this is not the only thing to watch out for when using getoptsinside a function. You also need to make sure that ${OPTIND}is local to the function by either unsetting it or declaring it local, otherwise you will encounter unexpected behaviour when invoking the function multiple times.
正如@Ansgar 指出的那样，您的选项的参数存储在 中${OPTARG}，但这并不是在getopts函数内部使用时唯一需要注意的事情。您还需要${OPTIND}通过取消设置或声明它来确保该函数是本地的local，否则在多次调用该函数时会遇到意外行为。
t.sh:
t.sh：
#!/bin/bash
foo()
{
foo_usage() { echo "foo: [-a <arg>]" 1>&2; exit; }
local OPTIND o a
while getopts ":a:" o; do
case "${o}" in
a)
a="${OPTARG}"
;;
*)
foo_usage
;;
esac
done
shift $((OPTIND-1))
echo "a: [${a}], non-option arguments: $*"
}
foo
foo -a bc bar quux
foo -x
Example run:
示例运行：
$ ./t.sh
a: [], non-option arguments:
a: [bc], non-option arguments: bar quux
foo: [-a <arg>]
If you comment out # local OPTIND, this is what you get instead:
如果你注释掉# local OPTIND，这就是你得到的：
$ ./t.sh
a: [], non-option arguments:
a: [bc], non-option arguments: bar quux
a: [bc], non-option arguments:
Other than that, its usage is the same as when used outside of a function.
除此之外，它的用法与在函数外使用时相同。
回答by kenorb
Here is simple example of getoptsusage within shell function:
以下是getoptsshell 函数中的简单用法示例：
#!/usr/bin/env bash
t() {
local OPTIND
getopts "a:" OPTION
echo Input: $*, OPTION: $OPTION, OPTARG: $OPTARG
}
t "$@"
t -a foo
Output:
输出：
$ ./test.sh -a bc
Input: -a bc, OPTION: a, OPTARG: bc
Input: -a foo, OPTION: a, OPTARG: foo
As @Adrian pointed out, local OPTIND(or OPTIND=1) needs to be set as shell does not reset OPTINDautomatically between multiple calls to getopts(man bash).
正如@Adrian 指出的那样，local OPTIND（或OPTIND=1）需要设置为 shell 不会OPTIND在多次调用getopts( man bash)之间自动重置。
The base-syntax for getoptsis:
的基本语法getopts是：
getopts OPTSTRING VARNAME [ARGS...]
and by default, not specifying arguments is equivalent to explicitly calling it with "$@" which is: getopts "a:" opts "$@".
默认情况下，不指定参数等效于使用“$@”显式调用它，即：getopts "a:" opts "$@"。
In case of problems, these are the used variables for getoptsto check:
如果出现问题，这些是用于getopts检查的变量：
OPTIND- the index to the next argument to be processed,OPTARG- variable is set to any argument for an option found bygetopts,OPTERR(not POSIX) - set to 0 or 1 to indicate if Bash should display error messages generated by thegetopts.
OPTIND- 要处理的下一个参数的索引，OPTARG- 变量设置为由getopts,找到的选项的任何参数OPTERR（不是 POSIX） - 设置为 0 或 1 以指示 Bash 是否应显示由getopts.
Further more, see: Small getopts tutorialat The Bash Hackers Wiki
更多信息，请参阅：The Bash Hackers Wiki 上的Small getopts 教程
回答by Ansgar Wiechers
The argument is stored in the varable $OPTARG.
参数存储在变量中$OPTARG。
function t() {
echo $*
getopts "a:" OPTION
echo $OPTION
echo $OPTARG
}
Output:
输出：
$ t -a bc
-a bc
a
bc