问题描述
我正在寻找一种确定函数是否正确的正确方法.符合POSIX的方式.
I'm after THE proper way to see if a function is defined or not. A POSIX compliant way.
__function_defined() {
FUNC_NAME=$1
d=$(declare -f $FUNCNAME)
if [ "${DISTRO_NAME_L}" = "centos" ]; then
if typeset -f $FUNC_NAME &>/dev/null ; then
echo " * INFO: Found function $FUNC_NAME"
return 0
fi
# Try POSIXLY_CORRECT or not
elif test -n "${POSIXLY_CORRECT+yes}"; then
if typeset -f ${FUNC_NAME} >/dev/null 2>&1 ; then
echo " * INFO: Found function $FUNC_NAME"
return 0
fi
else
# Arch linux seems to fall here
if $( type ${FUNC_NAME} >/dev/null 2>&1 ) ; then
echo " * INFO: Found function $FUNC_NAME"
return 0
fi
echo " * INFO: $FUNC_NAME not found...."
return 1
}
根据debian的 checkbashisms 脚本,以上所有内容均被视为bash'ism.
All of the above are considered bash'isms according to debian's checkbashisms script.
尝试grep脚本也不行.例如:
Trying to grep the script is also not ok. for example:
if [ "$(grep $FUNC_NAME $(dirname $0)/$(basename $0))x" != "x" ]; then
# This is really ugly and counter producing but it was, so far, the
# only way we could have a POSIX compliant method to find if a function
# is defined within this script.
echo " * INFO: Found function $FUNC_NAME"
return 0
fi
不起作用,因为脚本也应该像这样工作:
won't work because the script is also supposed to work like:
wget --no-check-certificate -O - http://URL_TO_SCRIPT | sudo sh
那么,执行此操作的正确的,符合POSIX的正确方法是什么?
So, what's the proper, POSIX compliant way to do this?
请简单使用sh,不要使用bash,ksh,不要使用其他shell,而应该使用普通sh.哦,也没有实际尝试运行该功能:)
Plain sh please, no bash, no ksh, no other shell, just plain sh. Oh, and without actually trying to run the function too :)
有可能吗?
if [ "$(command -v $FUNC_NAME)x" != "x" ]; then
echo " * INFO: Found function $FUNC_NAME"
return 0
fi
现在的问题,是否有更好的解决方案?
推荐答案
我找到了一种POSIX兼容方式
if [ "$(command -v $FUNC_NAME)x" != "x" ]; then
echo " * INFO: Found function $FUNC_NAME"
return 0
fi
现在的问题,是否有更好的解决方案?
这篇关于符合POSIX的方法来查看是否在sh脚本中定义了功能的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!