bash 脚本：如何测试列表成员资格
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bash scripting: How to test list membership
提问by kjetil b halvorsen
(This is debian squeeze amd64)
（这是 debian 挤压 amd64）
I need to test if a file is a member of a list of files. So long my (test) script is:
我需要测试一个文件是否是文件列表的成员。这么长时间我的（测试）脚本是：
set -x
array=$( ls )
echo $array
FILE=log.out
# This line gives error!
if $FILE in $array
then echo "success!"
else echo "bad!"
fi
exit 0
?Any ideas?
？有任何想法吗？
Thanks for all the responses. To clarify: The script given is only an example, the actual problem is more complex. In the final solution, it will be done within a loop, so I need the file(name) to be tested for to be in a variable.
感谢所有的回应。澄清：给出的脚本只是一个例子，实际问题更复杂。在最终的解决方案中，它将在循环中完成，因此我需要测试文件（名称）是否在变量中。
Thanks again. No my test-script works, and reads:
再次感谢。没有我的测试脚本有效，并读取：
in_list() {
local search=""
shift
local list=("$@")
for file in "${list[@]}" ; do
[[ "$file" == "$search" ]] && return 0
done
return 1
}
#
# set -x
array=( * ) # Array of files in current dir
# echo $array
FILE="log.out"
if in_list "$FILE" "${array[@]}"
then echo "success!"
else echo "bad!"
fi
exit 0
回答by moinudin
if ls | grep -q -x t1 ; then
echo Success
else
echo Failure
fi
grep -xmatches full lines only, so ls | grep -xonly returns something if the file exists.
grep -x仅匹配整行，因此ls | grep -x仅在文件存在时才返回内容。
回答by glenn Hymanman
If you just want to check if a file exists, then
如果你只是想检查一个文件是否存在，那么
[[ -f "$file" ]] && echo yes || echo no
If your array contains a list of files generated by some means other than ls, then you have to iterate over it as demonstrated by Sorpigal.
如果您的数组包含通过除 之外的其他方式生成的文件列表ls，那么您必须按照 Sorpigal 的演示对其进行迭代。
回答by sorpigal
How about
怎么样
in_list() {
local search=""
shift
local list=("$@")
for file in "${list[@]}" ; do
[[ $file == $search ]] && return 0
done
return 1
}
if in_list log.out * ; then
echo 'success!'
else
echo 'bad!'
fi
EDIT: made it a bit less idiotic.
编辑：让它不那么愚蠢。
EDIT #2: Of course if all you're doing is looking in the current directory to see if a particular file is there, which is effectively what the above is doing, then you can just say
编辑#2：当然，如果您所做的只是查看当前目录以查看是否存在特定文件，这实际上就是上面所做的，那么您可以说
[ -e log.out ] && echo 'success!' || echo 'bad!'
If you're actually doing something more complicated involving lists of files then this might not be sufficient.
如果您实际上正在做一些涉及文件列表的更复杂的事情，那么这可能还不够。
