bash 目录中每个文件夹的 for 循环，不包括其中一些
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for-loop for every folder in a directory, excluding some of them
提问by RafaelGP
Thank you very much in advance for helping!
非常感谢您的帮助！
I have this code in bash:
我在 bash 中有这个代码：
for d in this_folder/*
do
plugin=$(basename $d)
echo $plugin'?'
read $plugin
done
Which works like a charm. For every folders inside 'this_folder', echo it as a question and store the input into a variable with the same name.
这就像一个魅力。对于“this_folder”中的每个文件夹，将其作为问题回显并将输入存储到同名变量中。
But now I'd like to exclude some folders, so for example, it will ask for every folder in that directory, ONLY if they are NOT any of the following folders: global, plugins, and css.
但是现在我想排除一些文件夹，例如，它会询问该目录中的每个文件夹，前提是它们不是以下任何文件夹：global、plugins 和 css。
Any ideas how can I achieve this?
任何想法我怎样才能实现这一目标？
Thanks!
谢谢！
UPDATE:
更新：
This is how the final code looks like:
这是最终代码的样子：
base="coordfinder|editor_and_options|global|gyro|movecamera|orientation|sa"
> vt_conf.sh
echo "# ========== Base" >> vt_conf.sh
for d in $orig_include/@($base)
do
plugin=$(basename $d)
echo "$plugin=y" >> vt_conf.sh
done
echo '' >> vt_conf.sh
echo "# ========== Optional" >> vt_conf.sh
for d in $orig_include/!($base)
do
plugin=$(basename $d)
echo "$plugin=n" >> vt_conf.sh
done
回答by Thor
If you have a recent version of bash, you can use extended globs (shopt -s extglob):
如果您有最新版本的 bash，则可以使用扩展的 globs ( shopt -s extglob)：
for d in this_folder/!(global|plugins|css)/
do
plugin=$(basename "$d")
echo $plugin'?'
read $plugin
done
回答by choroba
You can use continueto skip one iteration of the loop:
您可以使用continue跳过循环的一次迭代：
for d in this_folder/*
do
plugin=$(basename $d)
[[ $plugin =~ ^(global|plugins|css)$ ]] && continue
echo $plugin'?'
read $plugin
done
回答by askmish
If you meant to exclude only the directories named global, css, plugins. This might not be an elegant solution but will do what you want.
如果您打算仅排除名为 global、css、plugins 的目录。这可能不是一个优雅的解决方案，但会做你想做的。
for d in this_folder/*
do
flag=1
#scan through the path if it contains that string
for i in "/css/" "/plugins/" "/global/"
do
if [[ $( echo "$d"|grep "$i" ) && $? -eq 0 ]]
then
flag=0;break;
fi
done
#Only if the directory path does NOT contain those strings proceed
if [[ $flag -eq 0 ]]
then
plugin=$(basename $d)
echo $plugin'?'
read $plugin
fi
done
回答by execjosh
You could use findand awkto build the list of directories and then store the result in a variable. Something along the lines of this (untested):
您可以使用find和awk来构建目录列表，然后将结果存储在变量中。类似的东西（未经测试）：
dirs=$(find this_folder -maxdepth 1 -type d -printf "%f\n" | awk '!match(while read -r d; do
# ...
done < <(gfind -maxdepth 1 -type d -printf "%f\n" | awk '!match(##代码##,/^(global|plugins|css)$/)')
,/^(global|plugins|css)$/)')
for d in $dirs; do
# ...
done
Update 2019-05-16:
2019-05-16 更新：
##代码##