我试图找出运行ls -al的每一行输出是文件还是
目录以及它是否被隐藏并计算每个目录的类型。
编辑：我必须不使用find。
#!/bin/bash
#declare four different regex statements that match files, hidden files, directories and hidden directories (excluding . and ..)
#based on the output of each line of running ls -al
re_file='^\-[rwx\-]{9}\s[0-9]+\s([a-z_][a-z0-9_]{0,30})\s([a-z_][a-z0-9_]{0,30})\s[0-9]+\s\w{3}\s[0-9]+\s[0-9]{2}:[0-9]{2}\s[^\.](\w|\.)*$'
re_hidden_file='^\-[rwx\-]{9}\s[0-9]+\s([a-z_][a-z0-9_]{0,30})\s([a-z_][a-z0-9_]{0,30})\s[0-9]+\s\w{3}\s[0-9]+\s[0-9]{2}:[0-9]{2}\s\.\w(\w|\.)*$'
re_directory='^d[rwx\-]{9}\s[0-9]+\s([a-z_][a-z0-9_]{0,30})\s([a-z_][a-z0-9_]{0,30})\s[0-9]+\s\w{3}\s[0-9]+\s[0-9]{2}:[0-9]{2}\s[^\.](\w|\.)*$'
re_hidden_directory='^d[rwx\-]{9}\s[0-9]+\s([a-z_][a-z0-9_]{0,30})\s([a-z_][a-z0-9_]{0,30})\s[0-9]+\s\w{3}\s[0-9]+\s[0-9]{2}:[0-9]{2}\s\.\w(\w|\.)*$'
#declare four different counters for each type
file_count=0
hidden_file_count=0
directory_count=0
hidden_directory_count=0
#read through the output of ls -al line by line, assigning x the value of each line
ls -al $1 | while read x; do
#test if each line matches each of the regex statements, if it does then increment the relevant counter
if [[ $x =~ $re_file ]] ; then
file_count+=1
elif [[ $x =~ $re_hidden_file ]] ; then
hidden_file_count+=1
elif [[ $x =~ $re_directory ]] ; then
directory_count+=1
elif [[ $x =~ $re_hidden_directory ]] ; then
hidden_directory_count+=1
else
echo "!!!"
fi
done
total=$((file_count + hidden_file_count + directory_count + hidden_directory_count))
echo "Files found: $file_count (plus $hidden_file_count hidden)"
echo "Directories found: $directory_count (plus $hidden_directory_count hidden)"
echo "Total files and directories: $total"
目前，脚本输出的!!!不匹配ls -al的每一行的任何正则表达式语句，并且所有计数器变量都保留在0。这是输入的一个示例（尽管Bash在完成正则表达式检查之前删除了用于填充的额外空格）。
drwx--x--x 37 username groupname 4096 Jan 8 14:37 .
drwxr-xr-x 235 root root 4096 Nov 15 12:16 ..
drwx------ 3 username groupname 4096 Oct 27 14:35 .adobe
-rw------- 1 username groupname 14458 Dec 5 20:24 .bash_history
-rw------- 1 username groupname 2680 Sep 30 16:12 .bash_profile
-rw------- 1 username groupname 1210 Oct 7 09:40 .bashrc
drwx------ 12 username groupname 4096 Dec 6 15:24 .cache
drwxr-xr-x 17 username groupname 4096 Jan 8 14:37 .config
drwx------ 4 username groupname 4096 Dec 5 17:51 dir1
drwx------ 2 username groupname 4096 Nov 23 12:26 dir2
...
我在online Regex checker上测试了正则表达式，他们按照我的意愿进行评估。我认为这是一个特定于Bash的问题。任何帮助表示赞赏。
答案 0 :(得分：2)
您不应解析find来获取文件。使用ls代替nul终止或通配。
问题是$ touch a$'\t'b
$ touch a$'\n'b
$ ls -l a*
-rw-r--r-- 1 andrew wheel 0 Jan 8 08:25 a?b
-rw-r--r-- 1 andrew wheel 0 Jan 8 08:26 a?b
会为文件名产生不明确的输出，否则这些文件名是合法的文件名。考虑：
\t \n和?的不可打印字符将替换为ls，并使$ touch "a b c "
$ touch "a b c "
$ ls -al a\ b*
-rw-r--r-- 1 andrew wheel 0 Jan 8 08:44 a b c
-rw-r--r-- 1 andrew wheel 0 Jan 8 08:44 a b c
中的文件不明确。
尾随空格会发生同样的情况：
find现在考虑使用$ find . -name "a*" -maxdepth 1 -print0 | xargs -0 printf "'%s'\n"
'./a b'
'./a
b'
'./a b c '
'./a b c '
：
$ for fn in a*; do printf "'%s'\n" "$fn"; done
'a b'
'a
b'
'a b c '
'a b c '
或者只是全球化：
file_count=0
hidden_file_count=0
regular_directory_count=0
hidden_directory_count=0
echo "=====regular files and directories:"
for fn in *; do
printf "'%s'\n" "$fn"
if [ -d "$fn" ]; then
regular_directory_count=$((regular_directory_count+1))
else
file_count=$((file_count+1))
fi
done
echo "====hidden files and direcotries:"
for fn in .*; do
printf "'%s'\n" "$fn";
if [ -d "$fn" ]; then
hidden_directory_count=$((hidden_directory_count+1))
else
hidden_file_count=$((hidden_file_count+1))
fi
done
printf "Regular files: %s regular directories: %s\n" $file_count $regular_directory_count
printf "Hidden files: %s hidden directories: %s\n" $hidden_file_count $hidden_directory_count
tf=$((hidden_file_count+file_count))
td=$((hidden_directory_count+regular_directory_count))
printf "Total files: %s total directories: %s\n" $tf $td
如果你想获得包含隐藏文件和目录的总目录和总文件，只需将其添加到你的glob模式中：
$ ls -la
total 0
drwxr-xr-x 9 andrew wheel 306 Jan 8 11:07 .
drwxrwxrwt 92 root wheel 3128 Jan 8 10:58 ..
drwxr-xr-x 2 andrew wheel 68 Jan 8 11:07 .hidden dir
-rw-r--r-- 1 andrew wheel 0 Jan 8 11:26 .hidden file
-rw-r--r-- 1 andrew wheel 0 Jan 8 11:26 a?b
-rw-r--r-- 1 andrew wheel 0 Jan 8 11:26 a?b
-rw-r--r-- 1 andrew wheel 0 Jan 8 11:26 a b c
-rw-r--r-- 1 andrew wheel 0 Jan 8 11:26 a b c
drwxr-xr-x 2 andrew wheel 68 Jan 8 11:07 regular dir
假设：
=====regular files and directories:
'a b'
'a
b'
'a b c '
'a b c '
'regular dir'
====hidden files and direcotries:
'.'
'..'
'.hidden dir'
'.hidden file'
Regular files: 4 regular directories: 1
Hidden files: 1 hidden directories: 3
Total files: 5 total directories: 4
运行它，你得到：
.如果要排除..和GLOBIGNORE=".:.."个隐藏目录，可以在使用.* glob模式之前设置id。
答案 1 :(得分：2)
我花了一段时间才开始工作。
我的方法：避免解析ls -l的输出。特别是在这里你不需要它。启用选项，以便*循环中的for看到隐藏的对象，并根据对象类型测试每个对象（使用shopt）。
另外：a+=1没有按照您的想法行事。它只是在字符串的末尾附加1！
#!/bin/bash
#declare four different regex statements that match files, hidden files, directories and hidden directories (excluding . and ..)
#based on the output of each line of running ls -al
re_hidden_file='^\..*'
#declare four different counters for each type
file_count=0
hidden_file_count=0
directory_count=0
hidden_directory_count=0
# enable hidden files/directories
shopt -s dotglob
#read through the output of ls -al line by line, assigning x the value of each line
for x in * ; do
#test if each line matches each of the regex statements, if it does then increment the relevant counter
if [ -d "$x" ] ; then
if [[ "$x" =~ $re_hidden_file ]] ; then
hidden_directory_count=$((hidden_directory_count+1))
else
directory_count=$((directory_count+1))
fi
else
if [[ "$x" =~ $re_hidden_file ]] ; then
hidden_file_count=$((hidden_file_count+1))
else
file_count=$((file_count+1))
fi
fi
done
total=$((file_count + hidden_file_count + directory_count + hidden_directory_count))
echo "Files found: $file_count (plus $hidden_file_count hidden)"
echo "Directories found: $directory_count (plus $hidden_directory_count hidden)"
echo "Total files and directories: $total"