我试图找出运行ls -al
的每一行输出是一个文件还是
目录以及它是否被隐藏,并计算每个目录的类型。
编辑:我必须不要使用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: $total"
当前,脚本为
!!!
的每一行输出不匹配任何regex语句的ls -al
,所有计数器变量保持在0
。下面是一个输入示例(尽管bash在完成regex检查之前删除了用于填充的额外空格)。drwx--x--x 37 username groupname 4096 Jan 8 14:37 .
drwxr-xr-x 235 root root 4096 Nov 15 12:16 ..
drwx------ 3 username groupname 4096 Oct 27 14:35 .adobe
-rw------- 1 username groupname 14458 Dec 5 20:24 .bash_history
-rw------- 1 username groupname 2680 Sep 30 16:12 .bash_profile
-rw------- 1 username groupname 1210 Oct 7 09:40 .bashrc
drwx------ 12 username groupname 4096 Dec 6 15:24 .cache
drwxr-xr-x 17 username groupname 4096 Jan 8 14:37 .config
drwx------ 4 username groupname 4096 Dec 5 17:51 dir1
drwx------ 2 username groupname 4096 Nov 23 12:26 dir2
...
我已经在online Regex checker上测试了regex,他们按照我的意愿进行评估。我认为这是bash特有的问题。如有任何帮助,我们将不胜感激。
最佳答案
我花了一段时间,但还是成功了。
我的方法:避免解析ls -l
的输出。特别是在这里你不需要它。启用选项,以便*
循环中的for
可以看到隐藏对象并根据对象类型测试每个对象(使用shopt
)。
还有:a+=1
并不像你想象的那样。它只是在字符串末尾附加1
!
#!/bin/bash
#declare four different regex statements that match files, hidden files, directories and hidden directories (excluding . and ..)
#based on the output of each line of running ls -al
re_hidden_file='^\..*'
#declare four different counters for each type
file_count=0
hidden_file_count=0
directory_count=0
hidden_directory_count=0
# enable hidden files/directories
shopt -s dotglob
#read through the output of ls -al line by line, assigning x the value of each line
for x in * ; do
#test if each line matches each of the regex statements, if it does then increment the relevant counter
if [ -d "$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"