查找不在数字列表中的文件

Question

I have a giant list of files that are all currently numbered in sequential order with different file extensions.

3400.PDF
3401.xls
3402.doc

There are roughly 1400 of these files in a directory. What I would like to know is how to find numbers that do not exist in the sequence.

I've tried to write a bash script for this but my bash-fu is weak.

I can get a list of the files without their extensions by using

FILES=$(ls -1 | sed -e 's/\\..*$//')

but a few places I've seen say to not use ls in this manner. (15 days after asking, I couldn't relocate where I read this, if it existed at all...)

I can also get the first file via ls | head -n 1 ls | head -n 1 but Im pretty sure I'm making this a whole lot more complicated that I need to.

Answer 1

Sounds like you want to do something like this:

shopt -s nullglob
for i in {1..1400}; do 
    files=($i.*)
    (( ${#files[@]} > 0 )) || echo "no files beginning with $i"; 
done

This uses a glob to make an array of all files 1.* , 2.* etc. It then compares the length of the array to 0. If there are no files matching the pattern, the message is printed.

Enabling nullglob is important as otherwise, when there are no files matching the array will contain one element: the literal value '1.*' .

Answer 2

根据已删除的答案，基本上是正确的：

for i in $(seq 1 1400); do ls $i.* > /dev/null 2>&1 || echo $i; done

Answer 3

ls [0-9]* \
| awk -F. '  !seen[$1]++ { ++N }
             END         { for (n=1; N ; ++n) if (!seen[n]) print n; else --N }
'

Will stop when it's filled the last gap, sub in N>0 || n < 3000 N>0 || n < 3000 to go at least that far.