Find missing files by their number? - bash

Question

I have a big list of ordered files with names like this
file_1.txt
file_2.txt
file_3.txt
file_6.txt
file_7.txt
file_8.txt
file_10.txt
In this case it is easy to see that files: file_4.txt,file_5.txt and file_9.txt are missing, but if i have a big list how can i find the missing files? i am just learning bash so i just know some simple examples. like this
for i in $(seq 1 1000) ;
do
if [i not in *.txt]; then
echo $i;
done
But this doesnt even work unless i erase the if [i not in *.txt];then line
so it just writes all the numbers between 1 and 1000.
I hope you can help me.
Thanks in advance.

Answer 1

If they are in a file then this should work
awk 'match($0,/([0-9]+)/,a){a[1]>max&&max=a[1];b[a[1]]++}
END{for(i=1;i<max;i++)if(!b[i])print "file_"i".txt"}' file
Output
file_4.txt
file_5.txt
file_9.txt

Answer 2

The suggestion from #user4453924 really helped me out. It does not have to be in a file, just pipe output from ls into his awk command, and you should be fine:
ls *.txt | awk 'match($0,/([0-9]+)/,a){a[1]>max&&max=a[1];b[a[1]]++}
END{for(i=1;i<max;i++)if(!b[i])print "file_"i".txt"}'
Outputs:
file_4.txt
file_5.txt
file_9.txt
Alternatively, if you prefer to do it in a two step fashion, it would be quite simple to pipe the output from ls into a file, and then use his command directly on the file, as it is:
ls *.txt > filelist.txt
awk 'match($0,/([0-9]+)/,a){a[1]>max&&max=a[1];b[a[1]]++}
END{for(i=1;i<max;i++)if(!b[i])print "file_"i".txt"}' filelist.txt

Answer 3

One way to do this is by
## TODO: You need to change the following path:
THELIST=/path/to/input-file
for i in $(seq 1 10);
do
FOUND=`grep "file_$i.txt" "$THELIST"` #look for file $i in $THELIST
#Note: double quotes were placed around $THELIST
# in case there is whitespace in the filename
[[ "$FOUND" == "" ]] && echo $i #if what you found is empty, then output $i
done
You can find info about [[ ... ]] here: What is the difference between single and double square brackets in Bash?
square-brackets