Shell Programming and Scripting

Hi team,

We have few files landing to our server based on sequence number. These files have to be processed in the sequence number order. Once the sequence number has reached its maximum, the files with sequence number 0000 has to be processed.
For example:

Code:

IN9997
IN9998
IN9999
IN0000
IN0001

Above is the correct order in which the files should be processed.

But 'ls' command processed the files in the wrong order as given below,

Code:

IN0000
IN0001
IN9997
IN9998
IN9999

Which is not recommended as there will be a seq number gap

ls | sort -rn also gives the wrong output.

Code:

IN9999
IN9998
IN9997
IN0000
IN0001

Please suggest ways to get the order as highlighted in red above.

Moderator's Comments:

Please use code tags next time for your code and data.

you can try by modified date as..

Code:

ls -ltr

The file time stamp might vary. The requirement is sort only based on the file name,.

try this....

Code:

ls | sed 's/IN00/INLS/g' |  sort | sed 's/INLS/IN00/g'

This can do your job....

but not the best solution.....

Code:

ls | awk 'NR==1{x=substr($1,3);i=1;a[i]=$1;i++;f=0;next}(f==0){y=substr($1,3);if(y-x>1){f=1}else{a[i]=$1;i++};x=y}(f==1){print}END{for(j=1;j<i;j++)print a[j]}'

---------- Post updated at 01:10 PM ---------- Previous update was at 01:09 PM ----------

output of ls should be

Code:

IN0000
IN0001
IN9997
IN9998
IN9999

Required output is --

Code:

IN9997
IN9998
IN9999
IN0000
IN0001

code output is this only

Code:

IN9997
IN9998
IN9999
IN0000
IN0001

the output i mentioned is for ls