I splitted the huge file and output is several files which start by x character.
I want to rename them and make a list which sorted by name like below:
part-1.gz
part-2.gz
part-3.gz …
I tried below CMD:
for (( i = 1; i <= 3; i++ )) ;do for f in `ls -l | awk '{print $9}' | grep '^x'`; do mv $f part-$i.gz ;done ; done;
for f in `ls -l | awk '{print $9}' | grep '^x'`; do for i in 1 .. 3 ; do mv -- "$f" "${part-$i}.gz" ;done ; done;
for i in 1 .. 3 ;do for f in `ls -l | awk '{print $9}' | grep '^x'`; do mv -- "$f" "${part-$i}.gz" ;done ; done;
for f in `ls -l | awk '{print $9}' | grep '^x'`; do mv -- "$f" "${f%}.gz" ;done
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Answer
Tip: don’t do ls -l if you only need the file names. Even better, don’t use ls at all, just use the shell’s globbing ability. x* expands to all file names starting with x.
Here’s a way to do it:
i=1; for f in x*; do mv $f $(printf 'part-%d.gz' $i); ((i++)); done
This initializes i to 1, and then loops over all file names starting with x in alphabetical order, assigning each file name in turn to the variable f. Inside the loop, it renames $f to $(printf 'part-%d.gz' $i), where the printf command replaces %d with the current value of i. You might want something like %02d if you need to prefix the number with zeros. Finally, still inside the loop, it increments i so that the next file receives the next number.
Note that none of this is safe if the input file names contain spaces, but yours don’t.