Removing the Last Token From a String in Bash with awk

Let’s say that you have some number of files for which you want to create a containing directory that is named with all but the last token of the file name, and you want to remove just the last token to create the name of the directory.

Much easier to explain with an example.  Given this list of files:

ls -1
foo_10_10_sometrash
foo_1_sometrash
foo_2_sometrash
foo_3_sometrash
foo_4_sometrash
foo_5_5_sometrash
foo_5_sometrash
foo_6_6_sometrash
foo_7_7_sometrash
foo_8_8_sometrash
foo_9_9_sometrash

You want to create a directory for each of the files as follows:

foo_5_sometrash should have a directory named foo_5.

Further, let’s assume that you have thousands, or hundreds of thousands of files.  In that case doing it via a script while you get a cup of coffee is the preferred solution.

The work will be done within a for i in loop iterating over the output from ls with a nested awk command.

for i in `ls -1`; do DIRNAME=$(echo $i | awk -F_ '{$NF=""; print $0}' | sed 's/ /_/g' | sed 's/_$//g'); mkdir $DIRNAME; done

Here is the command broken down:

DIRNAME=$(....)

will set the var $DIRNAME to the result of the code within the parenthesis.

awk -F_ '{$NF=""; print $0}

will set the field separator to a ‘_’, the character on which you will be ‘splitting’ your string.  $NF="" will set the last field to an empty string and then the print $0 will print the entire input line.

The following sed commands will replace the spaces generated by the awk command with the original separators, ‘_’, and then remove the spurious, trailing ‘_’.