Question

Write an awk script to print just the name and size of ordinary files (do not include directories), one on each line. The output must be aligned as shown in the example below. First, analyze a typical output of the ls -l command to see what differentiates an ordinary file from a directory and which fields contain the file name and size information. Then, determine what the field separator is and whether the default field separator of awk will work, or if you have to specify it using the FS= command. Insert this awk script in script1_tXX (preceded by ls -l |) and save it.

Answer 1

The output of "ls -l" looks like the following:

Here, the first character is 'd' for directories, and '-' for ordinary files. The fifth column is the size, and the ninth column is the file name.

Note that the separator for the fields is space.

Hence, the command is

ls -l | awk '{split($1, a, ""); if(a[1]=="-"){print $9, $5}else{}}'

The split function splits the first part which is the "drwxr-xr-x" into characters, and then the "if" condition matches the first character to "-". If it matches, it prints the filename and the size.

Here is the output

Comment in case of any doubts.