AWK and file names with space in it.
The issue here is parsing with ls. Consider to take a look here: Why you shouldn't parse the output of ls.
The reason why you shouldn't do it is since UNIX allows almost any character in a filename, including whitespace, newlines, commas, pipe symbols, and pretty much anything else you'd ever try to use as a delimiter except NUL. In its default mode, if standard output isn't a terminal, ls separates filenames with newlines. This is fine until you have a file with a newline in its name.
Oh my god, that's awful.
Your script uses bash; I suggest you do this instead:
#!/bin/bash
for i in *.wav; do mv "${i}" "${i%.wav}.ext"; done
See the Bash Guide for more details on parameter expansion.
You could try this.
awk '{print substr($0, index($0,$9))}'
For example this is the output of ls command:
-rw-r--r--. 1 root root 73834496 Dec 6 10:55 File with spaces 2
If you use simple awk like this
# awk '{print $9}'
It returns only
# File
If used with the full command
# awk '{print substr($0, index($0,$9))}'
I get the whole output
File with spaces 2
Here substr(s, a, b) : it returns b number of chars from string s, starting at position a. The parameter b is optional.
For example if the match is addr:192.168.1.133 and you use substr as follows
# awk '{print substr($2,6)}'
You get the IP i.e 192.168.1.133. Note the 6 is the character starting from a in addr
So in the proper command the $2 is $0 ( which prints whole line.) and index($0,$9) matches $9 and prints everything ahead of column 9. You can change that to index($0,$8) and see that the output changes to
# 10:55 File with spaces 2
`index(IN, FIND)' This searches the string IN for the first occurrence of the string FIND, and returns the position in characters where that occurrence begins in the string IN.
I hope it helps. Moreover if you are assigning this value to a variable in script then you need to enclose the variables in double quotes. Other wise you will get errors if you are doing some other operation for the extracted file name.