unicode - Using awk to remove the Byte-order mark

How would an awk script (presumably a one-liner) for removing a BOM look like?

Specification:

• print every line after the first (NR > 1)


• for the first line: If it starts with #FE #FF or #FF #FE, remove those and print the rest

Answer

Try this:

awk 'NR==1{sub(/^\xef\xbb\xbf/,"")}{print}' INFILE > OUTFILE

On the first record (line), remove the BOM characters. Print every record.

Or slightly shorter, using the knowledge that the default action in awk is to print the record:

awk 'NR==1{sub(/^\xef\xbb\xbf/,"")}1' INFILE > OUTFILE

1 is the shortest condition that always evaluates to true, so each record is printed.

Enjoy!

-- ADDENDUM --

Unicode Byte Order Mark (BOM) FAQ includes the following table listing the exact BOM bytes for each encoding:

Bytes         |  Encoding Form
--------------------------------------
00 00 FE FF   |  UTF-32, big-endian
FF FE 00 00   |  UTF-32, little-endian

FE FF         |  UTF-16, big-endian
FF FE         |  UTF-16, little-endian
EF BB BF      |  UTF-8

Thus, you can see how \xef\xbb\xbf corresponds to EF BB BF UTF-8 BOM bytes from the above table.