Question

I am trying to generate SQL filling in a date via command line, using awk’s printf. The code I am using is:

awk 'BEGIN{ printf " convert_tz(time,\047GMT\047,\047America/New_York\047) as timestamp , date_format(convert_tz(time,\047GMT\047,\047America/New_York\047), \047\045Y\045m\045d \045H\045i\045s\047) as dt from table where time >= convert_tz(%s,\047America/New_York\047,\047GMT\047) and time <= convert_tz(%s + interval 1 day ,\047America/New_York\047,\047GMT\047);", "2011-01-01", "2011-01-01"}'

I believe I have the escaping correct, but I get the following result:

awk: fatal: not enough arguments to satisfy format string

Does anyone have an idea of why the %s is not getting caught and populated?

The specific version of awk I’m using is GNU Awk 3.1.6.

Answer 1

Your escaping is a bit off, you can’t replace % with \045 since it’s replaced back to % before printf is called and makes it confused. The way to escape % in printf is to instead use %% and it will work well.

...\047%%Y%%m%%d %%H%%i%%s\047...