database debugging

database debugging

[Jude DaShiell]

The last field in each record ought to be a number from 0-6 inclusive 
followed by a );.  I may have one of those numbers wrong somewhere.  I was 
trying to use grep to show me whatever character immediately preceeds the 
); characters and couldn't figure out anything to pass to grep for any 
character that would return any results.  In dos, that would just be a ?, 
but that certainly doesn't work with grep.



jude <jdashiel@shellworld.net> Twitter: @jdashiel

Re: database debugging

[Geoff Shang]

On Wed, 22 Oct 2014, Jude DaShiell wrote:

> The last field in each record ought to be a number from 0-6 inclusive
> followed by a );.  I may have one of those numbers wrong somewhere.  I was
> trying to use grep to show me whatever character immediately preceeds the
> ); characters and couldn't figure out anything to pass to grep for any
> character that would return any results.  In dos, that would just be a ?,
> but that certainly doesn't work with grep.

First, grep will show yu the entire line.  You'd need to use something 
like cut or one of those other text manipulation commands to just show yu 
the end of the line.

You could get it to show you any offending lines by piping your output to:

grep -v '[0-6]);'

But it would surely be easier to use a select statement to find the 
offending entries for you.

SELECT <whatever> WHERE NOT REGEXP '[0-6]$';

See http://dev.mysql.com/doc/refman/5.1/en/regexp.html (you can click 
through to the version appropriate for your version of MySQL).

HTH,
Geff.

database debugging

[Jude DaShiell]

I found out what I can't do with grep, I can do with cut.  I used cut -b 
54-56 to get the last three characters of each record output to the 
terminal.



jude <jdashiel@shellworld.net> Twitter: @jdashiel