Super User is a question and answer site for computer enthusiasts and power users. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm attempting to use wget in a simple bash script to grab a jpeg image from an Axis camera. This script outputs a file named JPEGOUT, instead of the desired output, which should be a timestamp jpeg (ex: 201209292040.jpg) . Changing the variable in the wget statement from JPEGOUT to $JPEGOUT makes wget fail with "wget: missing URL" error.

The weird thing is wget parses the $IP vairable correctly. No luck on the output file name. I've tried single quotes, double quotes, parenthesis: all to no luck.

Here's the script



JPEGOUT= date +%Y%m%d%H%M.jpg

wget -O JPEGOUT http://$IP/axis-cgi/jpg/image.cgi?resolution=640x480&compression=25

Any ideas on how to get the output file name to parse correctly?

share|improve this question
up vote 4 down vote accepted

JPEGOUT= date +%Y%m%d%H%M.jpg throws an error. Try:



JPEGOUT=$(date +%Y%m%d%H%M.jpg)

wget -O $JPEGOUT http://$IP/axis-cgi/jpg/image.cgi?resolution=640x480&compression=25
share|improve this answer
That did it! Thanks! – Bill Westrup Sep 30 '12 at 2:08

Use command substitution to run the date command and grab the output:

JPEGOUT=`date +%Y%m%d%H%M.jpg`
share|improve this answer

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .