Change date format in first column using awk/sed

awk, linux, sed

Solution

It looks like you're using two different flavors (versions) of `date`. To check which versions you've got, I think `GNU date` accepts the `--version` flag whereas other versions, like `BSD/OSX` will not accept this flag.

Since you may be using completely different systems, it's probably safest to avoid `date` completely and use `perl` to print the current date:

perl -MPOSIX -e 'print POSIX::strftime("%d/%m/%y", localtime) . "\n"'

If you are sure you have `GNU awk` on both machines, you could use it like this:

awk 'BEGIN { print strftime("%d/%m/%y") }'

To fix the file you've got, here's my take using `GNU awk`:

awk '{ print gensub(/^(..\/)(..\/)..(..,)/, "\\2\\1\\3", "g"); next }1' file

Or using `sed`:

sed 's/^\(..\/\)\(..\/\)..\(..,\)/\2\1\3/' file

Results:

29/11/12,9654.80,194.32,2.01,7.19,-7.89,7.65,7.57,3.98,9625.27,160.10,1.66,4.90,-4.79,6.83,4.84,3.54                
03/12/12,5184.22,104.63,2.02,6.88,-6.49,7.87,6.67,4.10,5169.52,93.81,1.81,5.29,-5.45,7.87,5.37,4.10                
04/12/12,5183.65,103.18,1.99,6.49,-6.80,8.40,6.66,4.38,5166.04,95.44,1.85,6.04,-6.49,8.40,6.28,4.38                
07/11/12,5183.65,102.15,1.97,6.78,-6.36,8.92,6.56,4.67,5169.48,96.67,1.87,5.56,-6.10,8.92,5.85,4.67                
11/07/12,5179.39,115.57,2.23,7.64,-6.61,8.83,7.09,4.62,5150.17,103.52,2.01,7.01,-6.08,8.16,6.51,4.26                
26/11/12,5182.66,103.30,1.99,7.07,-5.76,7.38,6.37,3.83,5162.81,95.47,1.85,6.34,-5.40,6.65,5.84,3.44                
30/11/12,5180.82,95.19,1.84,6.51,-5.40,7.91,5.92,4.12,5163.98,91.82,1.78,5.58,-5.07,7.05,5.31,3.65

Problem

I have a shell script which is automatically ran each morning which appends that days results to a text file. The file should have todays date on the first column followed by results separated by commas. I use the command date +%x to get the day in the required format (dd/mm/yy). However on one computer date +%x returns mm/dd/yyyy ( any idea why this is the case?). I then sort the data in the file in date order. Here is a snippet of such a text file ``` 29/11/12,9654.80,194.32,2.01,7.19,-7.89,7.65,7.57,3.98,9625.27,160.10,1.66,4.90,-4.79,6.83,4.84,3.54 03/12/12,5184.22,104.63,2.02,6.88,-6.49,7.87,6.67,4.10,5169.52,93.81,1.81,5.29,-5.45,7.87,5.37,4.10 04/12/12,5183.65,103.18,1.99,6.49,-6.80,8.40,6.66,4.38,5166.04,95.44,1.85,6.04,-6.49,8.40,6.28,4.38 11/07/2012,5183.65,102.15,1.97,6.78,-6.36,8.92,6.56,4.67,5169.48,96.67,1.87,5.56,-6.10,8.92,5.85,4.67 07/11/2012,5179.39,115.57,2.23,7.64,-6.61,8.83,7.09,4.62,5150.17,103.52,2.01,7.01,-6.08,8.16,6.51,4.26 11/26/2012,5182.66,103.30,1.99,7.07,-5.76,7.38,6.37,3.83,5162.81,95.47,1.85,6.34,-5.40,6.65,5.84,3.44 11/30/2012,5180.82,95.19,1.84,6.51,-5.40,7.91,5.92,4.12,5163.98,91.82,1.78,5.58,-5.07,7.05,5.31,3.65 ``` Is it possible to change the date format for the latter four lines to the correct date format using awk or sed? I only wish to change the date format for those in the form mm/dd/yyyy to dd/mm/yy.

Original source