Shell Programming and Scripting

Hi...

I have an inputfile name as :- abc_test_20120213.dat (date in yyyymmdd format)

I need the output file name as abc_test_13022012.dat (date in ddmmyyyy format)

Please help me on this...


Thanks in advance.

Code:

x='abc_test_20120213.dat'
y=${x%_*}
z=$(date -d `echo ${x%.dat} | cut -d_ -f3` +%d%m%Y)
echo "${y}_$z.dat"

Code:

$ x="abc_test_20120213.dat"

$ y=`echo $x | sed 's/\(.*_\)\(....\)\(..\)\(..\)/\1\4\3\2/'`

$ echo $y
abc_test_13022012.dat

Guru.

Thanks Guru,

Could you please explain this pattern sed 's/\(.*_\)\(....\)\(..\)\(..\)/\1\4\3\2/'` , It will be very helpful.

Hi Kalpeer,

\( ..... \) :- this is called save operator, what ever you will put in side this like \(abc\), will be stored in unix system buffer(\1..\9). There are 9 system buffers present in unix which is used by shell to store some value for later use.

Here in this command:- sed 's/search/replace/'
1) sed 's/\(.*_\)\(....\)\(..\)\(..\)/\1\4\3\2/' :- the highlighted save operator will hold .*_ means "abc_test_" and will be stored in \1 (first system buffer)
2) sed 's/\(.*_\)\(....\)\(..\)\(..\)/\1\4\3\2/' :- the highlighted save operator will hold .... means 2012 and will be stored in \2 system buffer.

in this way month .. will be stored in \3 system buffer and day will be stored in \4 system buffer.

then you will replace the whole content with \1 means "abc_test_", \4 means day "13", \3 means month"02", \4 means year"2012". So finally you will get abc_test_13022012.dat

Please reply me if you r not clear abt save operator.... thnx

got it..Thanks a lot guru.