Remove commas and format dates - awk

Question

I have a large file with entries such as:
<VAL>17,451.26</VAL>
<VAL>353.93</VAL>
<VAL>395.00</VAL>
<VAL>2,405.00</VAL>
<DATE>31 Jul 2013</DATE>
<DATE>31 Jul 2013</DATE>
<DATE>31 Dec 2014</DATE>
<DATE>21 Jun 2002</DATE>
<DATE>10 Jul 2002</DATE>
<MOD>PL</MOD>
<BATCH>13382</BATCH>
<TYPE>Invoice</TYPE>
<REF1>13541/13382</REF1>
<REF2>671042638320</REF2>
<NOTES>a-07 final elec</NOTES>
<SNAME>EDF ENERGY ( Electricity )</SNAME>
<VAL>55.22</VAL>
</CLT>
<CLT>
<CHD>MAT-01</CHD>
<OPN>U5U1</OPN>
<PERIOD>07 2013</PERIOD>
<DATE>13 Jun 2013</DATE>
<DATE>10 Jul 2002</DATE>
<DATE>10 Jul 2002</DATE>
<DATE>21 Aug 2007</DATE>
<DATE>10 Jul 2002</DATE>
<VAL>-4,122,322.03</VAL>
I need to remove the commas in the VAL fields and change the dates to YYYY-MM-DD (e.g. 2013-07-31) in the DATE fields.
Looking for a quick (efficient) way of doing this.
Thanks

Answer 1

This should get you started:
awk -F"[<>]" 'BEGIN {split("Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec",month," ");for (i=1;i<=12;i++) mdigit[month[i]]=i} /<VAL>/ {gsub(/\,/,"")} /<DATE>/ {split($3,a," ");$0=sprintf("<DATE>%s-%02d-%02d</DATE>",a[3],mdigit[a[2]],a[1])}1' file
<VAL>17451.26</VAL>
<VAL>353.93</VAL>
<VAL>395.00</VAL>
<VAL>2405.00</VAL>
<DATE>2013-07-31</DATE>
<DATE>2013-07-31</DATE>
<DATE>2014-12-31</DATE>
<DATE>2002-06-21</DATE>
<DATE>2002-07-10</DATE>
<MOD>PL</MOD>
<BATCH>13382</BATCH>
<TYPE>Invoice</TYPE>
<REF1>13541/13382</REF1>
<REF2>671042638320</REF2>
<NOTES>a-07 final elec</NOTES>
<SNAME>EDF ENERGY ( Electricity )</SNAME>
<VAL>55.22</VAL>
</CLT>
<CLT>
<CHD>MAT-01</CHD>
<OPN>U5U1</OPN>
<PERIOD>07 2013</PERIOD>
<DATE>2013-06-13</DATE>
<DATE>2002-07-10</DATE>
<DATE>2002-07-10</DATE>
<DATE>2007-08-21</DATE>
<DATE>2002-07-10</DATE>
<VAL>-4122322.03</VAL>

Answer 2

sed '# init month convertor in holding buffer
1{h;s/.*/Jan01Fev02Mar03Apr04May05Jun06Jul07Aug08Sep09Oct10Nov11Dec12/;x;}
# change Val
/^<VAL>/ s/,//g
# Change Date
/^<DATE>/ {
# change month
G
s/[[:space:]]\{1,\}\([A-Z][a-z][a-z]\)[[:space:]]\{1,\}\(.*\)\n.*\1\([0-9][0-9]\).*/-\3-\2/
# reformat order
s/>\(.*\)-\(.*\)-\(.*\)</>\3-\2-\1</
}' YourFile
posix sed with not extra sub shell for dae conversion
reformat date take 2 s///here but could be merged in 1 s/// a bit more unreadeable (already very attractive regex like this)
could easily add some security feature about source date like bad date format

Answer 3

Your input seems like XML. I'd use a proper XML handling tool, e.g. XML::XSH2, a wrapper around Perl's XML::LibXML:
open file.xml ;
for //VAL set . xsh:subst(., ',', '','g') ;
perl { use Time::Piece } ;
for my $d in //DATE {
$t = $d/text() ;
set $d/text() { Time::Piece->strptime($t, '%d %b %Y')->ymd } ;
}
save :b ;

Answer 4

This might work for you (GNU sed & bash):
sed -r '/^<VAL>/s/,//g;/^(<DATE>)(.*)(<\/DATE>)$/s//echo "\1"$(date -d "\2" +%F)"\3"/e' file
This removes all commas on a line starting <VAL> and for those lines that contain date tags, uses the date utility and the evaluate flag in the substitution command to rearrange the date to YYYY-MM-DD.
An alternative solution, using only seds commands:
sed -r '/^<VAL>/s/,//g;/^<DATE>/!b;s/$/\nJan01Feb02Mar03Apr04May05Jun06Jul07Aug08Sep09Oct10Nov11Dec12/;s/^(<DATE>)(..) (...) (....)(<\/DATE>\n).*\3(..)/\1\4-\6-\2\5/;P;d' file
Appends a lookup to the end of the date line and uses regexp to rearrange the output.