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Need to replace the date inside a node of several rdf files


 
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# 1  
Need to replace the date inside a node of several rdf files

Hi,
I have a rdf zip file. This zip file consists of several *.rdf files.
I need to replace the date (this is different for each rdf) inside the node "Date_de_Publication_Periodique" of these rdf files.
e.g.,
Code:
 
awk '/Date_de_Publication_Periodique/ && /XMLSchema#date/' MM_NN-A1B1C1_ABC.rdf

gives:
Code:
 
 <j.1:Date_de_Publication_Periodique rdf:datatype="http://www.w3.org/2001/XMLSchema#date">2013-10-30</j.1:Date_de_Publication_Periodique>

I need to replace this date: 2013-10-30 with my desired date.
With the shell script, I want to unzip the file, take out the strings (from the rdfs) for the mentioned node containing the date and replace it with my desired date (fixed date) for all the rdf files and then zip it back.
Following are my failed attempts:

Code:
awk '/Date_de_Publication_Periodique/ && /XMLSchema#date/' MM_NN-A1B1C1_ABC.rdf | awk '{gsub(/\d{4}\-\d{2}-\d{2}/,"JJJJJJ"); print}'

Code:
 
awk '/Date_de_Publication_Periodique/ && /XMLSchema#date/' MM_NN-A1B1C1_ABC.rdf | awk '{sub(/\d{4}\-\d{2}-\d{2}/,"JJJJJJ"); print }'

Need suggestions.
# 2  
Try:

Code:
awk '/Date_de_Publication_Periodique/ && /XMLSchema#date/ {sub(/[[:digit:]]{4}-[[:digit:]]{2}-[[:digit:]]{2}/,"JJJJJJ"); print} MM_NN-A1B1C1_ABC.rdf

# 3  
Quote:
Originally Posted by Chubler_XL
Try:

Code:
awk '/Date_de_Publication_Periodique/ && /XMLSchema#date/ {sub(/[[:digit:]]{4}-[[:digit:]]{2}-[[:digit:]]{2}/,"JJJJJJ"); print} MM_NN-A1B1C1_ABC.rdf


Thanks for the reply but your code is not working:

Code:
 
awk '/Date_de_Publication_Periodique/ && /XMLSchema#date/ {sub(/[[:digit:]]{4}-[[:digit:]]{2}-[[:digit:]]{2}/,"JJJJJJ"); print}' MM_NN-A1B1C1_ABC.rdf

Code:
 
        <j.1:Date_de_Publication_Periodique rdf:datatype="http://www.w3.org/2001/XMLSchema#date">2013-10-30</j.1:Date_de_Publication_Periodique>

# 4  
Chubler_XL's solution works for me on AIX & Debian.

What's your OS? If you're running SunOS/Solaris then try with /usr/xpg4/bin/awk.
# 5  
I'm using linux and executing command in bash shell:

Quote:
bash-3.2$ uname -a
Linux SMP Mon Feb 21 05:52:39 EST 2011 x86_64 x86_64 x86_64 GNU/Linux
# 6  
Try with sed:
Code:
sed '/Date_de_Publication_Periodique.*XMLSchema#date/s/[0-9]\{4\}-[0-9]\{2\}-[0-9]\{2\}/JJJJJJ/' MM_NN-A1B1C1_ABC.rdf

This User Gave Thanks to Subbeh For This Post:
# 7  
Thank you subbeh and chubler_xl and carlom.
I figured out the working solution:
Quote:
awk '/Date_de_Publication_Periodique/ && /XMLSchema#date/' MM_NN-A1B1C1_ABC.rdf | sed 's/[0-9]\{4\}-[0-9]\{2\}-[0-9]\{2\}/JJJJJJ/'
I forgot to add another requirement that I need to pick only the node with
"Date_de_Publication_Periodique" and "XMLSchema#date" and then replace the date inside with the desired date
While
Quote:
awk '/Date_de_Publication_Periodique/ && /XMLSchema#date/' MM_NN-A1B1C1_ABC.rdf
is returning that line but with sed
Quote:
sed '/Date_de_Publication_Periodique.*XMLSchema#date/s/[0-9]\{4\}-[0-9]\{2\}-[0-9]\{2\}/JJJJJJ/' MM_NN-A1B1C1_ABC.rdf
it was returning multiple lines.

Last edited by Ribosome; 11-06-2013 at 08:35 AM.. Reason: corrected username

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