Replace a string pattern

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# 1  
Old 10-07-2013
Replace a string pattern


I have a CSV with following type of data and would like to replace the timestamp information with 'null' string. Can you please suggest me on same?

8,1,'1','1',11,'2013-08-12 18:34:17.0','null',1,'2013-08-12 18:34:17.0','null','PROMOTIONAL','12','1','11','11',11,'0'

Thanks for your time.
# 2  
Old 10-07-2013
try bit lengthy

sed 's/....-..-.. ..:..:..../null/g' your_filename

$ echo "8,1,'1','1',11,'2013-08-12 18:34:17.0','null',1,'2013-08-12 18:34:17.0','null','PROMOTIONAL','12','1','11','11',11,'0'"  | sed 's/....-..-.. ..:..:..../null/g'



Last edited by Akshay Hegde; 10-07-2013 at 07:05 AM..
# 3  
Old 10-07-2013

echo "8,1,'1','1',11,'2013-08-12 18:34:17.0','null',1,'2013-08-12 18:34:17.0','null','PROMOTIONAL','12','1','11','11',11,'0'" | perl -nle '{s/\d{4}-(0[1-9]|1[012])-(0[1-9]|[12][0-9]|3[01]) ([01][0-9]|2[0-3]):[012345][0-9]:[012345][0-9]\.\d?/NULL/g; print}'

# 4  
Old 10-07-2013
Thanks Akshay for replying.

Can you please suggest how can I include the "(" and ")" brackets inside this replacement string as some of the lines in CSV have date value inside braces.

('2013-08-12 18:34:17.0') should be set to 'NULL'
# 5  
Old 10-07-2013

$ echo "('2013-08-12 18:34:17.0')" | sed 's/[()]//g; s/....-..-.. ..:..:..../null/g'

# 6  
Old 10-07-2013
The "[()]" is replacing all braces in CSV. The negative values are mentioned within () and it must be retained.

echo "(1), ('2013-08-12 18:34:17.0')" | sed 's/[()]//g; s/....-..-.. ..:..:..../null/g'
1, 'null'

Sorry for posting multiple queries. I am totally new to this project and had to try with uploading CSV to find out possible errors. I hope you will understand.
# 7  
Old 10-07-2013
You can split file and then upload
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