Rename file using sed or awk


 
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# 1  
Old 06-07-2012
Rename file using sed or awk

I have a filename like 1_DATE_3_4.5_888 and I want to modify the date field (ie the last 4 digits ) alone and remove the last field.

Old filename:1_DATE_3_4.5_888
Given date (for eg):120606259532

modified date:120606259899

new filename:1_<modified date>_3.4.5

Last edited by zaxxon; 06-08-2012 at 07:19 PM.. Reason: No shouting!
# 2  
Old 06-07-2012
Hello, please edit your post above and change the all caps.

In case you forgot to read the forum rules, here is quick copy.

Quote:
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# 3  
Old 06-07-2012
Code:
FILE="1_120606259532_3_4.5_888"
NEW_FILE=$(echo "$FILE" | nawk -F_ -v n="120606259899" -v OFS=_ '$2=n')
echo $NEW_FILE

This User Gave Thanks to itkamaraj For This Post:
# 4  
Old 06-07-2012
last segment

thanks a lot.. but the output i'm getting is
Code:
echo $NEW_FILE
1_120606259899_3_4.5_888

the last segment is 888, which is not removed.

Last edited by zaxxon; 06-08-2012 at 07:20 PM.. Reason: code tags!
# 5  
Old 06-07-2012
Code:
 
NEW_FILE=$(echo "$FILE" | nawk -F_ -v n="120606259899" -v OFS=_ '$2=n;$NF="_"' | sed 's/__$//')

This User Gave Thanks to itkamaraj For This Post:
# 6  
Old 06-07-2012
thanks

thanks a lot,

but m getting 2 outputs,

Code:
$echo $NEW_FILE
1_120606259899_3_4.5_888 1_120606259899_3_4.5

---------- Post updated at 03:46 AM ---------- Previous update was at 03:35 AM ----------

suppose ther are 100 files. i need to remove the 888 segments for all the 100 at the end, and the date field is constant it always contains 12 digits,and i need 2 remove the last 4 digits in it, and replace it with "9899".
Is this possible with script??

Last edited by zaxxon; 06-08-2012 at 07:20 PM.. Reason: code tags!
# 7  
Old 06-07-2012
use this

Code:
 
NEW_FILE=$(echo "$FILE" | nawk -F_ -v n="120606259899" -v OFS=_ '{$2=n;$NF="_";print}' | sed 's/__$//')

This User Gave Thanks to itkamaraj For This Post:
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