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Print only one occurrence of a file

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Top Forums Shell Programming and Scripting Print only one occurrence of a file
# 1  
Old 11-15-2014
Print only one occurrence of a file
I have the below file

Code:
$cat sample.txt
one.zip#one.pdf#one.jpeg#1
two.zip#two.pdf#two.jpeg#1
two.zip#two.pdf#two.jpeg,three.jpeg#2
two.zip#two.pdf#two.jpeg,three.jpeg,four.jpeg#3
three.zip#three.pdf#three.jpeg#1


I need the output as

Code:
one.zip#one.pdf#one.jpeg#1
two.zip#two.pdf#two.jpeg,three.jpeg,four.jpeg#3
three.zip#three.pdf#three.jpeg#1

the logic is the file .zip should have only one entry and that should be the largest file name and I will not know the number of lines or no of files.
It is an automated script that will produce the file.

Please help.
Last edited by Don Cragun; 11-15-2014 at 03:37 AM.. Reason: Add CODE tags.
# 2  
Old 11-15-2014
Welcome to the forum.

Any efforts on your part?
# 3  
Old 11-15-2014
Try This..

Code:
awk -F"#" '{ arr[$1]=$0;}
END{ for(v in arr) print arr[v]}'  sample.txt

Output

Code:
three.zip#three.pdf#three.jpeg#1
one.zip#one.pdf#one.jpeg#1
two.zip#two.pdf#two.jpeg,three.jpeg,four.jpeg#3

This User Gave Thanks to bharat1211 For This Post:
Bhavi
# 4  
Old 11-15-2014
@Bhavi

You'll find out that bharat1211 post does not produce what you want.
Code:
arr[$1]=$0;

will overwrite the previous entry without regard if it has more files or less.

Please, let us know what you have tried.
Last edited by Aia; 11-15-2014 at 02:57 AM.. Reason: Adding why it will not work.
# 5  
Old 11-15-2014
Yes it is working for my requirement.


Thank you,
# 6  
Old 11-15-2014
Then run it on this input:
Code:
one.zip#one.pdf#one.jpeg#1
two.zip#two.pdf#two.jpeg,three.jpeg,four.jpeg#3
two.zip#two.pdf#two.jpeg,three.jpeg#2
two.zip#two.pdf#two.jpeg#1
three.zip#three.pdf#three.jpeg#1

# 7  
Old 11-15-2014
My previous requirement is satisfied and its working fine now.

Now i need another help for the below condition.

I have the below file

$cat sample.txt
Code:
one.zip#one.pdf#one.jpeg#1
two.zip#two.pdf#two.jpeg#1
two.zip#two.pdf#three.jpeg#1
two.zip#two.pdf#four.jpeg#
three.zip#three.pdf#three.jpeg#1

The file should look like below with all the jpeg files for the zip file in the same line.

Code:
one.zip#one.pdf#one.jpeg#1
two.zip#two.pdf#two.jpeg,three.jpeg,four.jpeg#3  --> count of jpeg files
three.zip#three.pdf#three.jpeg#1

I googled and found the below script

Code:
awk -F "," 's != $1 || NR ==1{s=$1;if(p){print p};p=$0;next}
{sub($1,"",$0);p=p""$0;}END{print p}' sample.txt

But its again printing the same. Please advice


Thanks,

Moderator's Comments:
Mod Comment Please use CODE tags when showing sample input, output, and code.
Please do not use one thread for two different problems; open a new thread instead of adding on to an existing thread.
Last edited by Don Cragun; 11-15-2014 at 01:45 PM.. Reason: Add CODE tags.
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