parse file AND parse column with diff delimiter

[AlbertAguirre]

Ok I need to parse a file but ALSO parse one of the columns in the file.

The file is delimited by | but the column I am parsing is delimited by '/'

Here is a sample record from the original file:

2008-08-28 23:08:36|218.165.182.133|image|78450882|/albums/rr29056/someusername/20080809/IMG_146701.jpg

My results must look like this:
2008-08-28 23:08:36|someusername

How do I accomplish this?

 

[feherke]

Hi

awk -F'[|/]' -vOFS='|' '{print $1,$8}' /input/file

[gray]# or ( gawk only )[/gray]

awk -F'|' -vOFS='|' '{split($5,a,"/");print $1,a[4]}' /input/file

[gray]# or[/gray]

awk -F'|' -vOFS='|' '{FS="/";split($5,a);FS="|";print $1,a[4]}' /input/file

Feherke.

http://rootshell.be/~feherke/

 

[dstxaix]

Albert

Assuming your filename is list1

this should work for your example

cat list1 | tr "/" "|" | awk -F"|" '{print $1"|"$8}'

 

[feherke]

Hi

Please avoid UUOC.

tr "/" "|" < list1 | awk -F"|" '{print $1"|"$8}'

Feherke.

http://rootshell.be/~feherke/

 

[booke]

You could:

awk -F"|" '{split($NF,a,"/"); print $1"|"a[4]}' your_filename