Shell Programming and Scripting

Hello

I have a abc.txt file which semicolon delimited. I need to read the first line of the abc.txt file where i need to extract specific string which at specific location.

i.e first line of the abc.txt file is as below
abc;123;xyz;345;678

my requirement is i need to get the 678. I was trying with awk but not successful

thanks for the help

Something tells me that it's not the whole of your input file ( and not the whole of your goal)...
getting the end of the only line you give as a sample is easy, but from an entire file, it might be more difficult...
if you're just looking for a line that ends with '678', `grep` could do the the job, couldn't it?
for any other pattern, please define the "delimiters" of the 'expression' you're looking for in the file...

please be more precise.

hi
Thanks for the reply.
It might not be the end of the line.Let say if i want to extrct the string out of 3 rd and 4th delimter; then which command should i use. if i want 345 then which command should i use.

Thnaks

---------- Post updated at 11:26 PM ---------- Previous update was at 11:17 PM ----------

Infact my aim is to read the first line of the file and out of the first line i need to read the string which is in between semicolons.

go on, refine your thinking...

if you have a fixed delimiter (kind of ':' or ';' or ',' ...), you should be able to use `cut`; why shouldn't you

$ cat abc.txt
abc;123;xyz;345;678
$ cat abc.txt | perl -e 'while (<>){ chomp; @fields = split(";"); last;} print "$_\n" foreach (@fields);'
abc
123
xyz
345
678

Cheers

one way..

Code:

-bash-3.2$ cat file
abc;123;xyz;345;678
dwad;231;3213;434;343
-bash-3.2$ awk -F";" '{print $1,$2,$3,$4}' file
abc 123 xyz 345
dwad 231 3213 434
-bash-3.2$ awk -F";" '{print $1,$2,$3}' file
abc 123 xyz
dwad 231 3213
-bash-3.2$

depend on your different request:

Code:

awk -F";" '{print $5}' abc.txt

awk -F";" '{print $NF}' abc.txt