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Thanks for quick reply and find the below inform.

Enterprise Linux Enterprise Linux Server release 5.5 (Carthage)
Red Hat Enterprise Linux Server release 5.5 (Tikanga)

and i tried above...

Hi All,

i am using the below command and once get the output and i need to keep the
first batch only.in this case how to do this one. please help me on thistop -b -n 5 >top.txt


Thanks,

Working fine. Thanks Rudic and Singh

Thanks it's working. please let me know "next" keyword why it is required.

Hi Team

I have file as below

empno,ename,sal
123,smith,1000
124,adams,2000

Required output: Using AWK

empno,ename,sal,deptno
123,smith,1000
124,adams,2000

Thanks,
Murali

Thanks All,it's working fine i use IFS option

Hi All,

I wrote below script

for i in `ls`
do
echo "list of files $i"
done

but i want display as like ls -ltr,it's displaying as one column

Thanks

Awesome @Don and good helping guyes

Thanks,if we use below data test1.txt with header created. but test2.txt only creating the header without data this one i don't want because there no data for this one.

"TYPE_ID" "ID" ...

Thanks ...
if i use this data not coming header row in the two files.
"TYPE_ID" "ID" "LIST_ID"
"18" "52010" "1059"
"18" "52010" "1010"
"18" "52010" "1059"
"18" "52010"...

Hi All,
i have script like ...

"TYPE_ID" "ID" "LIST_ID"
"18" "52010" "1059"
"18" "52010" "1059"
"18" "52010" "1059"
"18" "52010" "1059"

i am using the below code it's not...

Working Fine...

Thanks lot @vidyadhar85 and @Don (https://www.unix.com/members/302044378.html)

Hi All,
i want get the last 1 hours modified files.....But i want only filename without ./ like that..how can get only file name using FIND.

find . -type f -name "*.txt" -mmin -60
./test.txt...

Thank lot.. i will try..

---------- Post updated at 06:17 AM ---------- Previous update was at 02:29 AM ----------

@Don.. i tried with awk command not working.. it raising the below error..
...

Thanks.. Do you any idea.. how to do simply way

i tried with this one.. Now it's working ... can you suggest me how to do simply way..
cat test.txt | awk -F "," '{
month=substr($1,2,length($1))
year=substr($2,1,length($2)-1)
...

I am not get any idea..suggest me

Hi All,

I have file like

“April 10, 2013”,”raj”
“April 29, 2013”,”raj1”


Output :

“2013/04/10”,”raj”
“2013/04/29”,”raj1”

Please help me how to do...

i didn't find the below one in the above code...
ar_data_$mmyyyy

Can you post that portion of the script..

Post your code...

i didn't find the v_table_name and v_ret_days in the dbms_output.

---------- Post updated at 02:29 AM ---------- Previous update was at 02:28 AM ----------

Can run the manually for the...

Can you this options..
set termout off
set feedback off
set head on
set verify off
set echo off
set linesize 30000
set pagesize 0
set trimspool on

---------- Post updated at 07:39 AM...

you can try also
grep 'Business Date' test1.txt

Try like..
awk '{print $2}' test1.txt|grep '[0-9]'|awk '{sum+=$1} END {print sum}'