HI
can any one tell me how to replace a delimiter : with another delimiter '\001' it is a non printable octal character.
thanks in adv
spandu (4 Replies)
Hi everybody,
This time I am having one issue in perl.
I have to create comma separated file using the following type of information. The problem is the columns do not have any specific delimiter. So while using split I am getting different value. Some where it is space(S) and some where it is... (9 Replies)
arr_Ent_NameId variable holds 'Prakash pawar' 'sag' '23' '50000' this value
'Prakash pawar' 'sag' '23' '50000' I want to replace space( ) with comma (,)
There are 4 fields here. I don't want to replace first field with comma.
output should be: 'Prakash,pawar','sag','23','50000'
... (2 Replies)
I have a csv file and there is a problem which I need to resolve.
Column1,Column2,Colum3,Column4
,x,y,z
,d,c,v
t,l,m,n
,h,s,k
,k,,y
z,j, ,p
Now if you see column1 for row 1 and row 4 though they are null there is a space but in case of row2 and row 5 there is no space.
I want row... (3 Replies)
not sure if i'm doing this right i'm new tho this but i'm trying to use a space as a delimiter with the cut command
my code is
size=$( du -k -S -s /home/cmik | cut -d' ' -f1 )
i've also tried -f2 and switching the -d and -f around if that does anything (3 Replies)
Hi,
Extremely new to Perl scripting, but need a quick fix without using TEXT::CSV
I need to read in a file, pass any delimiter as an argument, and convert it to bar delimited on the output. In addition, enclose fields within double quotes in case of any embedded delimiters.
Any help would... (2 Replies)
Hi,
I am having a files in my directory like this:
2014 1049_file1.txt
2014 1050_file2.txt
2014 1110_file3.txt
2014 1145_file4.txt
2014 2049_file5.txt
I need to replace the above file names like this without changing the content of filename:
file1.txt
file2.txt
file3.txt... (10 Replies)
i have a file with following data.
{
EqName "Tan 1"
....
....
}
{
EqName "Sin 2"
...
...
}
I have to replace the value of EqName to Tan_1 and Sin_2 in file.Can i use sed or awk ?
cat file|grep EqName|awk '{print $2 $3}'|sed -i 's//_/g'
I tried with this but it... (2 Replies)
I have an input file as below
Emp1|FirstName|MiddleName|LastName|Address|Pincode|PhoneNumber
1234|FirstName1|MiddleName2|LastName3| Add1 || ADD2|123|000000000
Output :
1234|FirstName1|MiddleName2|LastName3| Add1 ,, ADD2|123|000000000
OR
1234,FirstName1,MiddleName2,LastName3, Add1 ||... (2 Replies)
Discussion started by: styris
2 Replies
LEARN ABOUT DEBIAN
dbswiss
DBSWISS(1) User Commands DBSWISS(1)NAME
dbSwiss - create DBM version of Swiss-Prot data
SYNOPSIS
/usr/share/librg-utils-perl/dbSwiss [OPTIONS]
/usr/share/librg-utils-perl/dbSwiss --datadir /data/swissprot --infile /data/swissprot/uniprot_sprot.dat
/usr/share/librg-utils-perl/dbSwiss [--help] [--man]
DESCRIPTION
dbSwiss creates DBM version of Swiss-Prot data. This procedure is to replace splitSwiss.pl. splitSwiss.pl saves Swiss-Prot records in
separate files resulting in over 13 million relatively tiny files that take very long to create and rsync. dbSwiss instead saves each
record into a DBM database that is optimized for fast retrieval.
OPTIONS -d, --datadir=path
directory of database files, default: '/mnt/project/rost_db/data/swissprot'
--debug
--nodebug
--first20
--nofirst20
process only first 20 records, for debugging
--help
-i, --infile=path
Swiss-Prot data flatfile, default: '/mnt/project/rost_db/data/swissprot/uniprot_sprot.dat'.
--man
--quiet
--noquiet
do not print progress status
--readback
--noreadback
read records back after storing and print them
--table
name of database table and consequently the base name of database files, default: 'dbswiss'
--version
-w, --workdir=path
Optional working directory. Automatically created and removed if not defined.
AUTHOR
Laszlo Kajan <lkajan@rostlab.org>
1.0.43 2011-11-28 DBSWISS(1)