File name in separate column

jeesun · 04-26-2010, 01:50 AM

Hi

I have some series of files, which actually named bellow

<day_month_yr>_<hh:mm:ss>.<host_name>.<IP.ip.ip.ip>.<log_name>.txt

I want to show this file name below the column may be

Date Time Host_Name IP_address Login_Name
----- ---- --------- ---------- ----------
<day_month_yr> <hh:mm:ss> host_name x.x.x.x log_name
...

...

Can anyone help

centosboy · 04-26-2010, 02:54 AM

Quote:

Originally Posted by jeesun

Hi

I have some series of files, which actually named bellow

<day_month_yr>_<hh:mm:ss>.<host_name>.<IP.ip.ip.ip>.<log_name>.txt

I want to show this file name below the column may be

Date Time Host_Name IP_address Login_Name
----- ---- --------- ---------- ----------
<day_month_yr> <hh:mm:ss> host_name x.x.x.x log_name
...

...

Can anyone help

Do you know perl? Format can do this and/or the printf command
See example here:

http://www.webreference.com/programming/perl/format/

druuna · 04-26-2010, 03:35 AM

Hi,

A sed solution:

sed 's/\([0-9][0-9_]*\)_\([0-9][0-9:]*\).\([a-z][a-z0-9]*\).\([1-9][0-9]*.[1-9][0-9]*.[1-9][0-9]*.[1-9][0-9]*\).\([a-z0-9][a-z0-9]*\)/\1 \2 \3 \4 \5/' infile

Example run:

Code:

cat infile 
01_01_2001_01:01:01.foo.1.1.1.1.log1.txt
05_05_2005_05:05:05.bar.5.5.5.5.log5.txt
10_10_2010_10:10:10.foobar.10.10.10.10.log10.txt


sed 's/\([0-9][0-9_]*\)_\([0-9][0-9:]*\).\([a-z][a-z0-9]*\).\([1-9][0-9]*.[1-9][0-9]*.[1-9][0-9]*.[1-9][0-9]*\).\([a-z0-9][a-z0-9]*\)..*/\1 \2 \3 \4 \5/' blaat 
01_01_2001 01:01:01 foo 1.1.1.1 log1
05_05_2005 05:05:05 bar 5.5.5.5 log5
10_10_2010 10:10:10 foobar 10.10.10.10 log10

Hope this helps.

jeesun · 04-26-2010, 03:54 AM

Thanks druuna for your valuable reply.

But not the cat command to format file content.

suppose the file name is 01_01_2001_01:01:01.foo.1.1.1.1.log1.txt and I want to reformat it's name actually

catkin · 04-26-2010, 04:50 AM

Here's a pure bash solution. Some of the bash-isms are pretty dense so it's structured for legibility rather than minimalism

Code:

# File name general format
# <day_month_yr>_<hh:mm:ss>.<host_name>.<IP.ip.ip.ip>.<log_name>.txt
# Assume date and time are fixed length

# Column titles
readonly ct_date='Date'
readonly ct_time='Time'
readonly ct_hostname='Host_Name'
readonly ct_ip='IP_address'
readonly ct_login_name='Login_Name'

# Simulate reading file names into an array
filename[0]='01_01_2001_01:01:01.foo.1.1.1.1.log1.txt'

# Parse variable length parts of file names
for (( i=0; i<${#filename[*]}; i++ ))
do
    buf=${filename[i]:20}
    hostname[i]=${buf%%.*}
    buf=${buf#*.}
    buf=${buf%.txt}
    login_name[i]=${buf##*.}
    ip=${buf%.$login_name}
done

# Determine required colum widths
cw_date=10
cw_time=8
cw_hostname=${#ct_hostname}
cw_ip=${#ct_ip}
cw_login_name=${#ct_login_name}
for (( i=0; i<${#filename[*]}; i++ ))
do
    [[ ${#hostname[i]} -gt ${#${cw_hostname[i]}} ]] && cw_hostname=${#hostname[i]}
    [[ ${#login_name[i]} -gt ${#${cw_login_name[i]}} ]] && cw_login_name=${#login_name[i]}
    [[ ${#ip[i]} -gt ${#${cw_ip[i]}} ]] && cw_ip=${#ip[i]}
done

# Build format string
fmt="| %${cw_date}s | %${cw_time}s | %${cw_hostname}s | %${cw_ip}s | %${cw_login_name}s |\n"

# Write table
printf "$fmt" "$ct_date" "$ct_time" "$ct_hostname" "$ct_ip" "$ct_login_name"
for (( i=0; i<${#filename[*]}; i++ ))
do
    printf "$fmt" \
        "${filename[i]:0:10}" \
        "${filename[i]:11:8}" \
        "${hostname[i]}" \
        "${ip[i]}" \
        "${login_name[i]}"
done

druuna · 04-26-2010, 05:27 AM

Quote:

Originally Posted by jeesun

Thanks druuna for your valuable reply.

But not the cat command to format file content.

suppose the file name is 01_01_2001_01:01:01.foo.1.1.1.1.log1.txt and I want to reformat it's name actually

I really don't understand what you are trying to say.

Have a look at catkin's example, maybe that will do whatever it is you want.