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Using records iteratively with increament

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# 1  
Old 04-28-2012
Using records iteratively with increament
I am very new to unix. Trying to use a for loop for this task

I have a file like this

Code:
abc.uml
ccc.uml
ddd.uml
rrr.uml
kkk.uml
.
.
.
n.uml ( more than 10000)

I want to read this file but just 5 names at a time and then again it should read next 5 names and agin next 5. this should go on till the end of file. What ever it read it should be directed in the output file.

The code I am using

Code:
 for i in `cat input_file` ; doecho $i done



The output should be

Code:
hello there abc.uml
                 ccc.uml
                 ddd.uml
                 rrr.uml
                 kkk.uml
hello there hhh.uml
                 yyy.uml
                 eee.uml
                 qqq.uml
                 fff.uml
hello there iii.uml
                 ttt.uml
                 aaa.uml
                 sss.uml
                 zzz.iml
                 nnn.uml
hello there adn.uml
.
.
.
(till 10000 file names)

Please help me with this friends
Last edited by Scrutinizer; 04-28-2012 at 01:39 PM.. Reason: code tags or data files
# 2  
Old 04-28-2012
Assuming you're using Kshell or bash, this would be one way:

Code:
i=0
while read buffer
do
    (( i++ ))
    case $i in
        1) printf "hello there ";;  # print "leader text" on first of each group
        5) i=0;;             # last line of the group, reset counter
        *) ;;                # no special action needed; do nothing when i is 2,3,4
    esac

    printf "%s\n" "$buffer"    # always print the buffer
done <input-file

# 3  
Old 04-28-2012
Hello agama,

Sorry for the misunderstanding created by me
The code is working fine, just small modification.

The output has to be in this format

Code:
hello there abc.uml,
                 ccc.uml,
                 ddd.uml,
                  rrr.uml,
                kkk.uml with delimiter '|';
hello there hhh.uml,
                 yyy.uml,
                  eee.uml,
                  qqq.uml,
                   fff.uml with delimiter '|';
hello there iii.uml,
                ttt.uml,
              aaa.uml,
                sss.uml,
                zzz.iml with delimiter '|';
hello there adn.uml,
                 nnn.uml


Please help me with this format.

Moderator's Comments:
Mod Comment Please click this link: How to use [code][/code] tags
Last edited by Scrutinizer; 04-28-2012 at 02:42 PM..
# 4  
Old 04-28-2012
Small change to accomplish that:

Code:
i=0
while read buffer
do
    (( i++ ))
    case $i in
        1)  printf "hello there %s,\n" "$buffer";;  # print first
        5)  printf "%s |\n" "$buffer"               # print last and reset
            i=0
            ;;
        *) printf "%s,\n" "$buffer";;               # print middles with commas
    esac
done <input-file



Didn't know if you need/want a space before the pipe sym, you can remove it easly if it's not needed.
# 5  
Old 04-29-2012
Perfectly working now
Thanks for sharing your great knowledge.

---------- Post updated 04-29-12 at 12:13 PM ---------- Previous update was 04-28-12 at 11:28 PM ----------

I am trying to modify the given script to add some text in the output.

The output will be as follows

Code:
hello there /dev/pts/data abc.uml on node0,
                 /dev/pts/data ccc.uml on node0,
                 /dev/pts/data ddd.uml on node0,
                 /dev/pts/data rrr.umlon node0,
                /dev/pts/data kkk.uml on node0with delimiter '|';
hello there /dev/pts/data hhh.uml on node0,
                 /dev/pts/data yyy.uml on node0,
                  /dev/pts/data eee.uml on node0,
                  /dev/pts/data qqq.uml on node0,
                   /dev/pts/data fff.uml on node0 with delimiter '|';
hello there /dev/pts/data iii.uml on node0,
                /dev/pts/data ttt.uml on node0,
               /dev/pts/data aaa.uml on node0,
                /dev/pts/data sss.uml on node0,
                /dev/pts/data zzz.iml on node0 with delimiter '|';
hello there /dev/pts/data adn.uml on node0,
                 /dev/pts/data nnn.uml on node0,
.
.
.
.
.

Please help me with this modification
Last edited by Scrutinizer; 04-29-2012 at 03:55 AM.. Reason: code tags
# 6  
Old 04-29-2012
I think that if you experiment a little with the printf statements you should be able to do this yourself. What have you tried so far?
These 2 Users Gave Thanks to Scrutinizer For This Post:
agama nnani
# 7  
Old 04-30-2012
I have done some modifications and I got the expected output. Thanks to all.
This User Gave Thanks to nnani For This Post:
agama
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