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Grep multiple lines and save to a file

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# 1  
Old 08-31-2009
Grep multiple lines and save to a file
Sir


I have a data file e.g. DATA31082009. This file consists of several data files appended to that file. The size of each data file is different. The first line of each file starts with "44". I want to grep data from "44" to the preceding line of next "44" and save it as a individual file. Thus I want to divide the main DATA31082009 into multiple files giving multiple names.
I am new to unix and shell scripting and highly grateful if anybody help me.
# 2  
Old 08-31-2009
Hi.

Can you post a sample of your input data, and what convention you want for the output filenames (meaning, how should the output filenames be named)?
# 3  
Old 08-31-2009
what have you tried and what problems have you had? (Hint: grep won't split your files)
# 4  
Old 08-31-2009
Code:
446003117FA&CAO/ICF, PERAMBUR,CHENNAI-38         ICFPF-  141207         6000010002320000005        0000006500000000001565000030122007                       
2260002500610            023037JOHN ASIRVATHAM    S                    6000010006003117FA&CAO/ICF/PER-CH-38122007815365 0000002410000                       
2260001904310   000000465483840HARIDOSS     T                          6000010006003117FA&CAO/ICF/PER-CH-38122007612111 0000006500000                       
2260002501210            028692SIVAKUMAR     PG                        6000010006003117FA&CAO/ICF/PER-CH-38122007599295 0000000300000                       
2260001505010   9173           BASKARAN     S                          6000010006003117FA&CAO/ICF/PER-CH-38122007673678 0000000120000                       
2260002501310            040558MOHANRAJ     A                          6000010006003117FA&CAO/ICF/PER-CH-38122007652990 0000000820000                       
2260002501510            072469GAJENDRA BABU    MT                     6000010006003117FA&CAO/ICF/PER-CH-38122007717061 0000000700000                       
2260002800610            015219VIJAYACHANDAR     T                     6000010006003117FA&CAO/ICF/PER-CH-38122007781118 0000004000000                       
2260001101410            030438SIGHAMANI     S                         6000010006003117FA&CAO/ICF/PER-CH-38122007699333 0000000800000

This is one sample file consists 8 records and called an individual file. Other files may have much more or less records. For every file the differentiator is "44" at the beginning.

I have used
Code:
grep "^44" data31082009 (combined file name pl.) | cut -c 3-9,126-133 > myfile

to obtain no. of files in the combined file.

With the help of your forum, I could come upto this. From now on I am not progressing.
I need to grep each file startiang with 44 and ending with preceding line of next file which starts again with 44. The name of the file should be 3-9 (number in "44" line followed by 126-133 (date in "44" field.
Sorry for lengthy reply - question.

---------- Post updated at 08:21 PM ---------- Previous update was at 08:21 PM ----------

Code:
446003117FA&CAO/ICF, PERAMBUR,CHENNAI-38         ICFPF-  141207         6000010002320000005        0000006500000000001565000030122007                       
2260002500610            023037JOHN ASIRVATHAM    S                    6000010006003117FA&CAO/ICF/PER-CH-38122007815365 0000002410000                       
2260001904310   000000465483840HARIDOSS     T                          6000010006003117FA&CAO/ICF/PER-CH-38122007612111 0000006500000                       
2260002501210            028692SIVAKUMAR     PG                        6000010006003117FA&CAO/ICF/PER-CH-38122007599295 0000000300000                       
2260001505010   9173           BASKARAN     S                          6000010006003117FA&CAO/ICF/PER-CH-38122007673678 0000000120000                       
2260002501310            040558MOHANRAJ     A                          6000010006003117FA&CAO/ICF/PER-CH-38122007652990 0000000820000                       
2260002501510            072469GAJENDRA BABU    MT                     6000010006003117FA&CAO/ICF/PER-CH-38122007717061 0000000700000                       
2260002800610            015219VIJAYACHANDAR     T                     6000010006003117FA&CAO/ICF/PER-CH-38122007781118 0000004000000                       
2260001101410            030438SIGHAMANI     S                         6000010006003117FA&CAO/ICF/PER-CH-38122007699333 0000000800000

This is one sample file consists 8 records and called an individual file. Other files may have much more or less records. For every file the differentiator is "44" at the beginning.

I have used
Code:
grep "^44" data31082009 (combined file name pl.) | cut -c 3-9,126-133 > myfile

to obtain no. of files in the combined file.

With the help of your forum, I could come upto this. From now on I am not progressing.
I need to grep each file startiang with 44 and ending with preceding line of next file which starts again with 44. The name of the file should be 3-9 (number in "44" line followed by 126-133 (date in "44" field.
Sorry for lengthy reply - question.
# 5  
Old 08-31-2009
Code:
#!/bin/sh
while read line
do
  if echo "$line" | grep "^44" > /dev/null
  then
    filename=`echo "$line" | cut -c 3-9,126-133`
  fi
  echo "$line" > "${filename}.txt"
done

Untested - try it someplace safe and expect to do a little debugging Smilie
# 6  
Old 08-31-2009
yes, the same shorter:
Code:
#!/bin/sh

while read line; do
   [ ${line:0:2} = 44 ] && filename=`cut -c 3-9,126-133 <<<"$line"`
   echo "$line" > "${filename}.txt"
done < data31082009

# 7  
Old 09-02-2009
Question
Quote:
Originally Posted by daPeach
yes, the same shorter:
Code:
#!/bin/sh
...
   [ ${line:0:2} = 44 ] && filename=`cut -c 3-9,126-133 <<<"$line"`
...

Is that bourne syntax?
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