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Full Discussion: Grep multiple lines and save to a file
Top Forums Shell Programming and Scripting Grep multiple lines and save to a file Post 302349376 by chssastry on Monday 31st of August 2009 09:21:42 PM
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.
 

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