Shell Programming and Scripting

Hi Guys,

How do I find a file that contains a string using the position number?

For example:

I have files in a directory X

Code:

cd X

ls -ltrh 

-rw-r--r--   1 username   group         3.5M Jun 27 21:09 file_1
-rw-r--r--   1 username   group         3.5M Jun 28 21:09 file_2
-rw-r--r--   1 username   group         3.5M Jun 29 21:09 file_3
...

Now I want to search for a file that contains the string test starting from position 184 in the file.

Please Help!!!!!!!!!!!

Thanks in advance...

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Moderator's Comments:

Please view this link: How to use [code]...[/code] tags

Assuming that there are not too many files in the directory and that these file-names do not contain spaces, you may try

Code:

cd dir
grep -lE '^.{183}test' *

Thanks for the reply...

The files do contain the spaces...

I meant the file names.

Then use the following:

Code:

cd dir
find . -type f -exec grep -lE '^.{183}test' {} \;

Thanks...

I tried the find but I get this error:

grep: illegal option -- E
Usage: grep -hblcnsviw pattern file . . .

On Solaris use /usr/xpg4/bin/grep rather than grep

Thanks a lot for the help...

I must be doing something wrong here...

the command

Code:

find . -type f -exec /usr/xpg4/bin/grep -lE '^.{183}test' {} \;

run fine now...but I dont get the file even if I use the string from a particular file.

My file contains data like these:

Code:

96021702970839602170297083        12345689test

[QOUTE]I want to search with the first 6 string and the string from position 184 and that would be test...[/QOUTE]

I meant to say I want to expand my search now and use the first six digit and the six digit from position 184.

Please Help....

---------- Post updated at 11:59 PM ---------- Previous update was at 11:48 PM ----------

I meant to say I want to expand my search now and use the first six digit and the six digit from position 184.

Please Help