Hi all,
If you look at the example below,I want to replace the 21st character (,) with a period (.). I have 1000 records in a file can someone help me how to do that. Thankyou all in advance.
"2008-07-15... (3 Replies)
I need to replace the character on 6th position. If this character is 1 I have to repleace it with A, if it is 2 than I have to replace it with B. If it is not 1 or 2 I should not repleace it.
input:
abcd defg
abcd 1efg
mnop weac
rstu 2bcd
i need:
abcd defg
abcd Aefg
mnop weac
rstu... (2 Replies)
Hello everyone this is my first post of many to come :)
I am writing a script and in this script at one point i need to replace a character in a particular position in a string for example:
in the string "mystery" i would need to replace the 3rd position to an "r" so the string becomes... (3 Replies)
Hi,
i want find the character '-' in a file from position 284-298, if it occurs i need to replace it with 'O ' for the position in the file. How to do that using SED command.
thanks in advance,
Sara (9 Replies)
Hi guyz i want to know nth position of character in string. For ex.
var="UK,TK,HK,IND,AUS"
now if we see 1st occurance of , is at 3 position, 2nd at 6,..4th at 13 position.
1st position we can find through INDEX, but what about 2nd,3rd and 4th or may be upto nth position. ?
In oracle we had... (2 Replies)
I have a requirement as below.
In one of my column, I have data which may or may not be separted with coma always. Now I need to validate the length of these text within the coma (if available) and if the length is more than 30 characters, I need to insert a coma either at 30 characters if its... (3 Replies)
In file, we have millions of records each of 1000 in length. And at specific position say 800 there is a space, we need to replace it with Character X if the ID in that row starts with 123.
So far i have used the below which is replacing space at that position to X but its not checking for... (3 Replies)
Discussion started by: Jagmeet Singh
3 Replies
LEARN ABOUT PLAN9
grep
GREP(1) General Commands Manual GREP(1)NAME
grep - search a file for a pattern
SYNOPSIS
grep [ option ... ] pattern [ file ... ]
DESCRIPTION
Grep searches the input files (standard input default) for lines (with newlines excluded) that match the pattern, a regular expression as
defined in regexp(6). Normally, each line matching the pattern is `selected', and each selected line is copied to the standard output.
The options are
-c Print only a count of matching lines.
-h Do not print file name tags (headers) with output lines.
-i Ignore alphabetic case distinctions. The implementation folds into lower case all letters in the pattern and input before interpre-
tation. Matched lines are printed in their original form.
-l (ell) Print the names of files with selected lines; don't print the lines.
-L Print the names of files with no selected lines; the converse of -l.
-n Mark each printed line with its line number counted in its file.
-s Produce no output, but return status.
-v Reverse: print lines that do not match the pattern.
Output lines are tagged by file name when there is more than one input file. (To force this tagging, include /dev/null as a file name
argument.)
Care should be taken when using the shell metacharacters $*[^|()= and newline in pattern; it is safest to enclose the entire expression in
single quotes '...'.
SOURCE
/sys/src/cmd/grep.c
SEE ALSO ed(1), awk(1), sed(1), sam(1), regexp(6)DIAGNOSTICS
Exit status is null if any lines are selected, or non-null when no lines are selected or an error occurs.
GREP(1)