Ubuntu – Using sed to edit any one of the occurrence of a matching pattern

command linesedtext processing

A line starting with Fred Flintstone shall be appended with some string.
Look for specified occurrence of Fred Flintstone and append it.

How do I use this command for anyone of the occurrences of such a pattern?
I tried

sed '/Fred Flintstone/ s/$/ someString/2' filename

Apparently the above one isn't working.
It works well for all occurrences, but not a specific occurrence.
(Say I want to replace first or second or third [any one of them])

Example File1:

Fred Flintstone 
Johnson Stone
Fred Flintstone
Fred Flintstone
Michael Clark

Desired Output File1:

Fred Flintstone 
Johnson Stone
Fred Flintstone someString
Fred Flintstone
Michael Clark

Best Answer

Although you've mentioned sed, these are sort of awk-y tasks:

awk -v pat="Fred Flintstone" '$0 ~ pat {count++;\
               if (count == 2) { $0 = $0" someString" ;} ;}; 1' file.txt

-v pat="Fred Flintstone" saves the Regex pattern to match as variable pat to be used inside awk expressions
$0 ~ pat checks the record against pat for a match; if matched, the count variable is increased by 1 and if count is 2 the record is reset as having the current content plus someString ({count++; if (count == 2) { $0 = $0" someString" ;} ;})
1 is an idiom; as it is truthy, all the records would be printed

Example:

% cat file.txt
Fred Flintstone
Johnson Stone
Fred Flintstone
Fred Flintstone
Michael Clark

% awk -v pat="Fred Flintstone" '$0 ~ pat {count++; if (count == 2) { $0 = $0" someString" ;} ;}; 1' file.txt
Fred Flintstone
Johnson Stone
Fred Flintstone someString
Fred Flintstone
Michael Clark

Related Solutions

Ubuntu – How to save matched pattern into variable using ‘sed’ command

Here is a single invocation of sed that writes the revised line to stdout while at the same time saving the removed text in shell variable var:

$ var=$(echo "This X is test Y. But X is not test Y." | sed -nr 'h;s/[^X]*X([^Y]*)Y.*/\1/;p;x;s/X[^Y]*Y/REPLACE/;w /dev/stderr') 2>&1
This REPLACE. But X is not test Y.

The value of var is:

$ echo "==$var=="
== is test ==

Explanation:

h

This command copies the current pattern to the hold space.
s/[^X]*X([^Y]*)Y.*/\1/;p

This removes everything from the pattern space except the text between the first X and Y including any spaces. This is then printed to stdout. This is the output that is captured by the shell and assigned to var.
x

This copies the hold space back to the pattern space. When this is done, the pattern space contains a copy of the original input line.
s/X[^Y]*Y/REPLACE/; w /dev/stderr

The substitution is made and the result is written to stderr.
2>&1

After the shell has captured stdout into var, this instructs the shell to copy stderr (which has the line with REPLACE) to stdout.

Aside on handling of variable `var`

The variable var includes the leading and trailing spaces. If the shell were to subsequently subject var to word-splitting, these spaces would be removed. To prevent that, when var is referenced, do so inside double-quotes, as in the example above.

Ubuntu – How to replace every nth occurrence of pattern using sed ONLY

Isn't that a basic function..

Obviously, it isn't. But it's still possible:

sed 's/a\([^a]*\($\|a[^a]*\($\|a[^a]*\($\|a[^a]*\($\|a\)\)\)\)\)/A\1/g'

More readable with -r where we can drop the backslashes:

sed -r 's/a([^a]*($|a[^a]*($|a[^a]*($|a[^a]*($|a)))))/A\1/g'

i.e. a, followed by any number of non-a, followed by either end of line, or another a, followed by any number of non-a, etc. Everything after the first a is remembered in \1.

BTW, if Perl is available, you can use:

perl -pe 'BEGIN { @A = qw(a A) } s/a/$A[not $i++ % 5]/g'

% is the modulo operator. $i is incremented with each match, negation of the modulo is 1, 0, 0, 0, 0, 1, etc., which used as the index into @A picks the correct a or A.

Best Answer

Related Solutions

Ubuntu – How to save matched pattern into variable using ‘sed’ command

Aside on handling of variable var

Ubuntu – How to replace every nth occurrence of pattern using sed ONLY

Related Question

Aside on handling of variable `var`