Join lines based on pattern

Question

I have the following file:

test
1
My
2
Hi
3

i need a way to use cat ,grep or awk to give the following output:

test1
My2
Hi3

How can i achieve this in a single command? something like

cat file.txt | grep ... | awk ...

Note that its always a string followed by a number in the original text file.

Answer 1

Here is the answer: cat file.txt | awk 'BEGIN { lno = 0 } { val=$0; if (lno % 2 == 1) {printf "%s\n", $0} else {printf "%s", $0}; ++lno}'

Answer 2

paste -d "" - - < filename

This takes consecutive lines and pastes them together delimited by the empty string.

Answer 3

sed 'N;s/\n//' file.txt

This should give the desired output when the content is in file.txt

Answer 4

awk '{printf("%s", $0);} !(NR%2){printf("\n");}' file.txt

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EDIT: I just noticed that your question requires the use of cat and grep. Both of those programs are unnecessary to achieve your stated aims. If you have some reason for including them that you haven't mentioned, try this (uselessly inefficient) version of the line I wrote immediately above:

cat file.txt | grep '^' | awk '{printf("%s", $0);} !(NR%2){printf("\n");}'

It is possible that this command uses features not present in the original awk program. You may need to invoke the new awk program, nawk instead.

Answer 5

If your input file is always 1 number then 1 string, and you only want the strings, all you have to do is take every other line.

If you only want the odd lines, you can do awk 'NR % 2' file.txt

If you want the evens, this becomes awk 'NR % 2==0' data