convert txt to columnated file - bash

Question

I need to convert test.txt file to a columnated file.
I know how to convert it with awk if the number of lines after each keyword are same but they are different in this example.
awk 'NR % 5 {printf "%s ", $0; next}1' test.txt
if the number of lines are same here is the code but this one will not work with this input file.
Anyway to convert this? Please advise.
test.txt
"abc"
4
21
22
25
"standard"
1
"test"
4
5
10
11
12
Expected Output:
"abc" 4 21 22 25
"standard" 1
"test" 4 5 10 11 12

Answer 1

$ awk '{printf "%s%s", (/^"/ ? ors : OFS), $0; ors=ORS} END{print ""}' file
"abc" 4 21 22 25
"standard" 1
"test" 4 5 10 11 12

Answer 2

A bit magic, but works in this case:
sed -z 's/\n"/\n\x01"/g' |
tr '\n' ' ' |
tr $'\x01' '\n'
Each "header" starts is a string between " ... ". So:
Using sed I put some delimter (I chose 0x01 in hex) between a newline and a ", everywhere in the file. Note that -z is a gnu extension.
Then I substitute all newlines for a space.
Then I substitute all 0x01 bytes for newlines.
This method is a little tricky, but is simply and works in cases where the header starts with some certain character on the beginning of the line.
Live version available at tutorialspoint.
One can get with sed without gnu extension by using for example:
sed '2,$s/^"/\x01"/'
ie. for lines greater then the second if the line starts with a ", then add the 0x01 byte on the beginning of the line.

Answer 3

with GNU awk
$ awk -v RS='\n"' '{$1=$1; printf "%s", rt $0; rt=RT}' file
"abc" 4 21 22 25
"standard" 1
"test" 4 5 10 11 12

Answer 4

POSIX awk:
$ awk '/^"/{if (s) print s; s=$0; next} {s=s OFS $0} END{print s}' file
"abc" 4 21 22 25
"standard" 1
"test" 4 5 10 11 12
Or with perl:
$ perl -0777 -lnE 'for (/^"[^"]+"\R(?:[\s\S]+?)(?=^"|\z)/mg) {tr /\n/ /; say} ' file
If your fields do not have spaces in them, you can use a simple tr and sed pipe:
$ cat file | tr '\n' ' ' | sed -E 's/ ("[^"]*")/\
\1/g'
Or GNU sed:
$ cat file | tr '\n' ' ' | sed -E 's/ ("[^"]*")/\n\1/g'

Answer 5

While an awk or sed solution is advisable, since the question is also tagged bash, you can do all that is needed with a simple read loop and a flag variable to control the newline output for the first iteration. Essentially, you are reading each line and using the string indexing parameter expansion to test whether the first character is a non-digit, and on the 1st iteration simply output the string, for all additional iterations, output the string preceded by a '\n'. If the line begins with a digit, simply output it with a space preceding.
For example:
#!/bin/bash
declare -i n=0 ## simple flag to omit '\n' on first string output
while read -r line; do ## read each line
[[ ${line:0:1} =~ [^0-9] ]] && { ## begins with non-digit
## 1st iteration, just output $line, rest output '\n$line'
((n == 0)) && printf "%s" "$line" || printf "\n%s" "$line"
} || printf " %s" "$line" ## begins with digit - output " $line"
n=1 ## set flag
done < "$1"
echo "" ## tidy up with newline
Example Use/Output
$ bash fmtlines test.txt
"abc" 4 21 22 25
"standard" 1
"test" 4 5 10 11 12
While awk and sed will generally be faster (as a general rule), here with nothing more than a while read loop and a few conditionals and parameter expansions, the native bash solution would not be bad by comparison.
Look things over and let me know if you have questions.