如何在单行上打印bash数组的输出

Question

My code is

arr=($(awk '/"*"/' output.txt)) 

And output.txt says has following content -

user  
{  
'xx123':  
comment
=>  
"John  
Smith",  
groups  
=>  
$groups,  
id  
=>  
'1234',  

This is to search all usernames in my target file.It searches everything enclosed with quotes and stores in array arr.

Current Output:

"John  
Smith",  
"Abc  
Xyz",  
"...  
...",

But i want to achieve this:

"John Smith"  
"Abc Xyz"

Currently using a for loop to print this and two other arrays.

How can I achieve this?

Answer 1

I'm making assumptions about the data from the small sample given, but we can start here and you can let me know if some of those assumptions weren't so good. The main idea is just to not print a new line after reading a line with an opening quote:

$  awk '/^"/ { printf "%s ", $0; next } sub(/",$/, "\"")' output.txt
"John Smith"
"Abc Xyz"

$ cat output.txt
user
{
'xx123':
comment
=>
"John
Smith",
groups
=>
$groups,
id
=>
'1234',
=>
"Abc
Xyz",
groups
}

Answer 2

While I agree the awk solution is probably the shortest way to handle the case, the same can be done using the bash built-in comparison, string-indexes and parameter expansion (substring removal). For example:

#!/bin/bash

first=""
while read -r last; do 
    [ "${last:0:1}" = '"' ] && first="$last"
    [ "${last:(-2)}" = '",' ] && printf "%s %s\n" "$first" "${last%,}"
done <"$1"

Will provide the same output for your given input file, eg:

$ bash merge2l.sh file.txt
"John Smith"
"Abc Xyz"

Answer 3

In my first try I missed your point. Here is my correction.

You can try this:

printf "%s %s\n" "${arr[@]}"

Example:

$ set|grep BASH_VERSINFO
BASH_VERSINFO=([0]="3" [1]="2" [2]="25" [3]="1" [4]="release" [5]="i686-redhat-linux-gnu")
$ printf "%s %s\n" "${BASH_VERSINFO[@]}"
3 2
25 1
release i686-redhat-linux-gnu

I hope this is what your are looking for...

If there is no space in the array elements, then ${BASH_VERSINFO[@]} or ${BASH_VERSINFO[*]} can be used, but "${BASH_VERSINFO[@]}" is safer.

On the other hand it is better to implement inside awk as jas mentioned. If you do some other manipulation with arr then this can be a proper solution.

If the input file is small it can be implemented inside bash .

I think the used pattern "*" is not what you are intended to use. This check for zero or more occurrences of " followed by " . I think you wanted to check for ".*" or "[^"]*" or something like these...