在 bash 中连接两个没有换行符的命令的输出

Question

What I need:

Suppose I have two commands, A and B , each of which returns a single-line string (ie, a string with no newline character, except possibly 1 at the very end). I need a command (or sequence of piped commands) C that concatenates the output of commands A and B on the same line and inserts 1 space character between them.

Example of how it should work:

For example, suppose the output of command A is the string between the quotation marks here:

"The quick"

And suppose the output of command B is the string between the quotation marks here:

"brown fox"

Then I want the output of command(s) C to be the string between the quotation marks here:

"The quick brown fox"

My best attempted solution:

In trying to figure out C by myself, it seemed that the follow sequence of piped commands should work:

{ echo "The quick" ; echo "brown fox" ; } | xargs -I{} echo {} | sed 's/\n//'

Unfortunately, the output of this command is

The quick
brown fox

Answer 1

You can use tr :

{ echo "The quick"; echo "brown fox"; } | tr "\n" " "

OR using sed:

{ echo "The quick"; echo "brown fox"; } | sed ':a;N;s/\n/ /;ba'

OUTPUT:

The quick brown fox 

Answer 2

echo "$(A)" "$(B)"

should work assuming that neither A nor B output multiple lines.

$ echo "$(echo "The quick")" "$(echo "brown fox")"
The quick brown fox

Answer 3

I'll try to explain the solution with another simple example

We've to concatenate the output of the following command:
"pwd" and "ls"

echo "$(pwd)$(ls)";

Output: 2 concatenated strings

Answer 4

$ commandA () { echo "The quick"; }
$ commandB () { echo "brown fox"; }
$ x="$(commandA) $(commandB)"
$ echo "$x"
The quick brown fox

Answer 5

$ { echo -n "The quick" ; echo -n " " ; echo "brown fox" ; }
The quick brown fox