How to write shell script to output two variables in one line without loop

Question

I want to write shell script to output two variables in one line.

My script is:

SRC = "192.168.1.1"
PORT = "22"

I want to make the output looks like

192.168.1.1,22

How to do this? Thank you.

Answer 1

To echo any number of variables just reference them in the same echo command. Also, note that the correct shell syntax for assignments must not have spaces around =.

#!/bin/sh
SRC="192.168.1.1"
PORT="22"
echo "$SRC,$PORT"

Result of running the script will be:

 $ ./test.sh
 192.168.1.1,22

Answer 2

You would typically use printf to output variables' data, like so:

#!/bin/sh

src=192.168.1.1
port=22

printf '%s,%s\n' "$src" "$port"

• I added the missing #!-line at the top. I chose the /bin/sh shell (a POSIX shell) as there is nothing in your existing code that requires a specific other shell like bash or zsh (see also Which shell interpreter runs a script with no shebang?).
• Assignments are done without spaces around the = character (see Spaces in variable assignments in shell scripts).
• I used lower case variable names to avoid accidentally clashing with existing environment variables or shell-specific variables (see Are there naming conventions for variables in shell scripts?).
• I opted for using printf as a precaution as I don't know whether your data might eventually arrive from some external source, and echo is known for sometimes having issues with some particular strings (see Why is printf better than echo?).

You may also find the following interesting, regarding quoting:

• When is double-quoting necessary?

---

Alternatively,

#!/bin/sh

src=192.168.1.1
port=22

string="$src,$port"

printf '%s\n' "$string"

The only difference here is that I've created a new variable with the comma-delimited values of the two first variables, and then output that. This may be what you want depending on what your use of that string may be later in the script.