[SOLVED] How can I use "read" to substitute user input inside a bash command substitution?

Issue

Was trying to do a quick-and-dirty curl call like so:

curl -u username:$(read -s -p "password: ") https://some.basic.auth.url.com

However, this fails every time. What's more, I attempted to see what's happening with something like:

echo you entered: $(read -p "enter some text: ")

However, the output is simply:

you entered:

I'm clearly missing something essential about the use of this command (or Bash in general). Can someone shed some light on:

1. Can this sort of thing even work? why?
2. If so, how could I change this to make it work without going to a script file?

Solution

echo "You entered: $(IFS= read -rp "Enter some text: "; printf '%s' "$REPLY")"

From help read:

read ... [name ...]

Reads a single line from the standard input ... The line is split into fields as with word splitting, and the first word is assigned to the first NAME, the second word to the second NAME, and so on, with any leftover words assigned to the last NAME. Only the characters found in $IFS are recognized as word delimiters.

If no NAMEs are supplied, the line read is stored in the REPLY variable.

• IFS= ...: Don't trim leading and trailing spaces
• -r: Don't mangle backslashes
• echo "$REPLY": If you don't supply any NAME, the line read is stored in the $REPLY variable. However, read doesn't print it so the command substitution expands to nothing. Consequently, you have to print it explicitly with, for example, printf

Note that if you use read inside $(...), the variable is lost as soon as you leave the substitution. Better approach:

IFS= read -rp "Enter some text: " var
echo "You entered: $var"

Answered By - PesaThe

Answer Checked By - Pedro (WPSolving Volunteer)