Why does 'read -p' only remember one variable?-CodePudding

I figured out through a tutorial that using read -p would allow me to take the response as a variable. So I did this madlibs kind of thing to test it, hoping that I could repeatedly have user input translated to a variable, then back into output.

Here's an exerpt:

read -p "What is your favorite fruit?" fruit
echo Oh, I see...
sleep 2s

read -p "And what is your mama's name?" mama
echo Amazing...
sleep2s

read -p "And how many years have you been a little bitch?" years
echo Understood...
sleep 2s

echo So let me get this straight...
echo Your favorite fruit is $fruit
echo Your mama's name is $mama
echo And you have been a little bitch for $years years.

Now let's say I answered Apples, Martha, 3. When I do this, my output comes out:

So let me get this straight...
Your favorite fruit is Apples.
Your mama's name is $mama
And you have been a little bitch for $years

Only the first variable is being output properly, the rest isn't. I have tried using different notations for the variables mama and years, changing them for mumuh and yuurs, but no alas.

Is there something obvious I am missing? Am I misunderstanding how the read command handles the created variables?

CodePudding user response：

This has nothing to do with read. your script is simply missing some quotes for the string sequences.

so your output lines should more look like this:

echo "So let me get this straight..."
echo "Your favorite fruit is" $fruit
echo "Your mama's name is" $mama
echo "And you have been a little bitch for" $years "years."

you could even wrap the variables in the string like this:

echo "And you have been a little bitch for ${years} years."

for a bit more details, you might want to read this question.

CodePudding user response：

The problem lies in this single line:

echo Your mama's name is $mama

because the shell finds an apostrophe (single quote) which must be matched (closed) by another, but there is none.

So, as the previous reply says, it is a problem of quoting, but only this line needs it because of the literal single quote which must be echoed literally and not interpreted.

A mean shell should complain that there is an unmatched quote; my bash says:

Your favorite fruit is Apple

-bash: test.sh: line 15: unexpected EOF while looking for matching `''

Modify that line so it reads:

echo "Your mama's name is " $mama

and you are set.

If it happens that a matching closing single quote is found, the text between the apostrophes is not expanded (no variable substitution), and it seems this happened to you: in fact, the output still contained the "$". Didn't you see an error message? Strange.