Reading user command line input with PHP with readline, but bash is not default shell

Question

I wrote a PHP script and I'm trying to read user input from the command line. I started of with

$myVar = exec('read myVar; echo $myVar;');

which worked fine, but when typing the input the arrow keys didn't work the way it should. I learned that this could be solved with the -e switch, so it became:

$myVar = exec('read -e myVar; echo $myVar;');

Now this worked like a charm in my development envionment but unfortunately it seems like in our production environment PHP doesn't use a bash shell and doesn't support the -e switch, and it's probably not easy to change this.

According to the third answer to this question one can force PHP to use bash, so it became:

$myVar = exec('/bin/bash -c "read -e myVar; echo $myVar"');

Now, unfortunately this doesn't work as the variable myVar is not set.

When I run the commands directly from the command line $myVar (the shell variable) is set with whatever my input is, and consequently echo'ed

However when using the -c option, either in a PHP-exec() or directly on the command line like this:

/bin/bash -c "read -e myVar; echo $myVar";

$myVar isn't set at all, and an empty string (or whatever was the previous value of $myVar) is echo'ed

What do I do wrong here?

Yes, I did. And quit trying after I found out that it wasn't supported on my development environment, however indeed it works fine on production, so I think I'll just have to make that work on development, or could even go with a 'function_exists()' switch for the time being. So thanks, kind of forgot about that. Still makes me wonder though how the above doesn't work. — btum81
– btum81, Commented Oct 1, 2015 at 13:41
Are you developing against a different version of PHP than what you have on production? — glenn jackman
– glenn jackman, Commented Oct 1, 2015 at 14:26

Dezza · Accepted Answer · 2015-10-01 13:52:19Z

0

The issue is that $myVar is being interpreted before being passed to bash. The following will work by escaping the $ before it's passed to the shell:

/bin/bash -c "read -e myVar; echo \$myVar"

Also you can shorten it a little by making use of $REPLY

/bin/bash -c "read -e; echo \$REPLY"

answered Oct 1, 2015 at 13:52

Dezza

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2 Comments

btum81 Over a year ago

That's just it! Thanks :-)

Dezza Over a year ago

No problem @btum81 :-)

LZeta · Accepted Answer · 2022-04-29 09:01:59Z

0

The command you are using is correct. The solution is to use the correct escape sequence:

$myVar = exec ('/bin/bash -c \'read -e myVar; echo $myVar \'');

answered Apr 29, 2022 at 9:01

LZeta

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