PHP Array to Bash not Working

Question

Sorry for the silly question here today.

I am passing a PHP array to a bash script using implode();.

To test, I am echoing the implode and I can see all array items there, but when I printf '%s\n' "${files[@]}" only the first element of the array is printed.

Am I missing something?

Here is more info:

PHP:

$files = $_POST['files'];    
$files2 = implode(" ", $files);
echo $files2   ## I can see full output here. 
shell_exec ("./sequential.sh $files2");

Bash:

files = $1
printf '%s\n' "${files[@]}" >> mytempfile.txt

Thanks for any guidance.

$files2 should be string and not an array

georoot
– georoot

2017-03-21 16:02:16 +00:00
Commented Mar 21, 2017 at 16:02 — georoot
– georoot, Commented Mar 21, 2017 at 16:02

axiac · Accepted Answer · 2017-03-21 16:03:01Z

1

files = $1

$1 is only the first argument. If you want all arguments then you can find them in $@:

printf '%s\n' "$@" >> mytempfile.txt

answered Mar 21, 2017 at 16:03

axiac

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7 Comments

fcamp Over a year ago

Ok, I tested that and got pretty close. It did print the full contents. I suppose I should have included this in my post but I edited out a lot of the script. I am passing 4 other variables from PHP to bash, so if I use $@ it prints ALL the variables, but I only need whats in the array. So now to figure that out.

axiac Over a year ago

$@ contains all the parameters passed to the shell script in the command line.

fcamp Over a year ago

I marked yours correct as it was correct for how I posed the question, but I think I need to figure it out further to get it handling all the parameters correctly. Thanks for your help.

Gordon Davisson Over a year ago

@fcamp Pass the other 4 arguments first, then use files=({@:5}") to make files an array containing the arguments starting with $5.

fcamp Over a year ago

@GordonDavisson the syntax is off a bit in your example, but it got me exactly where I needed to go! It should be files=${@:5}. This works.

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