Pass BASH associative arrays to PHP script

Question

Is it possible to pass BASH associative arrays as argv to PHP scripts?

I have a bash script, that collects some variables to a bash associative array like this. After that, I need to send it to PHP script:

typeset -A DATA
DATA[foo]=$(some_bash_function "param1" "param2")
DATA[bar]=$(some_other_bash_function)

php script.php --data ${DATA[@]}

From PHP script, i need to access the array in following manner:

<?php

    $vars = getopt("",array(
        "data:"
    ));

    $data = $vars['data'];

    foreach ($data as $k=>$v) {
          echo "$k is $v";
    }

?>

What I've tried

Weird syntax around the --data parameter follows advice from a great post about bash arrays from Norbert Kéri how to force passed parameter as an array:

You have no way of signaling to the function that you are passing an array. You get N positional parameters, with no information about the datatypes of each.

However this sollution still does not work for associative arrays - only values are passed to the function. Norbert Kéri made a follow up article about that, however its eval based solution does not work for me, as I need to pass the actual array as a parameter.

Is the thing I'm trying to achieve impossible or is there some way? Thank you!

Update: What I am trying to accomplish

I have a few PHP configuration files of following structure:

<?php
return array(
    'option1' => 'foo',
    'option2' => 'bar'
)

My bash script collects data from user input (through bash read function) and stores them into bash associative array. This array should be later passed as an argument to PHP script.

php script.php --file "config/config.php" --data $BASH_ASSOC_ARRAY

So instead of complicated seds functions etc. I can do simple:

<?php

    $bash_input = getopt('',array('file:,data:'));
    $data = $bash_input['data'];

    $config = require($config_file);
    $config['option1'] = $data['option1'];
    $config['option2'] = $data['option2'];

    // or

    foreach ($data as $k=>$v) {
         $config[$k] = $v;
    }

    // print to config file
    file_put_contents($file, "<?php \n \n return ".var_export($config,true).";");
?>

This is used for configuring Laravel config files

Could you echo the bash array to a temporary file and then read that file? (as ugly as that would be) — will
– will, Commented Aug 19, 2014 at 10:50
I am a bash novice, so I need to figure out how to "echo" or "pipe" to file first and how exactly it will help my case. :) However if you would have time to elaborate to an answer ... And yes, I can work with creating files, as I source scripts from the main bash. — Petr Cibulka
– Petr Cibulka, Commented Aug 19, 2014 at 10:54
@ZanderRootman: I'm afraid so, as I am collection the data from user input (read function) in terminal ... — Petr Cibulka
– Petr Cibulka, Commented Aug 19, 2014 at 11:00

Zander Rootman · Accepted Answer · 2014-08-19 13:31:51Z

4

Different Approach to @will's

Your bash script:

typeset -A DATA
foo=$(some_bash_function "param1" "param2")
bar=$(some_other_bash_function)

php script.php "{'data': '$foo', 'data2': '$bar'}"

PHP Script

<?php

    $vars = json_decode($argv[1]);

    $data = $vars['data'];

    foreach ($data as $k=>$v) {
          echo "$k is $v";
    }

?>

EDIT (better approach) Credit to @will

typeset -A DATA
DATA[foo]=$(some_bash_function "param1" "param2")
DATA[bar]=$(some_other_bash_function)

php script.php echo -n "{"; for key in ${!DATA[@]}; do echo - "'$key'":"'${DATA[$key]}'", | sed 's/ /,/g' ; done; echo -n "}"

edited Aug 19, 2014 at 13:31

answered Aug 19, 2014 at 11:29

Zander Rootman

2,2082 gold badges19 silver badges29 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

will Over a year ago

This is definately cleaner, but i think it would be much nicer if you made the json encoding automatic, something like echo -n "{"; for key in ${!assoc_array[@]}; do echo -n "'$key'":"'${assoc_array[$key]}'",; done; echo -n "}" ( i don't know if json minds the extra trailing comma at the end though, so you might need to trim that off...

Zander Rootman Over a year ago

Yeah the trailing comma would break the syntax. But a trim wont hurt anyone. Will update the answer quick

Petr Cibulka Over a year ago

This is pretty much the thing I was trying to accomplish with Zander's code at the moment, yet more elegant. :) Thanks guys, you both helped me a lot!

will · Accepted Answer · 2014-08-19 11:17:44Z

2

this does what you want (i think) all in one bash script. You can obviously move the php file out though.

declare -A assoc_array=([key1]=value1 [key2]=value2 [key3]=value3 [key4]=value4)

#These don't come out necesarily ordered
echo ${assoc_array[@]} #echos values
echo ${!assoc_array[@]} #echos keys

echo "" > tmp

for key in ${!assoc_array[@]}
do
echo $key:${assoc_array[$key]} >> tmp   # Use some delimeter here to split the keys from the values
done

cat > file.php << EOF
<?php

    \$fileArray = explode("\n", file_get_contents("tmp"));

    \$data = array();

    foreach(\$fileArray as \$line){
        \$entry = explode(":", \$line);
        \$data[\$entry[0]] = \$entry[1];
    }

    var_dump(\$data);

?>
EOF


php file.php

the escaping is necessary in the cat block annoyingly.

answered Aug 19, 2014 at 11:17

will

10.7k8 gold badges50 silver badges75 bronze badges

2 Comments

Petr Cibulka Over a year ago

Thank you! I will give that a try right now and get back in a minute.

Petr Cibulka Over a year ago

Will, your approach worked, although I will probably settle for (heavily modified) @ZanderRootman's approach. Thanks a lot, you taught me a lot about temporary bash data storage to files!

Collectives™ on Stack Overflow

Pass BASH associative arrays to PHP script

What I've tried

Update: What I am trying to accomplish

2 Answers 2

3 Comments

2 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

What I've tried

Update: What I am trying to accomplish

2 Answers 2

3 Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related