Storing PHP variable into JavaScript variable in a For loop

Question

I know how to pass a PHP variable into JavaScript variable one by one, but now I wish to pass a bunch of variable in a For loop and I just stuck.

<?php
        $m = new MongoClient();
        $db = $m->data_from_tweeter;
        $collection = $db->output;
        $cursor = $collection->find();
        foreach($cursor as $document){
            echo $document['place'] ;
            echo $document['fir'];
            echo $document['sec'];
        }
?>

I want to store 'place' as a String variable, 'fir' and 'sec' as two Num variables, and make them look like:

var num_1 {
    some_place : some_fir,
    some_place : some_fir,
    some_place : some_fir,
    ....
}

and

var num_2 {
    some_place : some_sec,
    some_place : some_sec,
    some_place : some_sec,
    ....
}

How can I make it? Thanks in advance!

Object keys can't be the same in the same object. Change your object structure. — Taha Paksu
– Taha Paksu, Commented May 7, 2018 at 7:17

Hưng hoàng · Accepted Answer · 2018-05-07 07:16:36Z

1

if $document['fir'] and $document['sec'] are array .You can do like this

<?php
$m = new MongoClient();
$db = $m->data_from_tweeter;
$collection = $db->output;
$cursor = $collection->find();
$num1=array();
$num2=array();
foreach($cursor as $document){
    echo $document['place'] ;
    $num1 = json_encode ($document['fir']);
    $num2 = json_encode ($document['sec']);
}
?>

You can use a JSON string as a object for javascript

var num_1  = <?php echo $num1;?>
var num_2  = <?php echo $num2;?>

edited May 7, 2018 at 7:16

answered May 7, 2018 at 7:07

Hưng hoàng

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Comments

e.dan · Accepted Answer · 2018-05-07 07:16:01Z

0

The following code snippet might help get you started:

$num_1 = "var num_1 = {\n";
$num_2 = "var num_2 = {\n";
foreach ($cursor as $document) {
        $num_1 .= "    {$document['place']} : {$document['fir']},\n";
        $num_2 .= "    {$document['place']} : {$document['sec']},\n";
}  
$num_1 .= "};\n";
$num_2 .= "};\n";

echo $num_1;
echo $num_2;

answered May 7, 2018 at 7:16

e.dan

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1 Comment

fornowfortest Over a year ago

Thanks! Make sense a lot to me!

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