Send json data into a php variable to be used in javascript

Question

I'm quite new to php and don't actually know if this is possible

Im currently outputting JSON code with php using the following code

echo  json_encode($output, JSON_NUMERIC_CHECK);

But what I want to do is have the above data inside a variable.

I tried

$JSONDATAX = json_encode($output, JSON_NUMERIC_CHECK);

But it doesn't seem to like it when I call $JSONDATAX.

The original echo way works completely fine.

edit ........

      $lrs = CDB::ExecuteQuery($sql);

            if($lrs) {
    $jsonData = convert($lrs);
}



function convert($lrs) {

    $intermediate = array();

      while ($vals = CDB::GetAssoc($lrs))  {
        $key = $vals['POS'];
        $x = $vals['CODE'];
        $y = $vals['COUNT'];
        $intermediate[$key][] = array('x' => $x, 'y' => $y);
    }


   $output = array();

    foreach($intermediate as $key => $values) {
        $output[] = array(
            "key" => $key,
            'values' => $values
        );
    }

  $data1 = json_encode($output, JSON_NUMERIC_CHECK);


}










?>
<script>

 var negative_test_data = <?php echo $data1; ?>;

var chart;
nv.addGraph(function() {
    chart = nv.models.multiBarChart()
    .color(d3.scale.category10().range())
      .rotateLabels(0)      //Angle to rotate x-axis labels.
      .transitionDuration(300)
       .showControls(true)   //Allow user to switch between 'Grouped' and 'Stacked' mode.
      .groupSpacing(0.24)    //Distance between each group of bars.

      ;

As you can see, I am using php just after var negative_test_data , but it doesn't produce anything.

I get an array back with the code above by echoing it, and this works fine with the chart i'm using. but I'm trying to say if I try make it a variable, it doesn't work. — Engl12
– Engl12, Commented Oct 9, 2014 at 14:28
I don't think I've explained this very well, I will edit post. — Engl12
– Engl12, Commented Oct 9, 2014 at 14:36

Roel Harbers · Accepted Answer · 2014-10-09 15:48:09Z

1

In your edited example, $data is a local variable inside the convert function, so it cannot be accessed outside that function. the result of json_encode should be returned:

$data1 = json_encode($output, JSON_NUMERIC_CHECK);

should be

return json_encode($output, JSON_NUMERIC_CHECK);

Then, the result of the convert function can be echoed:

var negative_test_data = <?php echo $data1; ?>;

should be

var negative_test_data = <?php echo convert($lrs); ?>;

(There should probably be a an additional if around that whole part, depending on what you want to happen when $lrs does not evaluate to true)

answered Oct 9, 2014 at 15:48

Roel Harbers

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1 Comment

Engl12 Over a year ago

- Thanks for the answer man, worked a treat! Although I had to do var negative_test_data = <?php echo convert($jsonData); ?>; rather than $lrs

Marc B · Accepted Answer · 2014-10-09 14:42:03Z

0

This should be all you really need:

$phpvar = json_encode($output);
echo "<script>var negative_test_data = {$phpvar};</script>";

answered Oct 9, 2014 at 14:42

Marc B

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2 Comments

Engl12 Over a year ago

i used, $phpvar = json_encode($output, JSON_NUMERIC_CHECK); and inside my <script> tags <?PHP echo "var negative_test_data = {$phpvar};"; ?> It didn't bring anything back, just blank.

Marc B Over a year ago

then do a var_dump($phpvar). if it says boolean false, then the json encoding failed, and you need to check json_last_error()

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