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up vote 3 down vote favorite

I want to print a statistical graph through my PHP file when it's function is called. I use Google Charts API for the GUI and print of the actual graphs.

Here is a snippet from code I got from Google API (javascript):

var data = google.visualization.arrayToDataTable([
      ['Year', 'Sales', 'Expenses'],
      ['2004',  1000,      400],
      ['2005',  1170,      460],
      ['2006',  660,       1120],
      ['2007',  1030,      540]
    ]);

What I want to do is to replace the above values with PHP variables (probably a 2D-array) but I have not managed to get it to work. It either gives me a PHP error, or prints nothing, depending on what I try.

This is what I tried:

var data = google.visualization.arrayToDataTable(
    $array[0][0],$array[0][1],$array[0][2]],
    $array[1][0],$array[1][1],$array[1][2]],
    $array[2][0],$array[2][1],$array[2][2]],
    $array[3][0],$array[3][1],$array[3][2]],
    $array[4][0],$array[4][1],$array[4][2]]
);

and the array I made:

$array = array(array('Year', 'Sales', 'Expenses'), array('2004', 1000, 400), ... );

Ideas anyone? :) Thanks.

share|improve this question
    
what are the errors? have you tried print_r on $array, $array[0] and $array[0][1]? –  TecHunter Apr 24 '14 at 16:44
    
You could try to convert to JSON - probably then it will be much easier to work with from both sides. –  Stan Apr 24 '14 at 16:45
    
You are missing an opening square bracket in your "tried" section for each line. But really you should just be able to call json_encode on the array. –  Jonathan Kuhn Apr 24 '14 at 16:46
    
so just $array = json_encode($array); or? –  Chris Apr 24 '14 at 16:49
    
and yes, that bracket was an accidental leftover from my last failed solution. :P –  Chris Apr 24 '14 at 16:50

1 Answer 1

up vote 4 down vote accepted

Your PHP array format matches the structure required for the chart, so you can simply JSON encode it:

var data = google.visualization.arrayToDataTable(<?php echo json_encode($array); ?>);

This works because JSON is valid plain old JavaScript when assigned to a variable or passed as a function argument.

Whenever you output PHP variables into the page, you need to use the PHP tags and echo or similar.

If the JavaScript is in a PHP string that you echo, you can just concatenate the JSON into it , for example

$js = '
    <script>
    var data = google.visualization.arrayToDataTable(
    ' . json_encode($array) . '
    );
    </script>
';
echo $js;
share|improve this answer
    
+1 Prefect solution. –  iambriansreed Apr 24 '14 at 16:52
    
Hmm, this did not work for me. I get no error, but nothing is printed on screen. –  Chris Apr 24 '14 at 17:01
    
I suspect it's because I am already echoing the html? –  Chris Apr 24 '14 at 17:01
    
Is the JavaScript in a PHP string? or do you close the PHP tag to output the JS? –  MrCode Apr 24 '14 at 17:03
    
The former way. –  Chris Apr 24 '14 at 17:07

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