CODEWITHSUNDEEP
javascriptphpmysqlchartsgoogle-visualization
SmashingJummy
SmashingJummy

Reputation: 448

Unexpected token error when using PHP with JavaScript

Currently I'm struggling to get data loaded from my database to Google Charts, I'm following this tutorial.

I'm trying to load 4 different charts in my webpage. However everytime PHP starts to execute a SQL statement I get this error:

Uncaught SyntaxError: Unexpected token <

Inspect element shows this: The errors that I get[2]

I expect it is the same problem with every chart, so I will only post the code of the first chart, that looks like this:

<script type="text/javascript">
google.load("visualization", "1", {packages:["corechart"]});
google.setOnLoadCallback(drawChart);
function drawChart() {
  var data = google.visualization.arrayToDataTable([

      ['Date', 'Visits'],
        <?php 
            $query = "SELECT count(ip) AS count, vdate FROM visitors GROUP BY vdate ORDER BY vdate";

            $exec = mysqli_query($con,$query);
            while($row = mysqli_fetch_array($exec)){

                echo "['".$row['vdate']."',".$row['count']."],";
            }
   ?>

    ]);

    var options = {
        title: 'Date wise visits'
    };
  var chart = new google.visualization.ColumnChart(document.getElementById("columnchart"));
  chart.draw(data, options);
}
</script>

The error will start at the same line as PHP starts, so I expect the problem is related to that. I already tries a few things with CDATA, but that dind't work out.

Changing <script type="text/javascript"> to <script type="text/html"> makes the errors disappear, but won't let the script working, so what am I doing wrong here?

Upvotes: 0

Views: 1913

Answers (1)

Funk Forty Niner
Funk Forty Niner

Reputation: 74216

how are you using this, as http://localhost or file:///? – Fred -ii-
@Fred-ii- thanks! The example you posted worked! Many thanks! – SmashingJummy

So the answer to this is that you need to run as localhost, not as file:/// directly in your browser, since file:/// will not parse PHP directives.

  • I had a feeling about that.

Upvotes: 3

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