Hi,

It looks like your issue is that your dates are poorly formatted. They
should be formatted as "Date(2014, 11, 06, 09, 48, 03)", instead of "new
Date(...)". Just to clarify: when representing dates as a string in your
JSON, you should not write "new" in the string.


Hallo,

Es sieht aus wie das Problem ist, dass Ihre Daten sind schlecht formatiert.
Sie sollten als "Date(2014, 11, 06, 09, 48, 03)" formatiert werden, statt
"new Date (...)". Nur um zu klären: bei der Darstellung von Daten als
String in Ihrem JSON, sollten Sie nicht zu schreiben "new" in der
Zeichenkette.

On Mon Dec 08 2014 at 7:25:57 AM Phillip Dörrschuck <
[email protected]> wrote:

> Hallo, ich benötige hilfe bei meinem Google Bar Chart. Ich hole per PHP
> die Daten aus meiner Datenbank mit json, die funktioniert auch einwandfrei,
> nur bekomme ich als Fehlermeldung: a[ec] is not a function. Ich habe
> dieses Beispiel genommen https://google-developers.
> appspot.com/chart/interactive/docs/gallery/columnchart.
>
> *Hier einmal meine bar.html:*
>
> <html>
>   <head>
>     <script type="text/javascript" src="https://www.google.com/jsapi";
> ></script>
>     <script type="text/javascript" src="//ajax.googleapis.com/
> ajax/libs/jquery/1.10.2/jquery.min.js"></script>
>     <script type="text/javascript">
>
>       google.load("visualization", "1.1", {packages:["bar"]});
>
>       google.setOnLoadCallback(drawChart);
>
>       function drawChart() {
>         var jsonData = $.ajax({
>           url: "code.php",
>           dataType:"json",
>           async: false
>           }).responseText;
>
>
>         var data = new google.visualization.DataTable(jsonData);
>
>
>         var chart = new google.charts.Bar(document.getElementById(
> 'columnchart_material'));
>
>         chart.draw(data, {width: 900, height:500} );
>       }
>     </script>
>   </head>
>   <body>
>     <div id="columnchart_material" style="width: 900px; height: 500px;"
> ></div>
>   </body>
> </html>
>
>
> *Hier meine code.php:*
>
> <?php
>     $username = "***";
>     $password = "****";
>     $host = "localhost";
>     $database="***";
>
>     $server = mysql_connect($host, $username, $password);
>     $connection = mysql_select_db($database, $server);
>
>     $myquery = "SELECT  `date`, `connectionsEstablished` FROM  `tbl`";
>     $query = mysql_query($myquery);
>
>     if ( ! $query ) {
>         echo mysql_error();
>         die;
>     }
>
>     $result = array();
>
>
>     $table = array();
>     $table['cols'] = array(
>     array('id' => "", 'label' => 'date', 'pattern' => "", 'type' =>
> 'datetime'),
>     array('id' => "", 'label' => 'connectionsEstablished', 'pattern' => ""
> , 'type' => 'number')
>     );
>     $rows = array();
>     while ($nt = mysql_fetch_assoc($query))
>     {
>     $temp = array();
> //print_r(date_create($nt['date'])->format("Y, m, d, H, i, s"));
>     $temp[] = array('v' => date_create($nt['date'])->format('\n\e\w
> \D\a\t\e\(Y, m, d, H, i, s\)'), 'f' =>NULL);
>     $temp[] = array('v' => $nt['connectionsEstablished'], 'f' =>NULL);
>     $rows[] = array('c' => $temp);
>     }
>     $table['rows'] = $rows;
>  $jsonTable = json_encode($table, JSON_NUMERIC_CHECK);
> echo $jsonTable;
>
>     mysql_close($server);
> ?>
>
>
> *Und hier mal ein Ausschnit von dem generierten Jsoncode: *
> {"cols":[{"id":"","label":"date","pattern":"","type":"datetime"},{"id":"",
> "label":"connectionsEstablished","pattern":"","type":"number"}],"rows":[{
> "c":[{"v":"new
>  Date(2014, 11, 06, 09, 48,
> 03)","f":null},{"v":23438,"f":null}]},{"c":[{"v":"new Date(2014, 11, 06,
>  09, 48, 11)","f":null},{"v":23440,"f":null}]}
>
>
> Ich hoffe mir kann jemand helfen.
>
> Liebe Grüße Phillip
>
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