In that PHP script, instead of putting the json in a file, why not return
the json in response to client request? There's no need to clutter your
server with json files.
On Friday, August 17, 2012 7:29:32 AM UTC-4, Chriggi wrote:
>
> Thanks for your solution with the cases, it worked very well.
>
> Actually I already have a PHP script which creates the JSON file. On the
> website I can select which columns I want to have and the script creates
> the JSON file based on the created query.
>
> At the end of the script I do this:
> $fp = fopen('resultsOwnTable.json', 'w');
> fwrite($fp, json_encode($responseTable));
>
> So basically I could just save do this?
> $table = json_encode($responseTable)
> And use this to populate my DataTable in Javascript?
>
> Am Freitag, 17. August 2012 05:09:10 UTC+2 schrieb asgallant:
>>
>> You might want to consider building a PHP script that will query your
>> database for you and output the data in the DataTable JSON format (or you
>> could try out the 3rd-party PHP
>> library<http://code.google.com/p/mc-goog-visualization/wiki/UserDocumentation>that
>> implements the Google Query interface).
>>
>> If you don't want to do either of those, you could make a quick-fix
>> change to the PHP file to make it accept a parameter via the url and select
>> the JSON file based on that:
>>
>> <?php
>> // get the index of the file to output
>> $index = $_GET['index'];
>>
>> // select the file based on $index
>> switch ($index) {
>> case 0:
>> echo file('/path/to/file_0.json');
>> break;
>> case 1:
>> echo file('/path/to/file_1.json');
>> break;
>> case 2:
>> echo file('/path/to/file_2.json');
>> break;
>> case 3:
>> echo file('/path/to/file_3.json');
>> break;
>> .
>> .
>> .
>> case n:
>> echo file('/path/to/file_n.json');
>> break;
>> }
>> ?>
>>
>> You then choose the file to grab like this:
>>
>> var jsonData = $.ajax({
>> url: "getData.php?index=3",
>> dataType: "json",
>> async: false
>> }).responseText;
>> var data = new google.visualization.DataTable(jsonData);
>>
>> The selector doesn't have to be "index" and it doesn't have to be numeric
>> - as long as you can map the request from the js end of things to a file on
>> your server, it will work.
>>
>> On Thursday, August 16, 2012 6:46:21 PM UTC-4, Chriggi wrote:
>>>
>>> Hello,
>>>
>>> I have a lot of json-files with different data from my database and want
>>> to populate my datatables with them. Currently I'm using the method
>>> described here:
>>> https://developers.google.com/chart/interactive/docs/php_example
>>>
>>> Basically I do this:
>>>
>>> var jsonData = $.ajax({
>>> url: "getData.php",
>>> dataType:"json",
>>> async: false
>>> }).responseText;
>>> var data = new google.visualization.DataTable(jsonData);
>>>
>>> In getData.php there are just these 2 lines:
>>> $string = file_get_contents("results.json");
>>> echo $string;
>>>
>>> The problem here is, with one .php-file I can only get one JSON. So if i
>>> have for example 10 different charts with 10 different datatables created
>>> from different JSONS I need 10 .php-files. Isn't there a better solution
>>> for this?
>>>
>>>
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