I know what you mean, but I'm not sure how to do this. So I just save this
in a variable json_encode($responseTable)and in Javascript I use
.responseText() on this variable and use it as a DataTable?
Am Freitag, 17. August 2012 17:24:11 UTC+2 schrieb asgallant:
>
> 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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