I've never seen any " (string) " message. Can you post a screenshot, your
javascript code, and a sample of the json returned by your query?
On Wednesday, December 12, 2012 12:55:49 PM UTC-5, Ian Haylock wrote:
>
> query solved !!!!!!!! thank you.
> now my only problem is that before the browsers shows the graphs i keep
> getting the " (string) " message and have to click on it to see the graph
> ... any ideas ??? i have changed the types, added options, etc and still
> nothing
>
> On Wednesday, December 12, 2012 11:38:06 AM UTC-6, asgallant wrote:
>>
>> This isn't causing your problem, but since your "capacidadMW" is a
>> decimal type, and not an integer, you should probably change (int)
>> $r['capacidadMW']); to (float) $r['capacidadMW']);
>>
>> That aside, I don't see anything in your code that would truncate the
>> results. Try running this and see what number is displayed:
>>
>> <?php
>>
>> include('connect-db.php');
>>
>> $sth = mysql_query("SELECT planta, capacidadMW FROM planta");
>> echo mysql_num_rows($sth);
>> ?>
>>
>> If that shows 55 records, then there is a problem in the PHP; if it shows
>> less than 55, then there is a problem with your connection to the database.
>>
>> On Wednesday, December 12, 2012 12:17:24 PM UTC-5, Ian Haylock wrote:
>>>
>>> just to let u know, the "planta" record in my table is type Varchar(45)
>>> and the "capacidadMW" is type decimal (9,2)
>>>
>>> On Wednesday, December 12, 2012 11:03:40 AM UTC-6, Ian Haylock wrote:
>>>>
>>>> thanks for the reply ..
>>>>
>>>> the query im using in phpmyadmin showing 55 records: SELECT planta,
>>>> capacidadMW FROM planta;
>>>>
>>>> the code using to encode using JSon (result is only 14 records):
>>>>
>>>> <?php
>>>>
>>>> include('connect-db.php');
>>>>
>>>> $sth = mysql_query("SELECT planta, capacidadMW FROM planta");
>>>> $rows = array();
>>>> $flag = true;
>>>>
>>>> $table = array();
>>>> $table['cols'] = array(
>>>>
>>>> array('label' => 'planta', 'type' => 'string'),
>>>> array('label' => 'capacidad en MW', 'type' => 'number')
>>>> );
>>>>
>>>> $rows = array();
>>>> while($r = mysql_fetch_assoc($sth)) {
>>>> $temp = array();
>>>> $temp[] = array('v' => (string)$r['planta']);
>>>> $temp[] = array('v' => (int) $r['capacidadMW']);
>>>> $rows[] = array('c' => $temp);
>>>> }
>>>>
>>>> $table['rows'] = $rows;
>>>>
>>>> $jsonTable = json_encode($table);
>>>>
>>>> echo $jsonTable;
>>>> ?>
>>>>
>>>>
>>>> On Wednesday, December 12, 2012 10:55:45 AM UTC-6, asgallant wrote:
>>>>>
>>>>> At a guess, I would say you have an error in the SQL in PHP. If you
>>>>> post your PHP code, I'll take a look at it.
>>>>>
>>>>> On Wednesday, December 12, 2012 11:41:25 AM UTC-5, Ian Haylock wrote:
>>>>>>
>>>>>> Hi asgallant, thank you very much for th taking the time to help
>>>>>> others. i have a question if you dont mind, i followed Diana´s sample
>>>>>> and
>>>>>> got it to work, well sort of, i used my onw db and managed to show the
>>>>>> graph and everithing, the only thing is that my mysql query running in
>>>>>> phpmyadmin shows 55 records and my json result shows only 14 records,
>>>>>> how
>>>>>> is that possible ???
>>>>>>
>>>>>> On Wednesday, December 12, 2012 12:54:06 AM UTC-6, Chrystopher Medina
>>>>>> wrote:
>>>>>>>
>>>>>>>
>>>>>>> ok im gonna start with the php information. and u know i have
>>>>>>> another problem how i cant charge two pie charts in the same place but
>>>>>>> each
>>>>>>> with different results.... u know i have two questions and i want to
>>>>>>> charge
>>>>>>> the results from them. in two pie charts..... because i read that just
>>>>>>> one
>>>>>>> pie chart can be charged in the same place.... i could be mistaken
>>>>>>
>>>>>>
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