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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