I changed it up a bit to fit whatever column structure you like:
$sqlCols = array(
'Matning',
'Top',
'In',
'Pre',
'Int'
);
$result124 = mysql_query("
SELECT Matningsnamn AS Matning,
Top,
In,
Pre,
Int
FROM myDB WHERE Id ='$first' OR Id ='$second'"
);
$rawRows = array();
$cols = array();
// set up domain column
$cols[] = array(
'type' => 'string',
'label' => 'Foo (rename as you like)'
);
while($r = mysql_fetch_assoc($result124)) {
$cols[] = array(
'type' => 'string',
'label' => $r[$sqlCols[0]]
);
$rawRows[] = $r;
}
$outRows = array();
for ($i = 1; $i < count($rawRows[0]); $i++) {
$temp = array();
$temp[0] = array('v' => $sqlCols[$i]);
for ($j = 0; j < count($rawRows); j++) {
if ($rawRows[$j]['Matning'] == $cols[1]['label']) {
$temp[1] = array('v' => $rawRows[$j][$sqlCols[$i]]);
}
else if ($rawRows[$j]['Matning'] == $cols[2]['label']) {
$temp[2] = array('v' => $rawRows[$j][$sqlCols[$i]]);
}
}
$outRows[] = array('c' => $temp);
}
$dataTable = json_encode(array(
'cols' => $cols,
'rows' => $outRows
));
//.... then in the javascript, use:
var data = new google.visualization.DataTable(<?php echo $dataTable; ?>);
Make sure you are using a DataTable constructor and not the
arrayToDataTable method. I tried out the JSON string you posted and it
works fine for me.
On Wednesday, July 11, 2012 3:23:00 PM UTC-4, Adam Hardarson wrote:
>
> That solved that problem but now i get Table has no columns.×
> if I print the json I get:
>
> {"cols":[{"type":"string","label":"Value
> #"},{"type":"string","label":"Test 1"},{"type":"string","label":"Test 2"}],
>
> "rows":[{"c":[{"v":"Value1"},{"v":"61"},{"v":"66"}]},{"c":[{"v":"Value2"},{"v":"90"},{"v":"83"}]},{"c":[{"v":"Value3"},{"v":"37"},{"v":"71"}]},{"c":[{"v":"Value4"},{"v":"48"},{"v":"84"}]}]}
>
>
> For this to work I changed my original code like this:
>
> $result124 = mysql_query("
> SELECT Matningsnamn AS Matning,
> Top AS Value1,
> In AS Value2 ,
> Pre AS Value3 ,
> Int ASValue4
> FROM myDB WHERE Id ='$first' OR Id ='$second'"
>
> So I realize that I wrote that the "columnnames" were Value1, Value2,
> Value3 and Value 4 but this was just to illustrate. The column names is
> totally different from that, will that screw the code up or is it easy to
> edit to display the real names? Or even better is it possible to enter the
> labels manually and just getting the values from the database? So that
> when I choose a new row from the dropdown only the values changes?
>
> Sorry for confusing everything!!!!
>
> Adam
>
> Den onsdagen den 11:e juli 2012 kl. 20:30:18 UTC+2 skrev asgallant:
>>
>> I forgot the $'s in front of the i's there, try this:
>>
>> for ($i = 1; $i < count($rawRows[0]); $i++) {
>>
>> On Wednesday, July 11, 2012 2:04:09 PM UTC-4, Adam Hardarson wrote:
>>>
>>> Hi again, thank you so much for your assistance!
>>>
>>> However it's not working so I the page for syntax error and got a Parse
>>> error: syntax error, unexpected T_INC, expecting ')' in test.php on line 37.
>>>
>>> That's where the for loop starts... I can't see what the problem might
>>> be?!
>>>
>>>
>>>> $result124 = mysql_query("
>>>> SELECT Matningsnamn AS Matning,
>>>> Value1,
>>>> Value2 ,
>>>> Value3 ,
>>>> Value4
>>>> FROM myDB WHERE Id ='$first' OR Id ='$second'"
>>>> );
>>>> $rawRows = array();
>>>> $cols = array();
>>>> // set up domain column
>>>> $cols[] = array(
>>>> 'type' => 'string',
>>>> 'label' => 'Value #'
>>>> );
>>>> while($r = mysql_fetch_assoc($result124)) {
>>>> $cols[] = array(
>>>> 'type' => 'string',
>>>> 'label' => $r['Matning']
>>>> );
>>>> $rawRows[] = $r;
>>>> }
>>>>
>>>> $outRows = array();
>>>> for ($i = 1; i < count($rawRows[0]); i++) {
>>>> $temp = array();
>>>> $temp[0] = array('v' => "Value$i");
>>>> for ($j = 0; j < count($rawRows); j++) {
>>>> if ($rawRows[$j]['Matning'] == $cols[1]['label']) {
>>>> $temp[1] = array('v' => $rawRows[$j]["Value$i"]);
>>>> }
>>>> else if ($rawRows[$j]['Matning'] == $cols[2]['label']) {
>>>> $temp[2] = array('v' => $rawRows[$j]["Value$i"]);
>>>> }
>>>> }
>>>> $outRows[] = array('c' => $temp);
>>>> }
>>>>
>>>> $dataTable = json_encode(array(
>>>> 'cols' => $cols,
>>>> 'rows' => $outRows
>>>> ));
>>>>
>>>> You can then output this into the DataTable constructor:
>>>>
>>>> var data = new google.visualization.DataTable(<?php echo $dataTable; ?>
>>>> );
>>>>
>>>
>>> Seeing the end of the tunnel... :)
>>>
>>> Adam
>>>
>>
> Den onsdagen den 11:e juli 2012 kl. 20:30:18 UTC+2 skrev asgallant:
>>
>> I forgot the $'s in front of the i's there, try this:
>>
>> for ($i = 1; $i < count($rawRows[0]); $i++) {
>>
>> On Wednesday, July 11, 2012 2:04:09 PM UTC-4, Adam Hardarson wrote:
>>>
>>> Hi again, thank you so much for your assistance!
>>>
>>> However it's not working so I the page for syntax error and got a Parse
>>> error: syntax error, unexpected T_INC, expecting ')' in test.php on line 37.
>>>
>>> That's where the for loop starts... I can't see what the problem might
>>> be?!
>>>
>>>
>>>> $result124 = mysql_query("
>>>> SELECT Matningsnamn AS Matning,
>>>> Value1,
>>>> Value2 ,
>>>> Value3 ,
>>>> Value4
>>>> FROM myDB WHERE Id ='$first' OR Id ='$second'"
>>>> );
>>>> $rawRows = array();
>>>> $cols = array();
>>>> // set up domain column
>>>> $cols[] = array(
>>>> 'type' => 'string',
>>>> 'label' => 'Value #'
>>>> );
>>>> while($r = mysql_fetch_assoc($result124)) {
>>>> $cols[] = array(
>>>> 'type' => 'string',
>>>> 'label' => $r['Matning']
>>>> );
>>>> $rawRows[] = $r;
>>>> }
>>>>
>>>> $outRows = array();
>>>> for ($i = 1; i < count($rawRows[0]); i++) {
>>>> $temp = array();
>>>> $temp[0] = array('v' => "Value$i");
>>>> for ($j = 0; j < count($rawRows); j++) {
>>>> if ($rawRows[$j]['Matning'] == $cols[1]['label']) {
>>>> $temp[1] = array('v' => $rawRows[$j]["Value$i"]);
>>>> }
>>>> else if ($rawRows[$j]['Matning'] == $cols[2]['label']) {
>>>> $temp[2] = array('v' => $rawRows[$j]["Value$i"]);
>>>> }
>>>> }
>>>> $outRows[] = array('c' => $temp);
>>>> }
>>>>
>>>> $dataTable = json_encode(array(
>>>> 'cols' => $cols,
>>>> 'rows' => $outRows
>>>> ));
>>>>
>>>> You can then output this into the DataTable constructor:
>>>>
>>>> var data = new google.visualization.DataTable(<?php echo $dataTable; ?>
>>>> );
>>>>
>>>
>>> Seeing the end of the tunnel... :)
>>>
>>> Adam
>>>
>>
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