Ok, after fixing somethings on my page I get the graph to show but without
any "content". And after digging around I think this is the issue:
{"c":[{"v":"Top"},{"v":"61"},{"v":"57"}]}
The values (61 and 57) have " " beside them. If I hardcode a variable
without the " " it works...
So, how do I get rid of them? :)
Cheers
Adam
Den torsdagen den 12:e juli 2012 kl. 18:35:24 UTC+2 skrev asgallant:
>
> Is the json string being echo'd properly? Take a look at the rendered
> HTML to see (open page, right-click, select "view source" or something
> similar, depending on your browser). If it shows up, post the output HTML
> here. If it doesn't show up, post your PHP script here.
>
> On Wednesday, July 11, 2012 7:31:26 PM UTC-4, Adam Hardarson wrote:
>>
>> Nice work!
>>
>> I am however still getting the Table has no columns.×
>>
>> I copied the code from a google example but there is still something
>> wrong!
>>
>> This is what the script looks like:
>> <script type="text/javascript" src="https://www.google.com/jsapi
>> "></script>
>> <script type="text/javascript">
>> google.load("visualization", "1", {packages:["corechart"]});
>> google.setOnLoadCallback(drawChart);
>> function drawChart() {
>> var data = new google.visualization.DataTable(<?php echo
>> $dataTable; ?>);
>>
>> var options = {
>> title: 'MP'
>> };
>>
>> var chart = new
>> google.visualization.AreaChart(document.getElementById('chart_div'));
>> chart.draw(data, options);
>> }
>> </script>
>> </head>
>>
>> take care
>> Adam
>>
>> Den onsdagen den 11:e juli 2012 kl. 22:41:05 UTC+2 skrev asgallant:
>>>
>>> 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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