1) I have tried to use DataSource URL as my PHP URL. But i got errror as " Uncaught Error: Missing query for request id: undefined ". shall i want to pass any parameter to query static json data from my PHP?
2) Also i try a simple table like below, *index.html:* <script type="text/javascript" src="https://www.google.com/jsapi";></script> > <script type="text/javascript" src="tablequerywrapper.js"></script> > <script type="text/javascript"> > google.load('visualization', '1', {'packages' : ['table']}); > google.setOnLoadCallback(init); > var dataSourceUrl = 'getData.php'; > var query, options, container; > function init() { > query = new google.visualization.Query(dataSourceUrl); > container = document.getElementById("table"); > options = {'pageSize': 2}; > sendAndDraw(); > } > function sendAndDraw() { > query.abort(); > var tableQueryWrapper = new TableQueryWrapper(query, container, > options); > tableQueryWrapper.sendAndDraw(); > } > > function setOption(prop, value) { > options[prop] = value; > sendAndDraw(); > } > </script> > </head> > <body> > <form action=""> > Number of rows to show: > <select onChange="setOption('pageSize', parseInt(this.value, 10))"> > <option value="0">0</option> > <option value="3">3</option> > <option selected=selected value="5">5</option> > <option value="8">8</option> > <option value="-1">-1</option> > </select> > </form> > <br /> > <div id="table"></div> > </body> *getData.php:* * * > <?php > $string = file_get_contents("sampleData.json"); > echo $string; > ?> * sampleData.json:* * * > { > "cols": [ > {"id":"","label":"Topping","pattern":"","type":"string"}, > {"id":"","label":"Slices","pattern":"","type":"number"} > ], > "rows": [ > {"c":[{"v":"Mushrooms","f":null},{"v":3,"f":null}]}, > {"c":[{"v":"Onions","f":null},{"v":1,"f":null}]}, > {"c":[{"v":"Olives","f":null},{"v":1,"f":null}]}, > {"c":[{"v":"Zucchini","f":null},{"v":1,"f":null}]}, > {"c":[{"v":"Pepperoni","f":null},{"v":2,"f":null}]} > ] > } This is also not working? On Thu, Apr 12, 2012 at 7:27 PM, asgallant <[email protected]>wrote: > Assuming that your JSON is properly formatted in the way that the > Visualization API expects it to be, then you can just use the url to your > data source in place of the url to the Google spreadsheet and it should > work fine. > > > On Thursday, April 12, 2012 2:30:43 AM UTC-4, Ram wrote: >> >> I want to create a Table like "Table Query Wrapper Example" (like >> https://developers.**google.com/chart/interactive/**docs/examples#** >> tablequerywrapper<https://developers.google.com/chart/interactive/docs/examples#tablequerywrapper> >> ). I >> have noticed that this Table has referred data from a Spread sheet. But Now >> I have populated data as Json on server side php for my Visualization >> Table. Is it possible to use my php url(which returns JSON data for Table) >> for creating like TableQueryWrapper suppoted Table. >> >> Also i have read I need a Google Account with already created Spread >> sheet for getting spread sheet URL.So How to create a Spreadsheet >> dynamically from a PHP. >> > -- > You received this message because you are subscribed to the Google Groups > "Google Visualization API" group. > To view this discussion on the web visit > https://groups.google.com/d/msg/google-visualization-api/-/N7Zfni2nhXkJ. > > To post to this group, send email to > [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/google-visualization-api?hl=en. > -- BY R.RAMPRASAD -- You received this message because you are subscribed to the Google Groups "Google Visualization API" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/google-visualization-api?hl=en.
