codeigniter - 使用 CodeIgniter 基于 JSON 数据创建 Google 饼图

Question

我尝试根据示例创建饼图 here

在我的 Controller (mycontroller/json) 中，我有以下代码:

public function json(){

$whatever = '{
"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}]}
  ]
}';
echo $whatever;
}

在我看来，我有以下代码(显示了部分代码)

var jsonData = $.ajax({
    url: "<?= base_url()?>index.php/bunker/json",
    dataType:"json",
    async: false
    }).responseText;

// Create and populate the data table.
var popularCategory = google.visualization.DataTable(jsonData);
new google.visualization.PieChart(document.getElementById('category')).draw(popularCategory,
    {title:"Popularity by Category"});

最后我有一个名为“category”的 id 的 div。但是，我总是收到一条消息说数据未定义。

使用 firebug 进行快速调试，我没有收到任何错误，$whatever 变量也被视为正确的 JSON 格式。我在这里犯了什么错误？

干杯，

Answer 1

代替:

var popularCategory = google.visualization.DataTable(jsonData);

你必须有:

var popularCategory = new google.visualization.DataTable(jsonData);