如何为JSON文件中的每个对象声明一个变量？

Question

我有一个包含4个对象的data.json文件：a，b，c，d。

手动

我手动声明了变量。

if name == "main":
    for N in range(6):
        a = array(N,N,N)
        for i in range(N):
            a[i,i,i] = 1
        for i in range(N):
            for j in range(N):
                for k in range(N):
                    if i==j==k and a[i,j,k] == 1:
                        print(" \nYay! *a[{},{},{}] = 1 * ".format(i,j,k))
                    if a[i,j,k] == 1 and not i==j==k:
                        print(" \n+++ Crap!!! a[{},{},{}] = 1  +++ ".format(i,j,k))

动态

现在，我想动态地创建它，以便它可以处理我的JSON文件中的所有对象，并且它不会中断 - 当我向JSON文件添加更多对象时。

所以在循环中，我尝试添加：

var chart_a = new google.visualization.PieChart(document.getElementById('sa-piechart-a'));
var chart_b = new google.visualization.PieChart(document.getElementById('sa-piechart-b'));
var chart_c = new google.visualization.PieChart(document.getElementById('sa-piechart-c'));
var chart_d = new google.visualization.PieChart(document.getElementById('sa-piechart-d'));  

我从Ajax调用中获取了对象。

chart[object] = new google.visualization.PieChart($('#sa-piechart-'+ object.toLowerCase() )); // <---- I try to add this line

结果

我不断获得var data = {}; var chart = {}; for (var object in objects) { var total = objects[object].danger + objects[object].warning + objects[object].success ; data[object] = google.visualization.arrayToDataTable([ ['Piechart' , 'Number of Skills'], ['danger' , ( objects[object].danger/total ) * 100 ], ['warning' , ( objects[object].warning/total ) * 100 ], ['success' , ( objects[object].success/total ) * 100 ], ]); //console.log( '#sa-piechart-' + object.toLowerCase() ); // Return #sa-piechart-a chart[object] = new google.visualization.PieChart($('#sa-piechart-'+ object.toLowerCase() )); // <---- I try to add this line } // var chart_a = new google.visualization.PieChart(document.getElementById('sa-piechart-a')); // var chart_b = new google.visualization.PieChart(document.getElementById('sa-piechart-b')); // var chart_c = new google.visualization.PieChart(document.getElementById('sa-piechart-c')); // var chart_d = new google.visualization.PieChart(document.getElementById('sa-piechart-d'));

任何提示/建议都将不胜感激！

更新

Answer 1

Container is not defined表示找不到评论中提到的DOM元素。您需要动态创建元素，然后将它们传递给谷歌饼图。这样的事情应该做。

var el = document.getElementById('sa-piechart-' + object);
if (!el) {
  el = document.createElement('div');
  el.id = object;
  document.body.appendChild(el); // <--- I added to body. Add to required element
}
chart[object] = new google.visualization.PieChart(el); 
chart[object].draw(data[object]);

这是一个演示 http://jsfiddle.net/6M2sH/443/

一个更加漂亮的解决方案（使用jQuery）

var $el = $('#sa-piechart-' + object).length ? $('#sa-piechart-' + object) : $('<div id="#sa-piechart-' + object+'"></div>').appendTo('body');
chart[object] = new google.visualization.PieChart($el[0]);

发烧友演示 http://jsfiddle.net/6M2sH/444/