java - Google Chart JSON 问题 - Java

Question

我正在使用 jquery 从 servlet 返回一个基本的 JSON 对象，它工作得非常好，但是我在格式化 JSON 以实际填充图表时遇到了问题。

这是调用 servlet 的代码:

google.load("visualization", "1", {packages:["corechart"]});
google.setOnLoadCallback(drawChart);
function drawChart() {

    $.getJSON('SearchServlet', {servletAction: 'getSummary'}, function(json) {
        var data = new google.visualization.DataTable(json);
          var chart = new google.visualization.PieChart(document.getElementById('summaryPieChart'));
          chart.draw(data, {backgroundColor:'#e9f1f4',width: '90%', height: '90%', legend:{position:'none'}, chartArea:{width:"90%",height:"90%"},colors:['#94d784','#d78484']});
    });

}

在 servlet 端，我用来创建 JSON 的代码是:

private int positive;
private int negative;
private int neutral;

public DataTable createResultsJSON(){
    DataTable data = new DataTable();
    ArrayList<ColumnDescription> cols = new ArrayList<ColumnDescription>();
    cols.add(new ColumnDescription("summary", ValueType.TEXT, "Summary"));
    cols.add(new ColumnDescription("result", ValueType.NUMBER, "Result"));

    data.addColumns(cols);

    try {
        data.addRowFromValues("positive", positive, true);
        data.addRowFromValues("negative", negative, true);
        data.addRowFromValues("neutral", neutral, true);
      } catch (TypeMismatchException e) {
        System.out.println("Invalid type!");
      }
      return data;
}

使用以下方法将其转换为 JSON:

String resultJson = new Gson().toJson(createResultsJSON());

结果 JSON 的示例是:

{
"columns":[
{"id":"summary","type":"TEXT","label":"Summary","pattern":""}, 
{"id":"result","type":"NUMBER","label":"Result","pattern":""}],
"columnIndexById":{"result":1,"summary":0},
"rows":[
{"cells":[{"value":{"value":"positive"}},{"value":{"value":362.0}}]},
{"cells":[{"value":{"value":"negative"}},{"value":{"value":302.0}}]},
{"cells":[{"value":{"value":"neutral"}},{"value":{"value":349.0}}]}],
"warnings":[]
}

但是，根据 Google Chart JSON 规范，它应该类似于以下内容:

{
"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}]}
  ]

我已经尝试了多种不同的方法来在服务器端正确格式化 JSON，但仍然卡住了，我相信这是一个快速简单的修复方法，因此我们将不胜感激。

当查询获取 JSON 并尝试处理它时，它显示“表没有列”

谢谢

Answer 1

创建字符串而不是 json。 该字符串放入请求范围并直接在 jsp(request.getAttribute()); 中获取该字符串;

var data = new google.visualization.DataTable(request.getAttribute());

我做了这个我得到了答案..