2016-11-19 85 views
0

我正在处理页面上的图表。自从我与工作的Django,我得的数据JSON传递到HTML,然后解析JSON来获取数据。JS/JQuery:无法访问json中的数组元素

的问题是,控制台说:

Uncaught TypeError: Cannot read property '0' of undefined(…)

function drawChart() { 
    var google_chart_json = {"MyKey": [[[2016, 11, 2], 156.0], [[2016, 11, 3], 69.0], [[2016, 11, 4], 126.0], [[2016, 11, 5], 67.0], [[2016, 11, 6], 97.0], [[2016, 11, 7], 193.0], [[2016, 11, 8], 96.0], [[2016, 11, 9], 64.0], [[2016, 11, 10], 117.0], [[2016, 11, 11], 190.0]]}; 

    $.each(google_chart_json, function (key, val) { 
     console.log(key); 
     $.each(val,function (scan) { 
      var year = scan[0][0] 
      console.log(year) 
     }) 
    }); 
    .... 

正如你可以看到var year应该是2016等

问题出在哪里?

回答

1

传递给$.each回调的第一个参数是一个键或数组索引。第二个参数是值。你在外面$.each有这个权利,但不是内在的。

修正:

function drawChart() { 
 
    var google_chart_json = { 
 
    "MyKey": [ 
 
     [[2016, 11, 2], 156.0], 
 
     [[2016, 11, 3], 69.0], 
 
     [[2016, 11, 4], 126.0], 
 
     [[2016, 11, 5], 67.0], 
 
     [[2016, 11, 6], 97.0], 
 
     [[2016, 11, 7], 193.0], 
 
     [[2016, 11, 8], 96.0], 
 
     [[2016, 11, 9], 64.0], 
 
     [[2016, 11, 10], 117.0], 
 
     [[2016, 11, 11], 190.0] 
 
    ] 
 
    }; 
 

 
    $.each(google_chart_json, function(key, val) { 
 
    console.log(key); 
 
    $.each(val, function(idx, scan) { 
 
     var year = scan[0][0] 
 
     console.log(year) 
 
    }) 
 
    }); 
 
} 
 

 
drawChart();
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

1

的问题是,在scan你的内心仅环是关键值(或在这种情况下,数组的索引,0-9),而不是完整的数组。

$.each(google_chart_json, function (key, val) { 
    console.log(key); 
    $.each(val,function (scan) { 
     console.log(google_chart_json[key][scan][0][0]) 
    }) 
});