通过jQuery的循环遍历问题。多个数组json文件

Question

我已经找到了高和低，并没有找到这个答案的运气。

我有一个json文件，我使用.each。看看在以.json格式 - 它有几个子阵列：

http://heykoolaid.com/sections2.json

这里是html的

http://heykoolaid.com/json_output.html

正如你所看到的，它只是写“[对象对象]“，因为它贯穿阵列。

我在做什么错？我如何访问子数组？如何访问子数组？

到现在为止，我已经差不多了我解决此shizz头，但是当涉及到访问子阵列数据

在这里，我还是输了是代码：（JSON）

{ "zones":[ { "id":"arcade", "zoneColor":"#ed08ef", "sections":[  {   "sectionId":"arcade-145",   "name":"Arcade 145",   "coords":[[-32.36140331527542,43.2861328125],[-31.98944183792288,46.34033203125],[-29.80251790576445,46.5380859375],[-30.259067203213018,42.86865234375]],   "markers":[]  },  {   "sectionId":"arcade-146",   "name":"Arcade 146",   "coords":[[-30.977609093348676,37.08984375],[-33.06392419812064,37.177734375],[-32.41706632846281,42.6708984375],[-30.334953881988564,42.29736328125]],    "markers":[]  },  {   "sectionId":"arcade-147",    "name":"Arcade 147",   "coords":[[-31.034108344903498,36.40869140625],[-33.174341551002065,36.58447265625],[-33.5230788089042,33.8818359375],[-33.9433599465788,33.59619140625],[-33.26624989076273,33.15673828125],[-31.203404950917385,34.7607421875]],   "markers":[]  } ] }, { "id":"view-infield", "zoneColor":"#0a1966", "sections":[  {   "sectionId":"view-infield-305",     "name":"View Infield 305",   "coords":[[-40.83043687764923,-40.0341796875],[-44.245199015221274,-36.18896484375],[-40.39676430557204,-29.6630859375],[-36.66841891894784,-33.37646484375]],   "markers":[]  },  {   "sectionId":"view-infield-307",     "name":"View Infield 307",   "coords":[[-36.52729481454623,-44.560546875],[-40.17887331434696,-40.71533203125],[-36.42128244364948,-34.69482421875],[-32.56533316084102,-38.51806640625]],   "markers":[]  },  {   "sectionId":"view-infield-308",    "name":"View Infield 308",   "coords":[[-31.933516761903675,-49.02099609375],[-27.839076094777802,-43.06640625],[-32.082574559545904,-38.9794921875],[-35.97800618085566,-45.0439453125]],   "markers":[]  } ] }, { "id":"view-outfield", "zoneColor":"#165604", "sections":[  {   "sectionId":"view-outfield-333",    "name":"View Outfield 333",    "coords":[[60.45721779774397,36.6943359375],[56.58369172128337,47.373046875],[52.5897007687178,42.890625],[55.5161921571789,35.2001953125]],   "markers":[]  },  {   "sectionId":"view-outfield-334",    "name":"View Outfield 334",   "coords":[[56.353077613860826,48.0322265625],[52.656393941988,57.216796875],[52.03897658307622,57.7880859375],[49.48240137826932,50.4052734375],[52.32191088594773,43.52783203125]],   "markers":[]  },  {   "sectionId":"view-outfield-335",    "name":"View Outfield 335",    "coords":[[51.60437164681676,58.16162109375],[47.010225655683485,61.962890625],[44.276671273775186,54.7119140625],[49.081062364320736,50.82275390625]],    "markers":[]  }, {  "sectionId":"view-outfield-336",   "name":"View Outfield 336",   "coords":[[46.52863469527167,62.33642578125],[43.8186748554532,64.423828125],[40.94671366508002,57.216796875],[43.75522505306928,55.107421875]],  "markers":[]  } ] }  ] } 

和HTML：

<script type="text/javascript"> 
$(function() { 
    $.getJSON("sections2.json", tickets); 

    function tickets(data) { 
     var htmlString = ""; 

     $.each(data.zones, function(index, value) { 
      htmlString += item.sections + "<br/>"; 
     }); 

     $('#test').html(htmlString); 

    } 
}); 

Answer 1

可以使用

$.each({arr},function(index, value){ 
     alert(index + ': ' + value); 

    } 
    ); 

他$.each()功能可用于任何集合遍历，无论是地图（JavaScript对象）或数组

Answer 2

请参考的的getJSON http://api.jquery.com/jQuery.getJSON/

我不认为使用您将数据传递给成功处理程序。如果您按照上面链接中的示例进行操作，则应该朝着正确的方向前进。

成功处理程序例如：

$.getJSON('ajax/test.json', function(data) { 
    $('.result').html('<p>' + data.foo + '</p>' 
    + '<p>' + data.baz[1] + '</p>'); 
}); 

你的票（数据）函数没有的getJSON完成后得到的数据。

Answer 3

如果你确切地知道你需要从数据结构中得到什么，那么你可以通过结构下降，例如。

data.zones[i].sections[j].name // Access known location in data structure 

如果您需要迭代，然后jQuery的每个（）函数是很方便：

$.each(data, function (i, zone) { 
    //do stuff with zone data eg. print zone.id 

    $.each(zone.sections, function (j, section) { 
     //do stuff with section data, eg. print section.name 
    }); 
}); 

或者只使用迭代的JavaScript：

for (i in data.zones) { 
    zone = data.zones[i]; 
    // print zone.id 
    for (j in zone.sections) { 
     section = zone.sections[j]; 
     // print section data 
    } 
}