对不起,我确定这有重复,但我是jQuery的新手。我有以下JSON字符串...
{
"totalNumEntries":2,
"pageType":"CampaignPage",
"totalBudget":{
"period":{
"value":"DAILY"
},
"amount":{
"comparableValueType":"Money",
"microAmount":0
},
"deliveryMethod":null
},
"entries":[
{
"id":733413,
"name":"Interplanetary Cruise #1345659006301",
"status":null,
"servingStatus":null,
"startDate":null,
"endDate":null,
"budget":null,
"biddingStrategy":null,
"conversionOptimizerEligibility":null,
"campaignStats":{
"startDate":null,
"endDate":null,
"network":{
"value":"ALL"
},
"clicks":null,
"impressions":null,
"cost":null,
"averagePosition":null,
"averageCpc":null,
"averageCpm":null,
"ctr":null,
"conversions":null,
"viewThroughConversions":null,
"statsType":"CampaignStats"
},
"adServingOptimizationStatus":null,
"frequencyCap":{
"impressions":0,
"timeUnit":null,
"level":null
},
"settings":null,
"networkSetting":null,
"forwardCompatibilityMap":null
},
{
"id":733414,
"name":"Interplanetary Cruise banner #1345659006387",
"status":null,
"servingStatus":null,
"startDate":null,
"endDate":null,
"budget":null,
"biddingStrategy":null,
"conversionOptimizerEligibility":null,
"campaignStats":{
"startDate":null,
"endDate":null,
"network":{
"value":"ALL"
},
"clicks":null,
"impressions":null,
"cost":null,
"averagePosition":null,
"averageCpc":null,
"averageCpm":null,
"ctr":null,
"conversions":null,
"viewThroughConversions":null,
"statsType":"CampaignStats"
},
"adServingOptimizationStatus":null,
"frequencyCap":{
"impressions":0,
"timeUnit":null,
"level":null
},
"settings":null,
"networkSetting":null,
"forwardCompatibilityMap":null
}
]
}
这是从我的应用程序中的/ google / getCampaigns返回的。我正在使用以下代码循环它,但页面保持空白...
<script>
$(document).ready(function() {
$.getJSON('google/getCampaigns', function(data) {
$.each(data.entries, function(index) {
$('span').append(index.name);
});
});
});
</script>
Loading...
<span />
谁能看到我做错了什么?
谢谢,
大卫
答案 0 :(得分:7)
每个陈述应该是这样的:
$.each(data.entries, function(index,element) {
$('span').append(element.name);
});
或其他方法
$.each(data.entries, function(index) {
$('span').append(data.entries[index].name);
});
希望这能解决您的问题。
答案 1 :(得分:1)
什么是传递任何其他甚至索引的必要性这将有助于你肯定
$.each(data.entries, function() {
$('span').append(this.name);
});