Javascript 从getJSON添加到数据
我在一个url上调用$.getJSON,该url包含以下格式的日历事件Javascript 从getJSON添加到数据,javascript,json,getjson,Javascript,Json,Getjson,我在一个url上调用$.getJSON,该url包含以下格式的日历事件 [{"guid":"53c2d70ae6014","date":"1408161600000","type":"event","title":"Weekend Classes Begin","description":"","url":"" }, {"guid":"53c2d70ae601c","date":"1408248000000","type":"event","title":"Freshman Orientati
[{"guid":"53c2d70ae6014","date":"1408161600000","type":"event","title":"Weekend Classes Begin","description":"","url":"" },
{"guid":"53c2d70ae601c","date":"1408248000000","type":"event","title":"Freshman Orientation","description":"","url":"" },
{"guid":"53c2d70ae6022","date":"1408334400000","type":"event","title":"Freshman Orientation","description":"","url":"" }]
它获取JSON并将其放入数组中,当我使用alert(data)时代码>它显示为
[object Object],[object Object],[object Object]...
我希望能够将一个或多个事件添加到从$.getJSON
调用接收的数据中
$.getJSON(eventsOpts.eventsjson + "?limit="+limit+"&year="+year+"&month="+month+"&day="+day, function(data) {
if(localStorage.getItem('agenda') === null)
{
}
else
{
var json = localStorage.getItem('agenda');
var jsonObj = JSON.stringify(json);
data.push(json);
alert(data);
}
flags.eventsJson = data; // save data to future filters
getEventsData(flags.eventsJson, limit, year, month, day, direction);
}).error(function() {
showError("error no internet connection.");
});
我想附加的json是{“guid”:“53c2e0a3d0680”,“日期”:“1418922000000”,“类型”:“事件”,“标题”:“2014年秋季毕业”,“描述”:“url”:”}
我得到的是
[object Object],[object Object],[object Object],{"guid":"53c2e0a3d0680","date":"1418922000000","type":"event","title":"Fall 2014 Graduation","description":"","url":"" }
不确定我做错了什么您必须从localStorage
重新解析对象,然后按下:
var json = localStorage.getItem('agenda'),
myObj = JSON.parse(json);
data.push(myObj);
您必须从localStorage
重新解析对象,然后推送:
var json = localStorage.getItem('agenda'),
myObj = JSON.parse(json);
data.push(myObj);
示例中的varjson
是一个字符串。将其传递给将JSON字符串转换为JavaScript对象的JSON.parse()
。varJSON
在您的示例中是一个字符串。将其传递给将JSON字符串转换为JavaScript对象的JSON.parse()
。显示如何追加,看起来像字符串而不是对象。不要使用alert()
检查变量。正如你所看到的,它不是很有用。改用浏览器控制台(例如console.log()
)。显示如何追加,看起来像字符串而不是对象。不要使用alert()
检查变量。正如你所看到的,它不是很有用。改用浏览器控制台(例如console.log()
)。