我试图从JSON响应中选取数据,但无法获取所需的所有值。JavaScript - 无法打印具有相同键名的元素
这里是JSON-响应体:
{
"status": "success",
"reservations": [
{
"id": "38177",
"subject": "subjectID",
"modifiedDate": "2017-05-16T12:46:17",
"startDate": "2017-05-30T08:00:00",
"endDate": "2017-05-30T22:00:00",
"resources": [
{
"id": "124",
"type": "room",
"code": "F407",
"parent": {
"id": "4",
"type": "building",
"code": "buildingF",
"name": "buildingName"
},
"name": " F407 (atk 34)"
}
],
"description": ""
},
{
"id": "38404",
"subject": "subjectID",
"modifiedDate": "2017-05-16T12:49:25",
"startDate": "2017-05-30T08:00:00",
"endDate": "2017-05-30T22:00:00",
"resources": [
{
"id": "128",
"type": "room",
"code": "F411",
"parent": {
"id": "4",
"type": "building",
"code": "buildingF",
"name": "buildingName"
},
"name": " F411 (atk 34)"
}
],
"description": ""
},
{
"id": "38842",
"subject": "subjectID",
"modifiedDate": "2017-05-30T06:03:13",
"startDate": "2017-05-30T08:00:00",
"endDate": "2017-05-30T22:00:00",
"resources": [
{
"id": "107",
"type": "room",
"code": "F211",
"parent": {
"id": "4",
"type": "building",
"code": "buildingF",
"name": "buildingName"
},
"name": " F211 (room 50)"
}
],
"description": ""
},
{
"id": "40186",
"subject": "subjectID",
"modifiedDate": "2017-05-26T08:45:50",
"startDate": "2017-05-30T09:00:00",
"endDate": "2017-05-30T14:00:00",
"resources": [
{
"id": "118",
"type": "room",
"code": "F312",
"parent": {
"id": "4",
"type": "building",
"code": "buildingF",
"name": "buildingName"
},
"name": " F312 (room 48)"
}
],
"description": ""
},
]
}
这样的想法是选择从每个受试者,其是下面房间代码和名称;
"code": "F407"
"name": "F407 (atk 34)"
"code": "F411"
"name": "F411 (atk 34)"
"code": "F211"
"name": "F211 (room 50)"
"code": "F312"
"name": "F312 (room 48)"
我试过用我自己的代码做这个,但是因为某种原因它跳过了一个房间名称。我使用for循环检查JSON响应,并在resources
中查找代码和名称,并将它们推入数组中;
var rooms = [];
for (var i = 0; i < json.reservations.length; i++) {
if (json.reservations[i].resources != null) {
for (var j = 0; j < json.reservations[i].resources.length; j++)
{
var reservation = json.reservations[i];
var resource = json.reservations[i].resources[j];
if (resource.type === "room") {
if (rooms.indexOf("code")) {
rooms.push(resource.code + resource.name);
}
}
}
}
}
document.getElementById("pageOne").innerHTML = rooms.join("<br/>")
输出如下它省去了"name": "F411 (atk 34)"
F407 F407 (atk 34)
F411
F211 F211 (room 50)
F312 F312 (room 48)
任何建议,为什么会这样?
粘贴代码为小提琴,它的工作原理,不被当场:https://jsfiddle.net/p474djan/ –
的方式,将JSON是无效的,你有一个不需要的“,”和你的代码在jsfiddle的作品 – ChaosPattern