当我尝试从以下 JSON 中获取"名称"的值时,我收到此错误:
{
"edges": [
{
"node": {
"Name": "Sunday River",
"Latitude": 44.4672,
"Longitude": 70.8472
}
},
{
"node": {
"Name": "Sugarloaf Mountain",
"Latitude": 45.0314,
"Longitude": 70.3131
}
}
]
}
这是我用来尝试访问这些值的代码片段,但我现在只是在测试获取"名称":
String[] nodes = stringBuilder.toString().split("edges");
nodes[1] = "{" + """ + "edges" + nodes[1];
String s = nodes[1].substring(0,nodes[1].length()-3);
Log.d(TAG, s);
JSONObject json = new JSONObject(s);
JSONArray jsonArray = json.getJSONArray("edges");
ArrayList<String> allNames = new ArrayList<String>();
ArrayList<String> allLats = new ArrayList<String>();
ArrayList<String> allLongs = new ArrayList<String>();
for (int i=0; i<jsonArray.length(); i++) {
JSONObject node = jsonArray.getJSONObject(i);
Log.d(TAG, node.toString(1));
String name = node.getString("Name");
Log.d(TAG, name);
}
我的输出如下所示:
{"edges":[{"node":{"Name":"Sunday River","Latitude":44.4672,"Longitude":70.8472}},{"node":{"Name":"Sugarloaf Mountain","Latitude":45.0314,"Longitude":70.3131}}]}}
{
"node": {
"Name": "Sunday River",
"Latitude": 44.4672,
"Longitude": 70.8472
}
}
org.json.JSONException: No value for Name
我知道我可以使用optString而不会收到错误,但这不会给我存储在每个节点中的数据。
下面是一个适用于未更改 JSON 的版本:
public static void main(String... args)
{
String json = "{"data":{"viewer":{"allMountains":{"edges":[{"node":{"Name":"Sunday River","Latitude":44.4672,"Longitude":70.8472}},{"node":{"Name":"Sugarloaf Mountain","Latitude":45.0314,"Longitude":70.3131}}]}}}}";
JSONObject obj = new JSONObject(json);
JSONObject data = obj.getJSONObject("data");
JSONObject viewer = data.getJSONObject("viewer");
JSONObject allMountains = viewer.getJSONObject("allMountains");
// 'edges' is an array
JSONArray edges = allMountains.getJSONArray("edges");
for (Object edge : edges) {
// each of the elements of the 'edge' array are objects
// with one property named 'node', so we need to extract that
JSONObject node = ((JSONObject) edge).getJSONObject("node");
// then we can access the 'node' object's 'Name' property
System.out.println(node.getString("Name"));
}
}