Question

这是我必须从网站上以JSON格式阅读文本。但我得到了错误

Java.lang.ClassCastException：org.json.simple.JSONObject不能强制转换为org.json.simple.JSONArray

这让我疯了。有人可以帮忙吗？我还需要为＆＃34;用户名＆＃34;的所有实例检查此字符串。并为每个人运行一些东西。

public class CommandCheck implements CommandExecutor {
private String username;
private static String host = "example.com";
private URL url;
private String apiKey = main.getNode("API-KEY");
@Override
public boolean onCommand(CommandSender sender, Command cmd, String label, String[] arg3) {
    try {
        this.url = new URL(CommandCheck.host);
        final URLConnection conn = this.url.openConnection();
        conn.setConnectTimeout(5000);

        if (this.apiKey != null) {
            conn.addRequestProperty("x-api-key", this.apiKey);
        }
        conn.addRequestProperty("User-Agent", main.USER_AGENT);

        conn.setDoOutput(true);

        final BufferedReader reader = new BufferedReader(new InputStreamReader(conn.getInputStream()));
        final String response = reader.readLine();
        sender.sendMessage(response); //Im just dumping the raw String for the person running the command to see Debug mostly
        final JSONArray array = (JSONArray) JSONValue.parse(response);

        if (array.isEmpty()) {
            sender.sendMessage("The Array appears to be empty");
            return false;
        }
        JSONObject latestUpdate = (JSONObject) array.get(array.size() - 1);
        username = (String) latestUpdate.get("Username");
        sender.sendMessage("whitelist add" + username);
        return true;
    } catch (final IOException e) {
        if (e.getMessage().contains("HTTP response code: 403")) {
            sender.sendMessage("I think there is an API key issue");
        } else {
            sender.sendMessage("Problem of unknown orign");
        }
        return false;
    }
}

Answer 1

尝试更改以下行：

final JSONArray array = (JSONArray) JSONValue.parse(response);

为：

final JSONObject jsObj = (JSONObject) JSONValue.parse(response);

你能提供你想要解析的JSON字符串吗？即response的价值？

Answer 2

JsonReader jsonReader = Json.createReader(new StringReader(response.readEntity(String.class)));
JsonArray jsonArray = jsonReader.readArray();
ListIterator l = jsonArray.listIterator();
while ( l.hasNext() ) {
      JsonObject j = (JsonObject)l.next();
      JsonObject ciAttr = j.getJsonObject("ciAttributes") ;

Answer 3

org.json.simple.JSONObject无法强制转换为org.json.simple.JSONArray意味着您正在尝试将json对象转换为json数组。如果您在json对象中的响应作为响应，那么您首先需要它在Json对象中进行转换。

转换后，你可以使用get（“key-name”）从json对象获取Json数组

JSONObject resObj = new JSONObject(responseString);
JSONArray resArray = resObj.getJSONArray("Username");
for (int i=0; i<resArray.length(); i++)
String resultString = resArray.getString(i);

它为您提供所有用户名。

我认为此代码可以帮助您解决问题。

JSON存储为字符串。似乎无法解析它 - Java

3 个答案: