我想从 JSON 文件中获取单个属性及其值,并将其打印在 excel 文件中。 我想从 http 响应方法中获取详细信息,但是我遇到了错误。
Exception in thread "main" Unexpected character (() at position 0.
at [email protected]/org.json.simple.parser.Yylex.yylex(Unknown Source)
at [email protected]/org.json.simple.parser.JSONParser.nextToken(Unknown Source)
at [email protected]/org.json.simple.parser.JSONParser.parse(Unknown Source)
at [email protected]/org.json.simple.parser.JSONParser.parse(Unknown Source)
at [email protected]/org.json.simple.parser.JSONParser.parse(Unknown Source)
try {
HttpResponse<String> response = null;
response = HttpClient.newHttpClient().send(request, HttpResponse.BodyHandlers.ofString());
int statusCode = response.statusCode();
System.out.println(statusCode);
System.out.println(response.body());
JSONObject object = (JSONObject) new JSONParser().parse(jsonString);
JSONArray users = (JSONArray) object.get("users");
JSONObject user0 = (JSONObject) users.get(0);
JSONObject user0data = (JSONObject) user0.get("data");
String empID = user0data.get("employeeid").toString();
System.out.println("Employee ID: "+empID);
System.out.println("\n");
}