我正在从字符串变量中的restful api获取数据,现在我想转换为JSON对象,但在转换时遇到问题,它抛出异常.Here是我的代码:
URL url = new URL("SOME URL");
HttpURLConnection conn = (HttpURLConnection) url.openConnection();
conn.setRequestMethod("GET");
conn.setRequestProperty("Accept", "application/json");
BufferedReader br = new BufferedReader(new InputStreamReader(
(conn.getInputStream())));
String output;
System.out.println("Output from Server .... \n");
while ((output = br.readLine()) != null) {
System.out.println(output);
}
conn.disconnect();
JSONObject jObject = new JSONObject(output);
String projecname=(String) jObject.get("name");
System.out.print(projecname);
我的字符串包含
{"data":{"name":"New Product","id":1,"description":"","is_active":true,"parent":{"id":0,"name":"All Projects"}}}
这是我在json中想要的字符串,但它在线程"main“中显示了异常。
java.lang.NullPointerException
at java.io.StringReader.<init>(Unknown Source)
at org.json.JSONTokener.<init>(JSONTokener.java:83)
at org.json.JSONObject.<init>(JSONObject.java:310)
at Main.main(Main.java:37)
发布于 2014-01-21 11:19:03
当while循环结束时,由于“NullPointerException”为null,您将得到输出。您可以将输出收集到某个缓冲区中,然后使用它,如下所示-
StringBuilder buffer = new StringBuilder();
String output;
System.out.println("Output from Server .... \n");
while ((output = br.readLine()) != null) {
System.out.println(output);
buffer.append(output);
}
output = buffer.toString(); // now you have the output
conn.disconnect();
发布于 2017-03-28 21:27:30
使用ObjectMapper对象将字符串转换为JsonNode:
ObjectMapper mapper = new ObjectMapper();
// For text string
JsonNode = mapper.readValue(mapper.writeValueAsString("Text-string"), JsonNode.class)
// For Array String
JsonNode = mapper.readValue("[\"Text-Array\"]"), JsonNode.class)
// For Json String
String json = "{\"id\" : \"1\"}";
ObjectMapper mapper = new ObjectMapper();
JsonFactory factory = mapper.getFactory();
JsonParser jsonParser = factory.createParser(json);
JsonNode node = mapper.readTree(jsonParser);
发布于 2013-11-19 19:42:47
您可以使用ObjectMapper将java对象转换为json字符串,而不使用JSONObject
ObjectMapper mapper = new ObjectMapper();
String requestBean = mapper.writeValueAsString(yourObject);
https://stackoverflow.com/questions/20070382
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