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java - JSONObject 未正确创建(可能是 JSON 格式错误?)

转载 作者:行者123 更新时间:2023-12-01 18:46:57 26 4
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我得到这个 JSON 字符串:

[
{
"id": 135,
"date": "2013-08-30 19:00:29",
"timestamp": "2013-08-30 19:00:29",
"lat": "54.328274",
"long": "-2.747215",
"strap": "annual International Festival of Street Arts",
"link": "http://dev.website.co.uk//?p=135",
"title": "Title"
}
]

我确信这是正确的 JSON 语法(在 iOS 应用程序中工作正常),但是当解析为 JSONObject 时,它会捕获错误。 Java:

public static JSONObject getJSONfromURL(String url){

//initialize
InputStream is = null;
String result = "";
JSONObject jArray = null;

//http post
try {

HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(url);
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();

is = entity.getContent();

} catch (Exception e) {
Log.e("log_tag", "Error in http connection "+e.toString());
}
//convert response to string

try {

BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8);
StringBuilder sb = new StringBuilder();
String line = null;

while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}

is.close();
result=sb.toString();

} catch (Exception e) {
Log.e("log_tag", "Error converting result "+e.toString());
}
//try parse the string to a JSON object

try {
Log.d("log_tag", "jresult: " + result + "finish");
jArray = new JSONObject(result);

} catch (JSONException e) {
Log.e("log_tag", "Error parsing data "+e.toString());
}
return jArray;
}

JSON 中是否有错误?

最佳答案

[表示json数组节点

{表示json对象节点

    JSONArray jArray  = new JSONArray(result);
return jArray;

此外,您可以使用一个 try block 而不是多个。

关于java - JSONObject 未正确创建(可能是 JSON 格式错误?),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17366459/

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