我在这里遇到的问题是从服务器:http://XXX.XXX.XXX.XXX:9090/appsapi/rest/services/getStudentPersonalDetails/123789读取一个json数组,这个json数组中没有json对象。它就像下面一样。:
{"password":"raj","address":"dfsaf","state":"tamilnadu","image":"Bindu Appalam.jpg",
"dept":"BE(Mech)","fname":"fasdf","mname":"saffd","dob":"14/12/1951","sex":"Male",
"email":"fdasd@gmail.com","contact":"212121212545855555545","city":"dsfadf",
"pin":"600106","sname":"fdjsf","studid":"123789"}
我想解析它..
我试图从httpclient请求中获取如下所示的json数组:
try {
System.out.println("url tt : "+url);
// http client
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpEntity httpEntity = null;
HttpResponse httpResponse = null;
System.out.println(" Urls Input to "+url);
// Checking http request method type
if (method == POST) {
HttpPost httpPost = new HttpPost(url);
// adding post params
if (params != null) {
httpPost.setEntity(new UrlEncodedFormEntity(params));
}
httpResponse = httpClient.execute(httpPost);
} else if (method == GET) {
// appending params to url
System.out.println(" Params Valyes "+params);
if (params != null) {
String paramString = URLEncodedUtils.format(params, "utf-8");
url += "?" + paramString;
System.out.println(" Params String "+paramString);
}
HttpGet httpGet = new HttpGet(url);
httpResponse = httpClient.execute(httpGet);
}
httpEntity = httpResponse.getEntity();
response = EntityUtils.toString(httpEntity);
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
但在回复时得到以下错误:
05-20 16:38:25.420: I/System.out(29983):
Output <!DOCTYPE html><html><head><title>Apache Tomcat/8.0.5 - Error report
</title><style type="text/css">H1 {font-family:Tahoma,Arial,sans-serif;color:white;background-color:#525D76;font-size:22px;}
H2 {font-family:Tahoma,Arial,sans-serif;color:white;background-color:#525D76;font-size:16px;}
H3 {font-family:Tahoma,Arial,sans-serif;color:white;background-color:#525D76;font-size:14px;}
BODY {font-family:Tahoma,Arial,sans-serif;color:black;background-color:white;}
B {font-family:Tahoma,Arial,sans-serif;color:white;background-color:#525D76;}
P {font-family:Tahoma,Arial,sans-serif;background:white;color:black;font-size:12px;}A {color : black;}A.name
{color : black;}.line {height: 1px; background-color: #525D76; border: none;}</style> </head><body><h1>HTTP Status 404 - Not Found
</h1><div class="line"></div><p><b>type</b> Status report</p><p><b>message</b> <u>Not Found</u></p><p><b>description</b>
<u>The requested resource is not available.</u></p><hr class="line"><h3>Apache Tomcat/8.0.5</h3></body></html>
任何身体,请帮助......
答案 0 :(得分:3)
在你的代码中,没有json数组,它是带有名称值对数据的json对象。因此,您必须解析JSONObject
而不是JSONArray
。
喜欢:
JSONObject json = new JSONObject(jsonString);
String password = json.getString("password");
同样,您可以获取对象的所有值。
答案 1 :(得分:1)
JSON对象以{开头,以{结尾},而JSON数组以[以#结尾]开头。
在您的情况下,请更改代码以改为使用JSONObject。
JSONObject json = new JSONObject(jsonString);
JSONArray jArray = json.getJSONArray("list");
System.out.println("*****JARRAY*****"+jArray.length());
for(int i=0;i<jArray.length();i++){
JSONObject json_data = jArray.getJSONObject(i);
Log.i("log_tag","_id"+json_data.getInt("account")+
", mall_name"+json_data.getString("name")+
", location"+json_data.getString("number")+
", telephone"+json_data.getString("url")+
",----"+json_data.getString("balance")+
",----"+json_data.getString("credit")+
",----"+json_data.getString("displayName")
);
}
答案 2 :(得分:1)