喜马拉雅中nodata value属性 for data怎么解决

点击联系发帖人 时间：2016-10-01 02:11

data value属性

Android 调用 JSON 编码出现 JSONException 错误如何解决？
我想用PHP实现安卓系统的远程数据库调用.出现了一个12-14 02:10:25.735: E/log_tag(377): Error parsing data org.json.JSONException: Value &br of type java.lang.String cannot be converted to JSONArray
这可能是那方面的问题
JSONArray jArray = new JSONArray(result);
改成JSONObject试试。如果不是唯一根节点，用JSONArray，如[{},{}]；如果是唯一根节点，如{,child:[]}用JSONObject。试试看，记得应该是这样的。
json 数据解析用gson（google的）试试挺简单的
我在android上解析json数据都用这个，你试试
String result = "";
//首先使用NameValuePair封装将要查询的年数和关键字绑定
ArrayList&NameValuePair& nameValuePairs = new ArrayList&NameValuePair&();
nameValuePairs.add(new BasicNameValuePair("year","1970"));
//使用HttpPost封装整个SQL语句
//使用HttpClient发送HttpPost对象
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("192.168.110.90/root1...");
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();
InputStream is = entity.getContent();
System.out.print("成功");
}catch(Exception e){
Log.e("log_tag", "Error in http connection "+e.toString());
//将HttpEntity转化为String
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("192.168.110.90/root1...");
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();
InputStream is = entity.getContent();
BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8);
StringBuilder sb = new StringBuilder();
String line =
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());
//将String通过JSONArray解析成最终结果
JSONArray jArray = new JSONArray(result);
for(int i=0;i&jArray.length();i++){
JSONObject json_data = jArray.getJSONObject(i);
Log.i("log_tag","id: "+json_data.getInt("id")+
", code: "+json_data.getInt("code")+
", sex: "+json_data.getInt("sex")+
", birthyear: "+json_data.getInt("birthyear")
}catch(JSONException e){
Log.e("log_tag", "Error parsing data "+e.toString());
mysql_connect("host","username","password");
mysql_select_db("PeopleData");
$q=mysql_query("SELECT * FROM people WHERE birthyear&'".$_REQUEST['year']."'");
while($e=mysql_fetch_assoc($q))
$output[]=$e;
print(json_encode($output));
mysql_close();
这个错误其实就是网上的那个答案.
出现的错误
12-14 02:10:25.735: E/log_tag(377): Error parsing data org.json.JSONException: Value &br of type java.lang.String cannot be converted to JSONArray如果我把php里的换成一个1983(一个数据库里有的year)去掉nameValuePairs.add(new BasicNameValuePair("year","1970"));
这句,会出现错误
12-14 06:58:43.379: E/log_tag(378): Error parsing data org.json.JSONException: No value for id 这是怎么回事,求高手!!!!晕了....MYSql里边的id是的i是大写的...呵呵
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