【发布时间】：2015-11-05 02:31:42
【问题描述】：
我得到了这个例外两天
"org.json.JSONException: java.lang.String 类型的值异常无法转换为 JSONArray",
我在网上搜索过，但无法解决。
它在浏览器中为我提供了理想的响应，但我不知道 android 有什么问题！
请帮助我。 这是我的代码
private class SigninActivity extends AsyncTask<String, Void, String> {
@Override
protected String doInBackground(String... arg0) {
String newString="";
try {
String tonTexte = loginDisplay.getText().toString();
String tonTexte1 = passDisplay.getText().toString();
String link = "http://192.168.56.1/projects/stage/atelier.php";
String data = URLEncoder.encode("username", "UTF-8") + "=" + URLEncoder.encode(tonTexte, "UTF-8");
data += "&" + URLEncoder.encode("password", "UTF-8") + "=" + URLEncoder.encode(tonTexte1, "UTF-8");
URL url = new URL(link);
URLConnection conn = url.openConnection();
conn.setDoOutput(true);
OutputStreamWriter wr = new OutputStreamWriter(conn.getOutputStream());
wr.write(data);
wr.flush();
BufferedReader reader = new BufferedReader(new InputStreamReader(conn.getInputStream(),"UTF-8"),8);
StringBuilder sb = new StringBuilder();
String line = null;
// Read Server Response
while ((line = reader.readLine()) != null) {
sb.append(line);
break;
}
return sb.substring(sb.indexOf("{"),sb.lastIndexOf("}") + 1);
}
catch (Exception e) {
return new String("Exception: " + e.getMessage());
}
}
@Override
protected void onPreExecute() {
super.onPreExecute();
}
@Override
protected void onPostExecute(String result) {
TextView a = (TextView) findViewById(R.id.aff);
JSONObject json_data=null;
TextView b = (TextView) findViewById(R.id.aff1);
//recuperation des donnees json
try{
//String jsonObjRecv = JSONGet.getJSONfromURL(URL_LIST);
if(!TextUtils.isEmpty(result)) {
JSONArray jArray = new JSONArray(result);
for (int i = 0; i < jArray.length(); i++) {
json_data = jArray.getJSONObject(i);
Log.i("log_tag", "cin: " + json_data.getString("nom"));
donnees.add(json_data.getString("nom"));
donnees1.add(json_data.getString("atelier"));
}
for (String elem : donnees) {
a.setText(elem);
}
for (String elem : donnees1) {
b.setText(elem);
}
}
else{
Log.w("json", "result is null");
}
}
catch(JSONException e){
Log.i("tagjsonexp",""+e.toString());
}
catch (ParseException e) {
Log.i("tagjsonpars", "" + e.toString());
}
}
}
这里是php代码
<?php
error_reporting(E_ALL ^ E_DEPRECATED);
$base = mysql_connect ('localhost', 'root', '');
mysql_select_db ('stage', $base) ;
$output=array();
$username = $_POST['username'];
$password = $_POST['password'];
$sql=mysql_query("SELECT CIN FROM authentification where login= '".$username."' AND password = '".$password. "'");
//$sql=mysql_query("SELECT CIN FROM authentification where login= 'amani' AND password = '123' ");
if (!$sql) {
die(mysql_error());
}
while($data = mysql_fetch_array($sql))
{
$a=$data['CIN'];
}
$sql1=mysql_query("SELECT nom,atelier FROM controlleur where CIN= '".$a."'");
while($row=mysql_fetch_assoc($sql1))
{
$output[]=$row;
//print(json_encode(array($output));
print(json_encode(array("NomTableau" => $output)));
mysql_close();
?>
【问题讨论】：
-
发布您从网络服务获得的响应
-
还有堆栈跟踪，至少 - 你已经发布了 lot 的 Java 代码，我们不知道错误在哪里。理想情况下，您应该自己缩小范围并展示一个简短但完整的程序，只是演示了问题......
-
如果响应包含在方括号中，则将其视为 JSONObject。而且我认为 JSONArray 有整数索引。