无法将类型为org.json.jsonarray的值[{" code"：" login_true"，" name"：" hhh"，" email"：" hhh"}]转换为jsonobject [英] Value[{"code":"login_true", "name":"hhh", "email":"hhh"}] of type org.json.jsonarray cannot be converted to jsonobject

问题描述

当编译器到达onPostExecute并尝试运行执行JSONArray时，行
 
 jsonArray = jsonObject.getJSONArray（  server_response）; 
 抛出异常：
 
    org.json.JSONException如：
  值[{代码 ： login_true  ， name < span class =code-string> ： hhh  ， email  ： hhh  }] 
类型org.json.JSONArray无法转换为JSONObject。 

我的尝试：



什么是正确的陈述？



我的代码：



protected String doInBackground（String ... params）

{

String call_type = params [0 ];

if（call_type.equals（login））

{

try {

URL url =新URL（login_url）;

HttpURLConnection httpURLConnection =（HttpURLConnection）url.openConnection（）;

httpURLConnection.setRequestMethod（POST）;

httpURLConnection.setDoOutput（true）;

httpURLConnection.setDoInput（true）;

OutputStream OS = httpURLConnection.getOutputStream（）;

BufferedWriter bufferedWriter = new BufferedWriter（new OutputStreamWriter（OS，UTF-8））;

String email，pass;

email = params [1];

pass = params [2];



字符串数据= URLEncoder.encode（email，UTF-8）+=+ URLEncoder .encode（电子邮件，UTF-8）+& +

URLEncoder.encode（pass，UTF-8）+=+ URLEncoder.encode（pass，UTF-8）;

< br $> b $ b bufferedWriter.write（data）;





InputStream IS = httpURLConnection.getInputStream（）;

BufferedReader BR = new BufferedReader（new InputStreamReader（IS））;

StringBuilder stringBuilder = new StringBuilder（）;

String line =;

while（（line = BR.readLine（））！= null）

{

stringBuilder.append（line +\ n）;

}

bufferedWriter.flush（）;

bufferedWriter.close（）;

OS.close（） ;

//IS.close（）;

httpURLConnection.disconnect（）;

Thread.sleep（500）;

Log.d（Test，Test 3 pass）;

返回stringBuilder.toString（）。trim（）;



} catch（UnsupportedEncodingException e）{

e.printStackTrace（）;

} catch（ProtocolException e）{

e.printStackTrace（）;

} catch（MalformedURLException e）{

e.printStackTrace（）;

} catch（IOException e）{

e.printStackTrace（）;

} catch（InterruptedException e）{

e.printStackTrace（）;

}

}

返回null;

}

**强文** / *我的onPostExecute方法如下; * / ***



@Override

protected void onPostExecute（String json）

{



尝试

{

progressD ialog.dismiss（）;



JSONObject jsonObject = new JSONObject（json）;

JSONArray jsonArray = jsonObject.getJSONArray（server_response） ;

******* //这里的异常来自org.json.JSONException：值[{code：login_true，name：hhh，email： org.json.JSONArray类型的hhh}]无法转换为JSONObject，编译器跳转到异常部分*******

JSONObject JO = jsonArray.getJSONObject（0）;

字符串代码= JO.getString（code）;

字符串消息= JO.getString（message）;



if（code.equals（reg_true））

{

ShowDialog（注册成功，消息，代码）;

}

else if（code.equals（reg_false））

{

ShowDialog（注册失败，消息，代码）;

}

else if（code.equals（login_true））

{

Intent intent = new Intent（activity，HomeActivity.class）;

intent.putExtra（message，message）;

activity.startActivity（intent）;

}

else if（code.equals（ login_false））

{

ShowDialog（登录错误，消息，代码）;

}





} catch（JSONException e）{

e.printStackTrace（）;

}

}

** / * php脚本如下：（login.php）* / **

需要init.php; < br $>


$ email = $ _ POST [email];

$ pass = $ _ POST [pass];



$ sql_query =选择姓名，来自beneficiary_details的电子邮件，其中包含电子邮件，如'。$ email。'并传递像'。$ pass。';

$ result = mysqli_query（$ con，$ sql_query）;

$ response = array（）;



$ rowcount = mysqli_num_rows（$ result）;

// print_r（$ rowcount ）;

if（$ rowcount> 0）

{



$ row = mysqli_fetch_row（$ result）;

$ name = $ row [ 0];

$ email = $ row [1];

$ code =login_true;



array_push（$ response，array（code=> $ code，name=> $ name，email=> $ email））;

echo json_encode（$ response） ）;

}

其他

{

$ code =login_false;

$ message =找不到用户，请再试一次..;

array_push（$ response，array（code=> $ code，message=> $ message） ）;

echo json_encode（$ response）;

}

mysqli_close（$ con）;

？> ;

** plz找出我在做错的地方???

解决方案

email =

< blockquote> _POST [email];

pass =

When compiler reaches onPostExecute and trying to run execute JSONArray the line

jsonArray=jsonObject.getJSONArray("server_response");
throws an exception:

org.json.JSONException like: 
    "Value[{"code":"login_true","name":"hhh","email":"hhh"}]
    of type org.json.JSONArray cannot be converted to JSONObject".

What I have tried:

What is the correct statement?

My code:

protected String doInBackground(String... params)
{
String call_type=params[0];
if(call_type.equals("login"))
{
try {
URL url = new URL(login_url);
HttpURLConnection httpURLConnection = (HttpURLConnection) url.openConnection();
httpURLConnection.setRequestMethod("POST");
httpURLConnection.setDoOutput(true);
httpURLConnection.setDoInput(true);
OutputStream OS = httpURLConnection.getOutputStream();
BufferedWriter bufferedWriter = new BufferedWriter(new OutputStreamWriter(OS, "UTF-8"));
String email,pass;
email=params[1];
pass=params[2];

String data = URLEncoder.encode("email", "UTF-8") + "=" + URLEncoder.encode(email, "UTF-8") + "&" +
URLEncoder.encode("pass", "UTF-8") + "=" + URLEncoder.encode(pass, "UTF-8");

bufferedWriter.write(data);


InputStream IS = httpURLConnection.getInputStream();
BufferedReader BR= new BufferedReader(new InputStreamReader(IS));
StringBuilder stringBuilder = new StringBuilder();
String line="";
while ((line=BR.readLine())!=null)
{
stringBuilder.append(line+"\n");
}
bufferedWriter.flush();
bufferedWriter.close();
OS.close();
//IS.close();
httpURLConnection.disconnect();
Thread.sleep(500);
Log.d("Test","Test 3 pass");
return stringBuilder.toString().trim();

} catch (UnsupportedEncodingException e) {
e.printStackTrace();
} catch (ProtocolException e) {
e.printStackTrace();
} catch (MalformedURLException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
return null;
}
**strong text**/*my onPostExecute method is as below;*/***

@Override
protected void onPostExecute(String json)
{

try
{
progressDialog.dismiss();

JSONObject jsonObject = new JSONObject(json);
JSONArray jsonArray=jsonObject.getJSONArray("server_response");
*******//here exception coming org.json.JSONException: Value [{"code":"login_true","name":"hhh","email":"hhh"}] of type org.json.JSONArray cannot be converted to JSONObject and compiler jumps to the exception part*******
JSONObject JO= jsonArray.getJSONObject(0);
String code=JO.getString("code");
String message=JO.getString("message");

if(code.equals("reg_true"))
{
ShowDialog("Registration Success",message,code);
}
else if (code.equals("reg_false"))
{
ShowDialog("Registration Fail",message,code);
}
else if(code.equals("login_true"))
{
Intent intent= new Intent(activity,HomeActivity.class);
intent.putExtra("message",message);
activity.startActivity(intent);
}
else if(code.equals("login_false"))
{
ShowDialog("Login Error,",message,code);
}


} catch (JSONException e) {
e.printStackTrace();
}
}
**/* php scrip is as follows: (login.php)*/**
require"init.php";

$email=$_POST["email"];
$pass=$_POST["pass"];

$sql_query="select name, email from beneficiary_details where email like '".$email."' and pass like '".$pass."' ";
$result=mysqli_query($con,$sql_query);
$response = array();

$rowcount=mysqli_num_rows($result);
//print_r( $rowcount);
if($rowcount > 0)
{

$row=mysqli_fetch_row($result);
$name=$row[0];
$email=$row[1];
$code="login_true";

array_push($response,array("code"=>$code,"name"=>$name,"email"=>$email));
echo json_encode($response);
}
else
{
$code="login_false";
$message="User not found, Please try again..";
array_push($response,array("code"=>$code,"message"=>$message));
echo json_encode($response);
}
mysqli_close($con);
?>
**plz find out where I am doing mistake???

解决方案

email=

_POST["email"];

pass=

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