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php - 从服务器发送字符串响应时,java.lang.Integer 无法转换为 JSONObject

转载 作者:行者123 更新时间:2023-11-30 00:08:59 25 4
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我正在通过 genAsync 库从 Android 应用程序向服务器发送数据。在服务器上,我正在插入并检查用户是否同时存在。如果用户已经存在,则我发送 json_encoded 数组,如果用户不存在,则我发送 消息,即用户插入成功,但当我发送时给应用程序的消息它给出了一个错误,即 java.lang.Integer cannot be converted to JSONObject

我还附上了 php 代码、android 代码和错误日志。

php代码

class Users extends CI_Controller
{

public function create(){

$data=$this->check_exist($this->input->post('user_id'));
if($data=='0')
{

$this->load->model('UsersModel');
$msg = $this->UsersModel->create();
if ($msg==true) {
echo 'user inserted successfully';
} else {
echo '0';
}
}
else{
$data["user_details"]=$data;
echo json_encode($data);
}
}
public function check_exist($user_id)
{
$this->load->helper('url');
$this->load->model('UsersModel');
$data=$this->UsersModel->user_exist($user_id);
if(!empty($data))
{
return($data);
}
else
{
return '0';
}
}
}

向服务器端发送数据的android代码

try {
final Database_Helper dg = new Database_Helper(getBaseContext());
ServerUrl obj =new ServerUrl();
HashMap postData = new HashMap();
postData.put("user_id",personId);
postData.put("user_name",personName);
postData.put("email",personEmail);
postData.put("profile_picture",String.valueOf(personPhoto));
final PostResponseAsyncTask readTask = new PostResponseAsyncTask(UserLogin.this, postData, new AsyncResponse()
{

public void processFinish(String mStringArray)
{
Log.w("data", mStringArray.toString());
try {
JSONObject json= null;
json = new JSONObject(mStringArray);
JSONArray jsonscoresarray=json.getJSONArray("user_details");
final JSONObject details = jsonscoresarray.getJSONObject(0);
if (jsonscoresarray.length() == 0)
{
//storing data to sqlite if user not exist
dg.user_Putrecord(dg, personId, personName, personEmail, String.valueOf(personPhoto));
}
else
{
//storing data to sqlite if user exist
dg.user_Putrecord(dg,details.getString("auth_id"),details.getString("full_name"),details.getString("email_id"),details.getString("profile_url"));
}
} catch (JSONException e) {
e.printStackTrace();
}
Toast.makeText(UserLogin.this, mStringArray.toString(), Toast.LENGTH_LONG).show();
}

});
obj.url("Users","create");
readTask.execute(obj.domain_url);
}
catch (Exception ex)
{
Toast.makeText(UserLogin.this, ex.getMessage().toString(), Toast.LENGTH_SHORT).show();
}

这是我的错误日志

01-29 13:49:33.765 4455-4455/com.example.ali.ansofexperts W/System.err: org.json.JSONException: Value 1 of type java.lang.Integer cannot be converted to JSONObject
01-29 13:49:33.765 4455-4455/com.example.ali.ansofexperts W/System.err: at org.json.JSON.typeMismatch(JSON.java:111)
01-29 13:49:33.765 4455-4455/com.example.ali.ansofexperts W/System.err: at org.json.JSONObject.<init>(JSONObject.java:160)
01-29 13:49:33.765 4455-4455/com.example.ali.ansofexperts W/System.err: at org.json.JSONObject.<init>(JSONObject.java:173)
01-29 13:49:33.765 4455-4455/com.example.ali.ansofexperts W/System.err: at com.example.ali.ansofexperts.UserLogin$4$1.processFinish(UserLogin.java:121)
01-29 13:49:33.765 4455-4455/com.example.ali.ansofexperts W/System.err: at com.kosalgeek.asynctask.PostResponseAsyncTask.onPostExecute(PostResponseAsyncTask.java:248)
01-29 13:49:33.765 4455-4455/com.example.ali.ansofexperts W/System.err: at com.kosalgeek.asynctask.PostResponseAsyncTask.onPostExecute(PostResponseAsyncTask.java:31)
01-29 13:49:33.766 4455-4455/com.example.ali.ansofexperts W/System.err: at android.os.AsyncTask.finish(AsyncTask.java:667)
01-29 13:49:33.766 4455-4455/com.example.ali.ansofexperts W/System.err: at android.os.AsyncTask.-wrap1(AsyncTask.java)
01-29 13:49:33.766 4455-4455/com.example.ali.ansofexperts W/System.err: at android.os.AsyncTask$InternalHandler.handleMessage(AsyncTask.java:684)
01-29 13:49:33.766 4455-4455/com.example.ali.ansofexperts W/System.err: at android.os.Handler.dispatchMessage(Handler.java:102)
01-29 13:49:33.766 4455-4455/com.example.ali.ansofexperts W/System.err: at android.os.Looper.loop(Looper.java:154)
01-29 13:49:33.766 4455-4455/com.example.ali.ansofexperts W/System.err: at android.app.ActivityThread.main(ActivityThread.java:6121)
01-29 13:49:33.766 4455-4455/com.example.ali.ansofexperts W/System.err: at java.lang.reflect.Method.invoke(Native Method)
01-29 13:49:33.766 4455-4455/com.example.ali.ansofexperts W/System.err: at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:889)
01-29 13:49:33.766 4455-4455/com.example.ali.ansofexperts W/System.err: at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:779)

我只想知道当从服务器发送字符串响应时如何处理这个异常。

最佳答案

我正在为初学者发布完整的答案。

我在下面编写时,您的 php 代码略有变化

public function create(){
$this->load->model('UsersModel');
$data=$this->UsersModel->check_exist($this->input->post('user_id'));
if($data==0){
$nData=$this->UsersModel->create();
if($nData!="0"){
$data=$this->UsersModel->check_exist($this->input->post('user_id'));
echo json_encode($data);
}else{
die("0");
}

}else{
$result["user_details"]=$data;
echo json_encode($result);
}

}

如您所见,我刚刚从 Controller 中删除了您的 check_exist 函数。因为您的 Controller 中不需要此功能。

所以我直接从模型中调用它,以减少代码行数。

现在android端编码发生变化

   try {
final Database_Helper dg = new Database_Helper(getBaseContext());
ServerUrl obj =new ServerUrl();
HashMap postData = new HashMap();
postData.put("user_id",personId);
postData.put("user_name",personName);
postData.put("email",personEmail);

postData.put("profile_picture",String.valueOf(personPhoto));
final PostResponseAsyncTask readTask = new PostResponseAsyncTask(UserLogin.this, postData, new AsyncResponse()
{
public void processFinish(String mStringArray)
{
Log.d("data", mStringArray);
if(mStringArray.equals("0"))
{
Toast.makeText(UserLogin.this, "Invalid login try again", Toast.LENGTH_LONG).show();

}
else
{
try {
JSONObject json = null;
json = new JSONObject(mStringArray);
JSONArray jsonscoresarray = json.getJSONArray("user_details");
final JSONObject details = jsonscoresarray.getJSONObject(0);
if (mStringArray.equals("0"))
{
Toast.makeText(UserLogin.this, "At if", Toast.LENGTH_LONG).show();
}
else
{
Toast.makeText(UserLogin.this, "At else", Toast.LENGTH_LONG).show();
}
Toast.makeText(UserLogin.this, "Login Success", Toast.LENGTH_LONG).show();

}
catch (JSONException e)
{
e.printStackTrace();
}
}

}

});
obj.url("Users","create");
readTask.execute(obj.domain_url);
}
catch (Exception ex)
{
Toast.makeText(UserLogin.this, ex.getMessage().toString(), Toast.LENGTH_SHORT).show();
}

希望对您有所帮助。

关于php - 从服务器发送字符串响应时,java.lang.Integer 无法转换为 JSONObject,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48497386/

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