我正在使用PHP，MYSQL，java Volley Library构建项目
问题是它不能同时工作所有3个出现此错误
E / catch ===：org.json.JSONException：SorrySignUpFirst没有值SorrySignUpFirst,SorrySignUpFirst,LoginPasswordWrong
如果我发表评论2，则运行我的模拟器，其工作原理如何解决此问题
PHP代码
<?php
if ($_SERVER['REQUEST_METHOD'] == "POST")
{
require_once("connection.php");
$data_array = array();
$login_email = $_POST['email'];
$login_email = strip_tags($login_email);
$login_email = str_replace(' ', '', $login_email); // remove spaces
$login_password = $_POST['password'];
$login_password = strip_tags($login_password);
$email = str_replace(' ', '', $login_password); // remove spaces
$db_email = mysqli_query($connection, "SELECT * FROM users WHERE email ='$login_email'");
if (mysqli_num_rows($db_email) == 0)
{
$data_array['SorrySignUpFirst'] = "1";
echo json_encode($data_array);
mysqli_close($connection);
}
else
{
$data = mysqli_fetch_array($db_email);
if (password_verify($login_password, $data['password']))
{
$data_array['LoginSuccessfull'] = "1";
$data_array['id'] = $data['id'];
$data_array['email'] = $data['email'];
echo json_encode($data_array);
mysqli_close($connection);
}
else
{
$data_array['LoginPasswordWrong'] = "1";
echo json_encode($data_array);
mysqli_close($connection);
}
}
}
?>
LoginActivity.java代码
final StringRequest LoginstringRequest = new StringRequest(Request.Method.POST, URLS.LOGIN_API, new Response.Listener<String>() {
@Override
public void onResponse(String response) {
try {
JSONObject jsonObject = new JSONObject(response);
String SorrySignUpFirst = jsonObject.getString("SorrySignUpFirst");
String LoginSuccessfull = jsonObject.getString("LoginSuccessfull");
String LoginPasswordWrong = jsonObject.getString("LoginPasswordWrong");
if (SorrySignUpFirst.contains("1")) {
Intent goToHomeScreenIntent = new Intent(LoginActivity.this, SignupActivity.class);
startActivity(goToHomeScreenIntent);
Toast.makeText(LoginActivity.this, "User Doen't Exist Sign Up", Toast.LENGTH_LONG).show();
}
if (LoginSuccessfull.contains("1")) {
String id = jsonObject.getString("id");
String fname = jsonObject.getString("email");
Intent goToHomeScreenIntent = new Intent(LoginActivity.this, HomeScreenActivity.class);
startActivity(goToHomeScreenIntent);
Toast.makeText(LoginActivity.this, "id = " + id + "email == " + fname, Toast.LENGTH_LONG).show();
}
if (LoginPasswordWrong.contains("1")) {
Toast.makeText(LoginActivity.this, "Password Wrong", Toast.LENGTH_LONG).show();
}
else {
Toast.makeText(LoginActivity.this, "if else if Error", Toast.LENGTH_LONG).show();
}
} catch (JSONException e) {
Toast.makeText(LoginActivity.this, "catch -- " + e.toString(), Toast.LENGTH_LONG).show();
Log.e("catch === ", e.toString());
}
}
}, new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
Toast.makeText(LoginActivity.this, "Error= 116" + error.toString(), Toast.LENGTH_LONG ).show();
Log.i("Catch error 116 ====", error.toString());
}
})
{
@Override
protected Map<String, String> getParams() throws AuthFailureError {
Map<String, String> loginParams = new HashMap<>();
loginParams.put("email", Email);
loginParams.put("password", Password);
return loginParams;
}
};
RequestQueue requestQueue = Volley.newRequestQueue(this);
requestQueue.add(LoginstringRequest);
最佳答案
在我看来，您的jsonObject上没有名为SorrySignUpFirst的属性。
尝试在此行上放置一个断点：
String SorrySignUpFirst = jsonObject.getString("SorrySignUpFirst");
检查jsonObject并检查是否存在SorrySignUpFirst。
关于java - 我面临着这个问题E/catch ===:org.json.JSONException:SorrySignUpFirst没有值，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/55852502/