【发布时间】：2020-03-26 17:33:39
【问题描述】：
我正在尝试登录，它给了我JSONExceptionError。下面是我的登录代码。如果我将localhost 放在我的 php 代码中，它可以工作，但是对于这个 IP 地址它不起作用。
public class MainActivity extends AppCompatActivity {
TextView textView;
Button login_button;
EditText Username,Password;
String username,password;
String login_url = "http://192.168.0.21/maps/login.php";
AlertDialog.Builder builder;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
textView = (TextView)findViewById(R.id.reg_txt);
textView.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View view) {
Intent intent = new Intent(MainActivity.this,Register.class);
startActivity(intent);
}
});
builder = new AlertDialog.Builder(MainActivity.this);
login_button = (Button)findViewById(R.id.bn_login);
Username = (EditText)findViewById(R.id.login_name);
Password = (EditText)findViewById(R.id.login_password);
login_button.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
username = Username.getText().toString();
password = Password.getText().toString();
if(username.equals("")||password.equals("")){
builder.setTitle("Something went wrong");
displayAlert("Enter a valid username and password");
}
else{
StringRequest stringRequest = new StringRequest(Request.Method.POST, login_url, new Response.Listener<String>() {
@Override
public void onResponse(String response) {
try {
JSONArray jsonArray = new JSONArray(response);
JSONObject jsonObject = jsonArray.getJSONObject(0);
String code = jsonObject.getString("code");
if(code.equals("login_failed")){
builder.setTitle("Error in login");
displayAlert(jsonObject.getString("message"));
}
else{
startActivity(new Intent(MainActivity.this,MapsActivity.class));
}
} catch (JSONException e) {
e.printStackTrace();
}
}
}, new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
Toast.makeText(MainActivity.this, error.getMessage(), Toast.LENGTH_LONG).show();
}
}){
@Override
protected Map<String,String> getParams() throws AuthFailureError{
Map<String,String> params = new HashMap<>();
params.put("user_name",username);
params.put("password",password);
return params;
}
};
MySingeleton.getInstance(MainActivity.this).addToRequestque(stringRequest);
}
}
});
}
public void displayAlert(String message){
builder.setMessage(message);
builder.setPositiveButton("OK", new DialogInterface.OnClickListener() {
@Override
public void onClick(DialogInterface dialog, int which) {
Username.setText("");
Password.setText("");
}
});
AlertDialog alertDialog = builder.create();
alertDialog.show();
}
}
下面是我的php连接代码。在浏览器中运行后，它给出了输出，但数据没有插入到数据库中。
<?php
header('Content-Type: application/json');
$host = "";
$user_name = "android";
$user_password = "^AndroidReg2019";
$db_name = "mapdb";
$conn = mysqli_connect($host,$user_name,$user_password,$db_name);
if($conn)
echo "Connection success..";
else
echo "Connection failed...";
?>
以下是我的登录 php 代码。
<?php
require "init.php";
$user_name = $_POST["user_name"];
$password = $_POST["password"];
$sql = "select user_name from user_info where user_name = '$user_name' and password = '$password';";
$result = mysqli_query($conn,$sql);
$response = array();
if(mysqli_num_rows($result)>0){
$row = mysqli_fetch_row($result);
$name = $row[0];
$code = "login_success";
array_push($response,array("code"=>$code,"user_name"=>$user_name));
echo json_encode($response);
}
else{
$code = "login_failed";
$message = "Please verify credentials and try login again.";
array_push($response,array("code"=>$code,"message"=>$message));
echo json_encode($response);
}
mysqli_close($conn);
?>
【问题讨论】：
-
链接 192.168.0.21/maps/login.php 是否在 POSTMAN/任何其他 REST 服务工具中工作？
-
是的。它的工作。
-
你能把你发送到链接的Json发给我吗？
-
不，我的意思是您通过邮件发送的值
-
name=a&email=a&user_name=a&password=a