我用php在android中注册时发送两次数据
我不熟悉Android和php,我尝试从服务器端用php在Android中注册,一切正常,我插入正确,但后来我决定我需要发送一封电子邮件,在做测试时,数据发送了两次,我跳了起来:用户已经存在,如果电子邮件到达并向我显示该消息,我已经复习了很长一段时间,因为我没有发现错误,我留下了脚本,以防我需要帮助,谢谢 PHP:我用php在android中注册时发送两次数据,php,android,Php,Android,我不熟悉Android和php,我尝试从服务器端用php在Android中注册,一切正常,我插入正确,但后来我决定我需要发送一封电子邮件,在做测试时,数据发送了两次,我跳了起来:用户已经存在,如果电子邮件到达并向我显示该消息,我已经复习了很长一段时间,因为我没有发现错误,我留下了脚本,以防我需要帮助,谢谢 PHP: Y este es mi codigo android: private void registerUser() { final String email = editT
Y este es mi codigo android:
private void registerUser() {
final String email = editTextEmail.getText().toString().trim();
final String username = editTextUsername.getText().toString().trim();
final String password = editTextPassword.getText().toString().trim();
progressDialog.setMessage("Registrando usuario...");
progressDialog.show();
StringRequest stringRequest = new StringRequest(Request.Method.POST,
Constants.URL_REGISTER,
new Response.Listener<String>() {
@Override
public void onResponse(String response) {
progressDialog.dismiss();
try {
JSONObject jsonObject = new JSONObject(response);
Toast.makeText(getApplicationContext(), jsonObject.getString("message"), Toast.LENGTH_LONG).show();
} catch (JSONException e) {
e.printStackTrace();
}
}
},
new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
progressDialog.hide();
Toast.makeText(getApplicationContext(), error.getMessage(), Toast.LENGTH_LONG).show();
}
}) {
@Override
protected Map<String, String> getParams() throws AuthFailureError {
Map<String, String> params = new HashMap<>();
params.put("username", username);
params.put("email", email);
params.put("password", password);
return params;
}
};
RequestHandler.getInstance(this).addToRequestQueue(stringRequest);
}
@Override
public void onClick(View view) {
if (view == buttonRegister)
registerUser();
if(view == textViewLogin)
startActivity(new Intent(this, LoginActivity.class));
}
private void registerUser(){
最终字符串email=editTextEmail.getText().toString().trim();
最终字符串用户名=editTextUsername.getText().toString().trim();
最终字符串密码=editTextPassword.getText().toString().trim();
progressDialog.setMessage(“registando usuario…”);
progressDialog.show();
StringRequest StringRequest=新StringRequest(Request.Method.POST,
Constants.URL\u寄存器,
新的Response.Listener(){
@凌驾
公共void onResponse(字符串响应){
progressDialog.disclose();
试一试{
JSONObject JSONObject=新JSONObject(响应);
Toast.makeText(getApplicationContext(),jsonObject.getString(“消息”),Toast.LENGTH\u LONG.show();
}捕获(JSONException e){
e、 printStackTrace();
}
}
},
新的Response.ErrorListener(){
@凌驾
公共无效onErrorResponse(截击错误){
progressDialog.hide();
Toast.makeText(getApplicationContext(),error.getMessage(),Toast.LENGTH_LONG.show();
}
}) {
@凌驾
受保护的映射getParams()引发AuthFailureError{
Map params=新的HashMap();
参数put(“用户名”,用户名);
参数put(“电子邮件”,电子邮件);
参数put(“密码”,密码);
返回参数;
}
};
RequestHandler.getInstance(this).addToRequestQueue(stringRequest);
}
@凌驾
公共void onClick(视图){
如果(视图==按钮注册表)
registerUser();
如果(视图==文本视图登录)
startActivity(新意图(这个,LoginActivity.class));
}
在phpI中用于发送电子邮件我想你只是读了标题…添加创建用户Php代码
private void registerUser() {
final String email = editTextEmail.getText().toString().trim();
final String username = editTextUsername.getText().toString().trim();
final String password = editTextPassword.getText().toString().trim();
progressDialog.setMessage("Registrando usuario...");
progressDialog.show();
StringRequest stringRequest = new StringRequest(Request.Method.POST,
Constants.URL_REGISTER,
new Response.Listener<String>() {
@Override
public void onResponse(String response) {
progressDialog.dismiss();
try {
JSONObject jsonObject = new JSONObject(response);
Toast.makeText(getApplicationContext(), jsonObject.getString("message"), Toast.LENGTH_LONG).show();
} catch (JSONException e) {
e.printStackTrace();
}
}
},
new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
progressDialog.hide();
Toast.makeText(getApplicationContext(), error.getMessage(), Toast.LENGTH_LONG).show();
}
}) {
@Override
protected Map<String, String> getParams() throws AuthFailureError {
Map<String, String> params = new HashMap<>();
params.put("username", username);
params.put("email", email);
params.put("password", password);
return params;
}
};
RequestHandler.getInstance(this).addToRequestQueue(stringRequest);
}
@Override
public void onClick(View view) {
if (view == buttonRegister)
registerUser();
if(view == textViewLogin)
startActivity(new Intent(this, LoginActivity.class));
}