我是Android开发的新手,我正在尝试使用volley https://www.simplifiedcoding.net/android-volley-tutorial/来关注此登录教程。我跟着,它完美无缺。但是,当我尝试登录时,它给我的错误从我的php webservice响应。
LoginActivity.class
private void loginUser( final String email, final String password) {
// Tag used to cancel the request
String cancel_req_tag = "login";
progressDialog.setMessage("Logging you in...");
showDialog();
StringRequest strReq = new StringRequest(Request.Method.POST,
URL_FOR_LOGIN, new Response.Listener<String>() {
@Override
public void onResponse(String response) {
Log.d(TAG, "Register Response: " + response.toString());
hideDialog();
try {
JSONObject jObj = new JSONObject(response);
boolean error = jObj.getBoolean("error");
if (!error) {
String user = jObj.getJSONObject("USER_ID").getString("LASTNAME");
// Launch User activity
Intent intent = new Intent(
LoginActivity.this,
CameraTestActivity.class);
intent.putExtra("username", user);
startActivity(intent);
finish();
} else {
//show error output in my login screen
String errorMsg = jObj.getString("error_msg");
Toast.makeText(getApplicationContext(),
errorMsg, Toast.LENGTH_LONG).show();
}
} catch (JSONException e) {
e.printStackTrace();
}
}
}, new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
Log.e(TAG, "Login Error: " + error.getMessage());
Toast.makeText(getApplicationContext(),
error.getMessage(), Toast.LENGTH_LONG).show();
hideDialog();
}
}) {
@Override
protected Map<String, String> getParams() {
// Posting params to login url
Map<String, String> params = new HashMap<String, String>();
params.put("USER_ID", email);
params.put("LASTNAME", password);
return params;
}
的login.php
<?php
require_once 'dbConnect.php';
$response = array("error"=> FALSE);
if(isset($_POST['USER_ID'])) {
echo "my echo response";
$username = $_POST['USER_ID'];
$password = md5($_POST['LASTNAME']);
$pdo = Database::connect();
$pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$query = $pdo->prepare("SELECT * FROM tbl_UserDetails WHERE USER_ID= ? AND LASTNAME= ?");
$query->execute(array($username,$password));
$row = $query->fetch(PDO::FETCH_ASSOC);
Database::disconnect();
}
if (!(empty($row))){
$response["error"]= FALSE;
$response["user"]["ID"]=$row["USER_ID"];
$response["user"]["LASTNAME"]=$row["LASTNAME"];
echo json_encode($response);
}
else{
$response["error"] = TRUE;
//It show this message even I correctly input the correct username and password
$response["error_msg"] = "Wrong credentials!";
echo json_encode($response);
}
?>
login.php回复:
{"error":true,"error_msg":"Wrong credentials!"}
你能帮助我登录吗?
答案 0 :(得分:0)
我认为您的 PDO 语句存在问题。
更改
$query = $pdo->prepare("SELECT * FROM tbl_UserDetails ");
$query->execute(array($username,$password));
到
$query = $pdo->prepare("SELECT * FROM tbl_UserDetails WHERE name= ? AND password= ?");
$query->execute(array($username, $password));
还有另一个问题。
if(isset($_GET['USER_ID']))
这种情况永远不会得到满足。
因为您正在从您的Android客户端发出POST
请求
所以把它改成
if(isset($_POST['USER_ID']))
注意:强>
在数据库中更改我的答案中的列名。
另请参阅this有关PDO prepare()