嗨,我正在使用以下代碼來檢查用戶名和密碼是否正確與否的PHP代碼。代碼總是打印不正確的用戶名密碼
問題是,即使我給出正確的細節,它總是給出不正確的用戶名或密碼。
String response = null;
try {
response = CustomHttpClient.executeHttpPost("http://192.168.3.27/abc/check.php", postParameters); //Enetr Your remote PHP,ASP, Servlet file link
String res=response.toString();
// res = res.trim();
res= res.replaceAll("\\s+","");
//error.setText(res);
error.setText(res);
if(res.equals("1")){
Intent myIntent = new Intent(LoginActivity.this, HomeActivity.class);
startActivity(myIntent);
}
else
error.setText("Sorry!! Incorrect Username or Password");
這是我的php代碼。
<?php
$un=$_POST['username'];
$pw=$_POST['password'];
//connect to the db
$user = ‘root’;
$pswd = ‘root’;
$db = ‘mylogin’;
$conn = mysql_connect(‘localhost’, $user, $pswd);
mysql_select_db($db, $conn);
//run the query to search for the username and password the match
$query = 「SELECT * FROM userpass WHERE login_id = ‘$un’ AND login_pass = ‘$pw’」;
$result = mysql_query($query) or die(「Unable to verify user because : 」 . mysql_error());
//this is where the actual verification happens
if(mysql_num_rows($result) > 0)
echo 1; // for correct login response
else
echo 0; // for incorrect login response
?>
嘗試調試;' –
是否PHP代碼返回正確的響應? – cartina
嘗試在php代碼中回顯1。這有助於您查找問題出在php代碼還是java代碼中。 –