我在這裏有這行代碼。登錄驗證到遠程mysql數據庫
我的Java代碼:
btnLogin.setOnClickListener(new View.OnClickListener() {
@Override
$public void onClick(View v) {
ArrayList<NameValuePair> postParameters = new ArrayList<NameValuePair>();
postParameters.add(new BasicNameValuePair("username", txtUsername.getText().toString()));
postParameters.add(new BasicNameValuePair("password", txtPassword.getText().toString()));
//String valid = "1";
String response = null;
try {
response = CustomHttpClient.executeHttpPost("http://www.sampleweb.com/imba.php", postParameters);
String res=response.toString();
// res = res.trim();
res= res.replaceAll("\\s+","");
//error.setText(res);
if(res.equals("1")){
txtError.setText("Correct Username or Password");
//Intent i = new Intent(CDroidMonitoringActivity.this, MenuClass.class);
//startActivity(i);
}
else {
txtError.setText("Sorry!! Incorrect Username or Password");
}
} catch (Exception e) {
txtUsername.setText(e.toString());
}
}
});
}
我的PHP腳本代碼:
<?php
$un=$_POST['username'];
$pw=$_POST['password'];
$user = ‘bduser’;
$pswd = ‘dbpwd’;
$db = ‘phplogin’;
$conn = mysql_connect("localhost","root","");
mysql_select_db($db, $conn);
$query = mysql_query("SELECT * FROM user WHERE username = '$un' AND password = '$pw'");
$result = mysql_query($query) or die("Unable to verify user because : " . mysql_error());
if(mysql_num_rows($result) == 1)
echo 1; // for correct login response
else
echo 0; // for incorrect login response
?>
我有這個代碼的問題。在我的android代碼中,當我嘗試將res.equals更改爲包含。它總是說正確的密碼,但如果我不改變它,它說不正確的密碼。我不知道我的java代碼或我的php代碼中有什麼問題。真的需要幫助。
你能給我一個PHP示例代碼先生嗎? – ching