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我寫了PHP代碼以從Database中獲取數據。它獲得登錄值。 （用戶名爲&密碼）。我通過mysql server/localhost連接它。當我運行這個PHP代碼時，它總是顯示「0」。這意味着它不會從數據庫獲取數據。這是爲什麼？在Android中使用PHP登錄表單
這裏我的PHP代碼：
<?php
$hostname_localhost ="localhost";
$database_localhost ="gpsvts_geotrack";
$username_localhost ="root";
$password_localhost ="";
$localhost = mysql_connect($hostname_localhost,$username_localhost,$password_localhost)
or
trigger_error(mysql_error(),E_USER_ERROR);
mysql_select_db($database_localhost, $localhost);
$username = $_POST['uname'];
$password = $_POST['passwd'];
$query_search = "select 'uname' & 'passwd' from user_master where uname = '.$username.' AND passwd = '.$password.'";
$query_exec = mysql_query($query_search) or die(mysql_error());
$rows = mysql_num_rows($query_exec);
//echo $rows;
if($rows == 0) {
echo "No Such User Found";
}
else {
echo "User Found";
}
?>
我把這個在我的WAMP服務器的WWW folder.when我運行這個PHP文件WAMP的服務器本地主機總是說「沒有發現這樣的用戶」。
我用這個php文件從數據庫中獲取數據來連接android登錄表單。它包含兩個字段。這是用戶名爲&的密碼。
這裏我給我的android登錄代碼。
b = (Button)findViewById(R.id.Button01);
et = (EditText)findViewById(R.id.username);
pass= (EditText)findViewById(R.id.password);
tv = (TextView)findViewById(R.id.tv);
b.setOnClickListener(new OnClickListener() {
@Override
public void onClick(View v) {
dialog = ProgressDialog.show(AndroidPHPConnectionDemo.this, "",
"Validating user...", true);
new Thread(new Runnable() {
public void run() {
login();
}
}).start();
}
});
}
void login(){
try{
httpclient=new DefaultHttpClient();
httppost= new HttpPost("http://10.0.2.2//new/nuwan1.php"); // make sure the url is correct.
//add your data
nameValuePairs = new ArrayList<NameValuePair>(2);
// Always use the same variable name for posting i.e the android side variable name and php side variable name should be similar,
nameValuePairs.add(new BasicNameValuePair("username",et.getText().toString().trim())); // $Edittext_value = $_POST['Edittext_value'];
nameValuePairs.add(new BasicNameValuePair("password",pass.getText().toString().trim()));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
//Execute HTTP Post Request
response=httpclient.execute(httppost);
// edited by James from coderzheaven.. from here....
ResponseHandler<String> responseHandler = new BasicResponseHandler();
final String response = httpclient.execute(httppost, responseHandler);
System.out.println("Response : " + response);
runOnUiThread(new Runnable() {
public void run() {
tv.setText("Response from PHP : " + response);
dialog.dismiss();
}
});
if(response.equalsIgnoreCase("User Found")){
runOnUiThread(new Runnable() {
public void run() {
Toast.makeText(AndroidPHPConnectionDemo.this,"Login Success", Toast.LENGTH_SHORT).show();
}
});
startActivity(new Intent(AndroidPHPConnectionDemo.this, UserPage.class));
}else{
showAlert();
}
}catch(Exception e){
dialog.dismiss();
System.out.println("Exception : " + e.getMessage());
}
}
public void showAlert(){
AndroidPHPConnectionDemo.this.runOnUiThread(new Runnable() {
public void run() {
AlertDialog.Builder builder = new AlertDialog.Builder(AndroidPHPConnectionDemo.this);
builder.setTitle("Login Error.");
builder.setMessage("User not Found.")
.setCancelable(false)
.setPositiveButton("OK", new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialog, int id) {
}
});
AlertDialog alert = builder.create();
alert.show();
}
});
}
總是我沒有這樣的用戶發現& PHP響應alart是沒有這樣的用戶發現
這是爲什麼？ 請幫助我。
我用下面的PHP代碼
<?php
$un=$_POST['uname'];
$pw=$_POST['passwd'];
//connect to the db
$host="localhost"; // Host name
$user="root"; // Mysql username
$pswd=""; // Mysql password
$db="gpsvts_geotrack"; // Database name
$tbl_name="user_master"; // Table name
$conn = mysql_connect($host, $user, $pswd);
mysql_select_db($db, $conn);
//run the query to search for the username and password the match
$query = "SELECT * FROM $tbl_name WHERE uname = '$un' AND passwd= '$pw'";
//$query = "SELECT uid FROM $tbl_name WHERE uname = '$un' AND passwd = '$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($result) > 0)
echo mysql_result($result,0); // for correct login response
else
echo 0; // for incorrect login response
?>
然後返回uid作爲響應。但不驗證用戶。有PHP代碼錯誤或Android代碼錯誤。我想匹配的值用戶輸入&數據庫得到。這是否發生在這裏。 如果不給我正確的事情。
如果這回答了你的問題，請把它標記爲接受的答案:) – Nick
尼克的回答將致力於改變「密碼」「psswd」從你的PHP和Android需要匹配每一個參數。 –