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我写了PHP代码以从Database中获取数据。它获得登录值。 （用户名为&密码）。我通过mysql server/localhost连接它。当我运行这个PHP代码时，它总是显示“0”。这意味着它不会从数据库获取数据。这是为什么？在Android中使用PHP登录表单
这里我的PHP代码：
<?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";
}
?>
我把这个在我的WAMP服务器的WWW folder.when我运行这个PHP文件WAMP的服务器本地主机总是说“没有发现这样的用户”。
我用这个php文件从数据库中获取数据来连接android登录表单。它包含两个字段。这是用户名为&的密码。
这里我给我的android登录代码。
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();
}
});
}
总是我没有这样的用户发现& PHP响应alart是没有这样的用户发现
这是为什么？ 请帮助我。
我用下面的PHP代码
<?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
?>
然后返回uid作为响应。但不验证用户。有PHP代码错误或Android代码错误。我想匹配的值用户输入&数据库得到。这是否发生在这里。 如果不给我正确的事情。
如果这回答了你的问题，请把它标记为接受的答案:) – Nick
尼克的回答将致力于改变“密码”“psswd”从你的PHP和Android需要匹配每一个参数。 –