Question

php代码

$userName=$_POST['userName'];
$userPassword=$_POST['userPassword'];
.
.
$result=MYSQL_QUERY("SELECT * FROM USER WHERE UserName='$userName' and   Password='$userPassword' ");

$count=mysql_num_rows($result);

if($count==1){

session_start();
$_SESSION["userName"] = "$userName";
echo 1;
header('Location: http://www.naviwi.com/points.php');
}
else {
header('Location: http://www.naviwi.com/signUp.php');
}
ob_end_flush();

?>

java代码。我从两个TexitEdit IO获取用户名和密码值。

public void userLogin(View view){


        HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost("http://naviwi.com/checkLogin");

        List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
        nameValuePairs.add(new BasicNameValuePair("userName", username));
        nameValuePairs.add(new BasicNameValuePair("userPassword", password));
        try {
            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
        }catch(UnsupportedEncodingException e){
            et.append("UnsupportedEncodingException");
        }

        try {
            response = httpclient.execute(httppost);
        }catch (Exception e) {
            et.append("Server not Responding"+ response);
        }

        et.append(" Server respond "+response);
    }

所以问题是我总是得到null。我想从PHP中的响应方法中获取1。

Answer 1

SQL查询使用了错误的字段。密码字段称为“userPassword”，因此您应该查询

$result=MYSQL_QUERY("SELECT * FROM USER WHERE UserName='$userName' and   userPassword='$userPassword' ");

警告：您在数据库中存储了明文密码，这是一种非常糟糕的做法。你应该hash them with a salt

无法使用HttpPost从我的Android应用程序登录

1 个答案: