Question

这个问题可能重复，但我无法在问题中找到任何解决方案。

这是一种简单的登录方式。

基本上我从用户那里获取两个参数并将它们发送到服务器，服务器搜索数据库并返回与我发送的参数匹配的结果。如果没有匹配的结果，方法只返回null，所以我可以要求用户更正用户名和密码。

以下是我用来发送参数的代码;

    @Override
    protected User doInBackground(Void... params) {

        ArrayList<NameValuePair> dataToSend = new ArrayList<>();
        dataToSend.add(0,new BasicNameValuePair("username",user.username));
        dataToSend.add(1,new BasicNameValuePair("password",user.password));



        HttpParams httpRequestParams = new BasicHttpParams();
        HttpConnectionParams.setConnectionTimeout(httpRequestParams,CONNECTION_TIMEOUT);
        HttpConnectionParams.setSoTimeout(httpRequestParams, CONNECTION_TIMEOUT);

        HttpClient client = new DefaultHttpClient();
        HttpPost post = new HttpPost("http://www.okanyakit.com/tez/FetchUserData.php");

        User returnedUser = null;

        try {
            post.setEntity(new UrlEncodedFormEntity(dataToSend));
            HttpResponse httpResponse = client.execute(post);

            HttpEntity entity = httpResponse.getEntity();
            String result = EntityUtils.toString(entity);
            JSONObject jObject = new JSONObject(result);

            if (jObject.length() == 0){
                returnedUser = null;
            }else {

                String username = jObject.getString("username");
                String password = jObject.getString("password");
                String email = jObject.getString("email");
                String  phonenumber = jObject.getString("phonenumber");
                String bloodtype = jObject.getString("bloodtype");
                String  birthday = jObject.getString("birthday");
                String address = jObject.getString("address");


                returnedUser = new User(username,password,email,phonenumber,bloodtype,birthday,address);

            }


        }catch (Exception e){
            e.printStackTrace();
        }


        return returnedUser;
    }

在这里，我可以将用户名和密码都放到列表中，但是当它连接到服务器时，我收到一个错误，它说我只发送了一个参数，而期望有两个。

我不认为我的PHP代码有问题，但我不确定，所以在这里;

    <?php   
$con=mysqli_connect("89.19.30.210","okan_tez","okan_tez","okan_tez_andro");

$username = $_POST["username"];
$password = $_POST["password"];




mysqli_close($con);
$statement = mysqli_prepare("SELECT * FROM user WHERE username = ? AND password = ? ");
mysqli_stmt_bind_param($statement, "ss",$username, $password);
mysqli_stmt_execute($statement);
mysqli_stmt_store_result($statement);
mysqli_stmt_bind_result($statement, $userid, $username, $password, $email, $phonenumber, $bloodtype, $birthday, $address);

$user = array();
while (mysqli_stmt_fetch($statement)){
    $user [username] = $username;
    $user [password] = $password;
    $user [email] = $email;
    $user [phonenumber] = $phonenumber;
    $user [bloodtype] = $bloodtype;
    $user [birthday] = $birthday;
    $user [address] = $address;

}
echo json_encode($user);
mysqli_stmt_close($statement);


?>

这是调试截图;

enter image description here

这里是用绿色字体说的

警告：mysqli_prepare（）需要2个参数，1 /home/okanyakit.com/httpdocs/tez/FetchUserData.php 在 11 行中给出>

警告：mysqli_stmt_bind_param（）要求参数1为mysqli_stmt，在 /home/okanyakit.com/httpdocs/tez/FetchUserData.php 12 <中给出null / b>

警告：mysqli_stmt_execute（）要求参数1为mysqli_stmt，在 /home/okanyakit.com/httpdocs/tez/FetchUserData.php 13 中给出null / b>

警告：mysqli_stmt_store_result（）要求参数1为mysqli_stmt，在 /home/okanyakit.com/httpdocs/tez/FetchUserData.php 中 14 < / b>

警告：mysqli_stmt_bind_result（）要求参数1为mysqli_stmt，在 /home/okanyakit.com/httpdocs/tez/FetchUserData.php 15 < / b>

警告：mysqli_stmt_fetch（）要求参数1为mysqli_stmt，在 /home/okanyakit.com/httpdocs/tez/FetchUserData.php 18 < / b>
[]
警告：mysqli_stmt_close（）要求参数1为mysqli_stmt，在 /home/okanyakit.com/httpdocs/tez/FetchUserData.php 29 <中给出null / b>

那么问题是什么，我该如何解决呢？请项目截止日期为两天，我还有很多问题。我对注册部分有一个类似的问题。我想如果我能解决这个问题，我也能解决这个问题。感谢。

Answer 1

你不应该在这里关闭连接：

mysqli_close($con);
$statement = mysqli_prepare("SELECT * FROM user WHERE username = ? AND password = ? ");

并且mysqli_prepare需要$con参数（http://php.net/manual/fr/mysqli.prepare.php）：

$statement = mysqli_prepare($con,"SELECT * FROM user WHERE username = ? AND password = ? ");

Android：通过post方法

1 个答案: