Question

所以我正在为我的游戏编写一个帐户登录区域，并在我的网站上使用json get action调用log.php。 json操作看起来像http://domain.com/log.php?fname='"&myUserName&"'&fpass='"&myPassword&"'

log.php内容如下。

<?php header('Access-Control-Allow-Origin: *'); 

    $username = $_GET['fname'];

    $password = $_GET['fpass'];

    $connect = mysqli_connect("localhost","user","pass","db") or die("Could not connect");

    $sql = "SELECT id FROM members WHERE username = '$username' and password = '$password'";

    $result =  mysqli_query($connect, $sql);

    if($result === FALSE) { 
            die(mysql_error());
    }   

    while ($row = mysqli_fetch_array($result)) {

        echo $row['id'];

    }

mysql_close($connect);

?>

我没有得到任何错误，但它没有返回id，假设在游戏中设置变量“userID”。它仍然是空白的。任何想法如何让这个工作？

由于

Answer 1

更改为

＆＃13;

$username = $_GET['fname'];

＆＃13;

Answer 2

您需要使用json_encode（）函数生成JSON响应。

$username = $_GET['fname'];

    $password = $_GET['fpass'];

    $connect = mysqli_connect("localhost","user","pass","db") or die("Could not connect");

    $sql = "SELECT id FROM members WHERE username = '$username' and password = '$password'";

    $result =  mysqli_query($connect, $sql);

    if($result === FALSE) { 
            die(mysql_error());
    }   

    while ($row = mysqli_fetch_array($result)) {
        $res = array("userID"=> $row['id']);
    }
    mysql_close($connect);
    echo $res_json = json_encode($res);
    ?>

在您的客户端，您将使用解析JSON＆amp; amp; data.userID

Answer 3

没关系，我在编码伙伴的帮助下解决了这个问题。如果你们有兴趣寻找，可以解决这个问题。

<?php 
header('Access-Control-Allow-Origin: *'); 
error_reporting(0);
require 'connection.php';

$username = $_GET["fname"];
$password = $_GET["fpass"];

$safeuser = $db->real_escape_string(md5($username));
$safepass = $db->real_escape_string(md5($password));

if($result = $db->query("SELECT * FROM `members` WHERE username='".$safeuser."' AND password='".$safepass."'")){
    if ($count = $result->num_rows){

        while($row = $result->fetch_assoc()){
            echo $row['id'];

        }


    }
}   

  $db->close();
 ?>

用php检索表值

3 个答案: