看起来我对这段代码有一些问题。我需要通过$uid和$password来检查用户。我以为我做对了，但它仍然不起作用，我想不通。我是编程新手，非常感谢您的帮助！
更新。。
下面的函数工作得很好，但当我按照下面的方式修改它时，它会给我带来一些错误。。
public function getUser($uid, $password) {
$result = mysql_query("SELECT * FROM users WHERE id = '$uid' AND pswd = '$password'") or die(mysql_error());
$no_of_rows = mysql_num_rows($result);
if ($no_of_rows > 0) {
$result = mysql_fetch_array($result);
return $result;
} else {
return false;
}
}
user_function.php
<?php
class DB_Functions {
private $db;
function __construct() {
require_once 'db_connect.php';
$this->db = new DB_Connect();
$this->db->connect();
}
public function getUser($uid, $password) {
$stmt = $db->prepare("SELECT * FROM users WHERE id=? AND pswd=?");
$stmt->execute(array($uid, $password));
return $stmt->fetch();
}
}
?>
index.php
<?php
if (isset($_POST['tag']) && $_POST['tag'] != '') {
$tag = $_POST['tag'];
require_once 'include/db_functions.php';
$db = new DB_Functions();
$response = array("tag" => $tag, "success" => 0, "error" => 0);
if ($tag == 'login') {
// check for user
$user = $db->getUser($_POST['id'], $_POST['pswd']);
if ($user != false) {
$response["success"] = 1;
$response["user"]["id"] = $user["id"];
echo json_encode($response);
} else {
$response["error"] = 1;
$response["error_msg"] = "Incorrect email or password!";
echo json_encode($response);
}
我认为您的数据库连接不在范围内；尝试：
public function getUser($uid, $password) {
$stmt = $this->db->prepare("SELECT * FROM users WHERE id=? AND pswd=?");
$stmt->execute(array($uid, $password));
return $stmt->fetch();
}
但是，请参阅有关检查返回值和错误消息的注释，以查看数据库实际返回的内容。
在类中，属性$db没有被正确引用。尝试更改
public function getUser($uid, $password) {
$stmt = $db->prepare("SELECT * FROM users WHERE id=? AND pswd=?");
$stmt->execute(array($uid, $password));
return $stmt->fetch();
}
至
public function getUser($uid, $password) {
$stmt = $this->db->prepare("SELECT * FROM users WHERE id=? AND pswd=?");
$stmt->execute(array($uid, $password));
return $stmt->fetch();
}
我不知道如何获得DB连接句柄，但您可能需要从类属性引用您的DB，该属性存储如下：
$this->db->prepare("SELECT * FROM users WHERE id=? AND pswd=?");