Php MySql 选择计数，准备好的语句返回始终为 1

Question

我正在使用此方法来登录Web服务:

function redeem() {
    if (isset($_POST["user"]) && isset($_POST["pass"]) && isset($_POST["computer"])) {
        $user = $_POST['user'];
        $pass = $_POST['pass'];
        $computer = $_POST['computer'];

        $galNumb = "SELECT COUNT(*) FROM Useres WHERE username = ? AND password = ?";
        $stmt = $this->db->prepare($galNumb);
        $stmt->bind_param('ss', $user, $pass);
        $gNumb = $stmt->execute();

        $result = array(
            "success" => "true",
        );
        $this->sendResponse(200, $gNumb);
        return true;
    }
    $this->sendResponse(400, 'Invalid request');
    return false;
}

问题在于，即使 sql 表不包含用户名 和密码，$gNumb 也始终返回 1。知道可能是什么问题吗？

Answer 1

您忘记获取结果:

...
 $stmt->bind_param('ss', $user, $pass);
 if ($stmt->execute()) {
      $stmt->bind_result($gNumb);
      $stmt->fetch();
 } else {
      $gNumb = 0;
 }
...