mysqli_query如何使用null参数？

Question

希望一切都很好。

我有两个环境，

舞台和制作。

在暂存中，代码不起作用，但是在生产中起作用。

这是我的登录代码：

ini_set("session.cookie_httponly", "True");
header('x-powered-by:');
session_start();
include "includes/header.php";
error_reporting(0);
$subflag = $_POST['subflag'];
//$flag="true";
$myval   = "true";
if ($subflag == "true") {
    $user_answer = $_POST[user_answer];
    $answer      = $_POST[answer];
    if (substr(md5($user_answer), 5, 10) === $answer) {
        $captcha_valid = "true";
        $myval         = "true";
    } else {
        $captcha_valid = "false";
        $myval         = "false";
    }
    if ($myval == "false") {
        $errortext .= "Verification Failed";
        $subflag = "false";
    }
    if ($myval == "true") {
        $uname     = $_POST[username];
        $pwd       = $_POST[pwd];
        //$uname     = trim(str_replace("|", "'", filter_var(str_replace("\'", "|", $uname), FILTER_SANITIZE_STRING)));
        $pwd       = trim(str_replace("|", "'", filter_var(str_replace("\'", "|", $pwd), FILTER_SANITIZE_STRING)));
        $pwd       = mysqli_real_escape_string($conresult, stripslashes(trim($pwd)));
        $pwd       = md5($pwd);
        //$uname     = mysqli_real_escape_string($conresult, $uname);
        $queryStmt = mysql_query("LOCK TABLES loan_login WRITE");
        $result1   = mysqli_query($conresult, "SELECT * FROM '' WHERE lgn_uname = '".$uname."' && lgn_pwd = '$pwd'");
        $num_rows1 = mysqli_num_rows($result1);
        if ($num_rows1 != 0) {
            $row                  = mysqli_fetch_array($result1);
            $_SESSION[cmsauth]    = $row[lgnid];
            $_SESSION[loginttype] = $row[lgn_typ];
            $_SESSION[cmsvar]     = "MNH2010ADMLG";
            $_SESSION[lastlogin]  = $row[lgn_lastupdt];
            $result1              = mysqli_query($conresult, "update loan_login set lgn_lastupdt = NOW() where lgnid=" . $row[lgnid]);
            mysqli_close($conresult);
            $queryStmt = mysql_query("UNLOCK TABLES");
            ?>
            <script>

            </script>
            <?php
        } else {
            $flag = "false";
            //echo mysql_error();
        }
    } //flage ends
} //subflag ends

我同时回显$ conresult和$ result1，它们都为空？

有人可以帮我吗，mysqli_query如何在生产中工作，而空$ conresult却不能在产品中工作。

谢谢。