我错过了什么导致：命令不同步；你现在不能运行这个命令？

Question

$recruitcheck/$badgecheck 确实有效，它 return 是正确的，并且它提供了经过验证的回显，但我假设我对 store/bind/close 做了一些不正确的事情，因为我我在弄清楚它的工作原理时遇到了一些问题。

$emailcheck/$namecheck 出现了问题，因为它没有准确地 return 查询，并且它最终绕过了 $register 阶段，在那里它 return 了错误echo，所以 $emailcheck/$namecheck 总是 false/0.

我以前让它工作过，但我正在尝试实现一些安全功能，因此我添加了准备好的语句，在添加这些语句之前我几乎不知道我在做什么，所以如果这个问题不合适，我深表歉意，或者信息不够。如果您发表评论，我会编辑 post 使其更相关。请给我机会

$recruitcheck = mysqli_prepare($maindb, "SELECT aurapass, recruitbadge FROM auras WHERE auraname = ?");
            mysqli_stmt_bind_param($recruitcheck, "s", $recruitername);
            mysqli_stmt_execute($recruitcheck);
            mysqli_stmt_bind_result($recruitcheck, $passcheck, $badgecheck);
            mysqli_stmt_fetch($recruitcheck);
            if($passcheck == $recruitpass){
                if($badgecheck == "valid"){
                    mysqli_close($recruitcheck);
                    echo "<script>alert('Recruiter badge verified.')</script>";
                    $emailcheck = mysqli_prepare($maindb, "SELECT * FROM auras WHERE email = ?");
                    mysqli_stmt_bind_param($emailcheck, "s", $email);
                    mysqli_stmt_execute($emailcheck);
                    mysqli_stmt_fetch($emailcheck);
                    $namecheck = mysqli_prepare($maindb, "SELECT * FROM auras WHERE auraname = ?");
                    mysqli_stmt_bind_param($namecheck, "s", $auraname);
                    mysqli_stmt_execute($namecheck);
                    mysqli_stmt_fetch($namecheck);
                    if(mysqli_num_rows($emailcheck)>0 || mysqli_num_rows($namecheck)>0){
                        die("<script>alert('New auraname/email is already taken.')</script>");
                    }
                    else{
                        mysqli_close($emailcheck);
                        mysqli_close($namecheck);
                        if($email == $emailconfirm){
                            $register = mysqli_prepare($maindb, "INSERT INTO auras (recruitername, recruitmethod, email, birthday, country, gender, auraname) VALUES (?, ?, ?, ?, ?, ?, ?)");
                            mysqli_stmt_bind_param($register, "sssisss", $recruitername, $recruitmethod, $email, $birthday, $country, $gender, $auraname);
                            mysqli_stmt_execute($register);
                            mysqli_close($register);
                            if(!$register) {
                                echo mysqli_error($maindb);
                                die("<script>alert('Error returned during registration: See bottom of page.')</script>");
                            }
                            else{
                            die("<script>alert('New aura registered successfully!')</script>");

$passcheck/$badgecheck 只是简单的绑定变量，它们唯一被调用的地方就是你在这里看到它们的地方。否则：

$recruitpass = cleaninput($_POST["recruitpass"]);

和

function cleaninput($info) {
            $info = trim($info);
            $info = stripslashes($info);
            $info = htmlspecialchars($info);
            return $info;}

recruitcheck 确实有效，因为它会随着输入的错误与正确的变化而变化，但代码无法正确地继续进行下一次检查。

Answer 1

您先为 $emailcheck 和 $namecheck 查询调用 mysqli_close()，然后再对它们调用 mysqli_num_rows()。

我错过了什么导致：命令不同步；你现在不能运行这个命令？

What am I missing that is causing: Commands out of sync; you can't run this command now?

php

mysqli

procedural-programming

我错过了什么导致：命令不同步；你现在不能 运行 这个命令？

What am I missing that is causing: Commands out of sync; you can't run this command now?

php

mysqli

procedural-programming

我错过了什么导致：命令不同步；你现在不能运行这个命令？