一个奇怪的情况，即php中是否有其他陈述故障

Question

我已经在这个问题上工作了几个小时，真的希望我不会错过任何基本的东西。我知道它主要是代码，但所有信息都在代码中的注释中。

我有一个我的函数，我可以在我的情况下喂养： $db是带有访问信用的数组 $uid在这种情况下为无效 $email是一个有效的电子邮件地址 $companyid是一个有效的特定数字值，无关紧要 $fname, $lname, $mname, $role-有效字符串

现在功能本身：

function hire_employee($db, $uid, $email, $companyid, $fname, $lname, $mname, $role){
    try{    
        $srvr = new PDO("mysql:host=".$db['server'].";dbname=".$db['db'], $db['mysql_login'], $db['mysql_pass'], array(PDO::MYSQL_ATTR_INIT_COMMAND => 'SET NAMES utf8'));
        $srvr->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
        if (is_numeric($uid)){ //uid is null in our case, so we skip this piece of code
            echo "<br>trace 1";//does not output
            $set=$srvr->prepare("INSERT into roles (uid, companyid, role, assigned_date) VALUES ((SELECT uid FROM users WHERE uid=:uid and active=1), :companyid, :role, CURDATE());");
            $set->bindParam(":uid", $uid);
            $set->bindParam(":companyid", $companyid);
            $set->bindParam(":role", $role);
            if ($set->execute()){
                return true;
            }
        } elseif (isset($email)) {// this is our case, we have a valid email
            echo "<br>trace 2";//does output
            $email = filter_var($email, FILTER_SANITIZE_EMAIL);
            if (!filter_var($email, FILTER_VALIDATE_EMAIL) === false) {
                //is a valid email address
                //first trying to add existing user (if exists)
                echo "<br>trace 2.1";//does output
                $check=$srvr->prepare("INSERT into roles (uid, companyid, role, assigned_date) VALUES ((SELECT uid FROM users WHERE email=:email and active=1), :companyid, :role, CURDATE());");
                $check->bindParam(":companyid", $companyid);
                $check->bindParam(":role", $role);
                $check->bindParam(":email", $email);
                //NOW THE WEIRD PART!!!-----------------------
                if ($check->execute()){//checking if statement executes successfully and returns true
                    echo "<br>trace 2.1.1";//does not output, a proper thing in our case
                    return true;
                } else { //nope user does not exist, we need to create a user
                    echo "<br>trace 2.1.2";//this does not output either!!!
                    //dear stackoverflow member can skip the rest of the code below
                    echo "<br>traceYO!!! $uid, $email, $companyid, $fname, $lname, $mname, $role";
                    $newuserid=register_user($db, $fname, $mname, $lname, "NULL", "NULL", "NULL", "NULL", true, $email, "NULL");
                    echo $newuserid;
                    if (!is_numeric($newuserid) AND ($newuserif > 0)) {
                        echo "<br>trace 2.1.2.1";
                        return false;
                    } else {
                        echo "<br>trace 2.1.2.2";
                        $insert=$srvr->prepare("INSERT into roles (uid, companyid, role, assigned_date) VALUES (:uid, :companyid, :role, CURDATE());");
                        $insert->bindParam(":uid", $newuserid);
                        $insert->bindParam(":companyid", $companyid);
                        $insert->bindParam(":role", $role);
                        if ($insert->execute()){
                            echo "<br>trace 2.1.2.2.1";
                            return true;
                        }
                    }
                }
            } else {
                //is not a valid email address
                return false;
            }
        } else { 
            echo "<br>trace 3";
            return false;
        }
    }
    catch(PDOException $e) {
        report_error($db, $_SESSION['uid'], $companyid, 'hire_employee', "companyid: $companyid, fname: $fname, mname: $mname, lname: $lname, email: $email, uid: $uid, role:$role", $e->getMessage());
        return false;
    }
}

现在我到达这里的唯一输出：

跟踪2跟踪2.1

以某种方式无法跟踪2.1.1或2.1.2。

任何帮助都将不胜感激。

Answer 1

我得到了它。该行if ($check->execute())生成了错误的异常，该异常立即获得了CAUHGT，从而阻止了该功能进一步执行。实际上，我看到错误的地方添加了错误，但我认为我希望它们成为一个菜鸟，但除了例外，我没有将两个和两个放在一起。我将ATTR_ERRMODE更改为PDO::ERRMODE_WARNING，似乎可以正常工作。我收到警告，execute()不返回true，并且功能按预期进行。

如果有人向我展示了如何捕捉警告，我真的很感激。

无论如何，谢谢大家！

upd_1：

允许生成任何警告是一种非常糟糕的做法，我正在对功能和或声明进行相应的调整。