为什么是SQLSTATE [HY000]：一般错误？

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问题描述

这是一个代码，用于注册群组的头，然后注册他的合作伙伴。

This is a code to register the head of a group and then his partner(s).

<?
session_start();
require_once('db.php');
$db = new DB();
$db->newHead($_POST['fname'], $_POST['lname'], $_POST['school'], $_POST['day'], $_POST['month'], $_POST['year'], $_POST['email']);
$hId = $db->getId($_POST['email'], 1);
if ($_POST['pnum'] === 1)
{
    $db->newPartner($hId, $_POST['fname1'], $_POST['lname1'], $_POST['day1'], $_POST['month1'], $POST['year1'], $_POST['email1']);
    $pId = getId($_POST['email1'], 0);
    $db->setHeadId($hId, $pId, 1);
}
if ($_POST['pnum'] === 2)
{
    $db->newPartner($hId, $_POST['fname1'], $_POST['lname1'], $_POST['day1'], $_POST['month1'], $POST['year1'], $_POST['email1']);
    $p1Id = getId($_POST['email1'], 0);
    $db->setHeadId($hId, $p1Id, 1);

    $db->newPartner($hId, $_POST['fname2'], $_POST['lname2'], $_POST['day2'], $_POST['month2'], $POST['year2'], $_POST['email2']);
    $p2Id = getId($_POST['email2'], 0);
    $db->setHeadId($hId, $p2Id, 2);
}
header("Location:Register.php");
?>

这是db.php的一部分：

and this is part of db.php:

    public function getId($email, $type)
{
    if ($type)
        $query = "SELECT * FROM t_head where h_email = '$email';";
    else
        $query = "SELECT * FROM t_partner where p_email = '$email';";
    return $this->query($query);
}

public function newPartner($hId, $fname, $lname, $day, $month, $year, $email)
{
    $query = "INSERT INTO t_partner (p_headid, p_fname, p_lname, p_day, p_month, p_year, p_email)
    VALUES ('$hId', '$fname', '$lname', '$day', '$month', '$year', '$email');";
    return $this->query($query);
}

public function newHead($fname, $lname, $hschool, $day, $month, $year, $email)
{
    $query = "INSERT INTO t_head (h_fname, h_lname, h_school, h_day, h_month, h_year, h_email)
    VALUES('$fname', '$lname', '$hschool', '$day', '$month', '$year', '$email');";
    return $this->query($query);
}

    public function setHeadId($hId, $pId, $num)
    {
        if ($num === 1)
            $query = "UPDATE t_head SET h_p1 = '$pId' WHERE h_id = '$hId';";
        else
            $query = "UPDATE t_head SET h_p2 = '$pId' WHERE h_id = '$hId';";
        $this->execute($query);
    }

有什么问题？
DB是与PDO一起使用的类。（我不知道这是什么，但我总是复制它然后我写的功能，然后它的底部，它总是工作正确。）但是知道这个错误是什么意思？

What's the problem?DB is class that is worked with PDO. (I don't know what's this! but I always copy it then I write the functions at then bottom of it and it always worked correctly.) But know What does this error mean?

对不起，如果我粘贴所有的代码！我不知道 SQLSTATE [HY000]：一般错误与哪个部分相关，因此我无法粘贴任何特定的部分/ func。

Excuse me If I pasted all of the code! I don't know SQLSTATE[HY000]: General error is related to which part, so that I couldn't paste any specific part/func for you.

我的网站已复制到另一个服务器，并且DNS在3小时前更改为一个目的。如果这个细节可以帮助你...

My site copied to another server and the DNS was changed near 3 hours ago for a purpose. If this detail can help you ...

xampplite

为什么是SQLSTATE [HY000]：一般错误？

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