如何检查邮件是否存在于mySQL的表列中


How to check if the email exists in Table Column in mySQL PHP

如何检查邮件是否存在于MySQL…?



 public function AddReseller($name,$businessName,$businessPageName,$contactNumber,$fullAddress,$email)
    {
        $db = new Mysqlidb();
        if(!$db) die("Database error");
        $data = Array(
            'Name' => $name,
            'BusinessName' => $businessName,
            'Address' => $fullAddress,
            'ContactNumber'=>$contactNumber,
            'Email'=>$email,
            'PageName'=>$businessPageName
        );
        if($email > 0)
         {
            echo "Email Exists Already";
         }
        $id = $db->insert ('tenterprise', $data);
        return $id;
    }
 

这是代码和$db->是插入查询表…!帮帮我……!

哇,你是相当糟糕的问问题:)你正在检查$email参数,你发送作为参数的函数,但你想检查如果该电子邮件已经在数据库中,你试图将此数据数组添加到表之前。在这种情况下,很明显,您需要首先在表上使用已发送参数$email执行select查询,如下所示:

 $checkquery =   "select email from tenterprise where email = ".$email;

那么您应该检查此查询是否有任何结果。如果result为空,则表示该电子邮件以前未被使用过。

当你问问题的时候,试着更具体地说明你想要达到的目标,这样别人就可以专注于正确的事情来帮助你!

代码可能是这样的:为了获得这项工作,您必须连接到数据库。

<?php 
$result =mysql_query("SELECT email FROM table WHERE `email` = 'test@sample.com'");
if ($result && mysql_num_rows($result) > 0){
  echo 'Email Found'; 
}else{
  echo 'Email not Found'; 
}
?>

我看到你正在使用mysqlidb。对于mysqlidb,你可以使用where ()->get()来避免写sql.

public function AddReseller($name,$businessName,$businessPageName,$contactNumber,$fullAddress,$email)
        {
            $db = new Mysqlidb();
            if(!$db) die("Database error");
            $data = Array(
                'Name' => $name,
                'BusinessName' => $businessName,
                'Address' => $fullAddress,
                'ContactNumber'=>$contactNumber,
                'Email'=>$email,
                'PageName'=>$businessPageName
            );
            if($user = $db->where("email", $email)->get ('tenterprise')) > 0)
             {
                echo "Email Exists Already";
                return $user['id'];
             }
            $id = $db->insert ('tenterprise', $data);
            return $id;
        }