Jump to content

RobPuckett

New Members
  • Posts

    1
  • Joined

  • Last visited

RobPuckett's Achievements

Newbie

Newbie (1/5)

0

Reputation

  1. $sql = "SELECT * FROM leads WHERE accesstoken = '".$_POST["userAc"]."'"; $result = mysql_query($sql) or trigger_error("Query Failed: " . mysql_error()); if(mysql_num_rows($result)>0) { while($row = mysql_fetch_array($result)) { $id= $row['id']; $fullname= $row['fullname']; $email= $row['email']; $to = $email; $message .= $_POST["userMessage"]; $subject = $_POST["userSubject"]; $headers = "From: " . $_POST["userEmail"] . "\r\n"; $headers .= "Reply-To: ". $_POST["userEmail"] . "\r\n"; $headers .= "MIME-Version: 1.0\r\n"; $headers .= "Content-Type: text/html; charset=ISO-8859-1\r\n"; if (mail($to, $subject, $message, $headers)) { $output = json_encode(array('type'=>'error', 'text' => '<p>Your Message was sent successfully!</p>')); die($output); } else { $output = json_encode(array('type'=>'error', 'text' => '<p>Your Message was not sent!</p>')); die($output); } } } I have another table called accounts that I want to join with this one, I need to select from ref to get the email from account table also! be something like this $sql = "SELECT * FROM accounts WHERE ref = '".$_POST["userAc"]."'"; $result = mysql_query($sql) or trigger_error("Query Failed: " . mysql_error());
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.