How can I check if email exists in db and if so, give error?

[kymberm]

I'm writing a survey, trying to enter data from a form into my database. I want to set it up so people can only enter the survey once and when they submit it i'll email a thank you. My code isnt' working, I can enter data into the database, but I added a conditional to check for if the email exists and if so, tell user bye:

// conect to my mysql database
$con = mysql_connect("sql02.mysite.com", "myuser", "mypass")or die("Connect Error: ".mysql_error());
$db = ("grp");
mysql_select_db($db, $con) or die (mysql_error());
//check for previous entry keyed to email
$chkEmail = mysql_query("SELECT * from survey where Email = '.$Email.'", $con) or die ("cannot get Database open");
if ($chkEmail)
{
printf ("<center><h2>You have already done this survey");
printf ("<center><h2>thanks!");
exit;
}
// insert data into the database grp
$myquery = "INSERT INTO survey (Email, Q1, Q2) VALUES
('".$Email."', '".$Q1."','".$Q2."')";

$result = mysql_query($myquery);
if ($result)
{
//Once the data is entered, redirect the user to give them visual confirmation
header("location: thanks.php");
exit;
}

[LordShryku]

1) Use the php tags. So much easier to read the code
2) Missing double quotes here
= '.$Email.'"
3) You need to fetch or count the row to see if it exists. It will always query if you're connected. See below

$chkEmail = mysql_query("SELECT * 
                         FROM survey 
                         WHERE Email = '".$Email."'");

$rows = mysql_num_rows($chkEmail);
if($rows > 0) {
    printf ("<center><h2>You have already done this survey");
    printf ("<center><h2>thanks!");
    exit;
}
else {
    $myquery = "INSERT INTO survey (Email, Q1, Q2) VALUES ('".$Email."', '".$Q1."','".$Q2."')";

$result = mysql_query($myquery);
if($result) {
   header("location: thanks.php");
   exit;
}
else {
   echo "Cannot insert data";
   // good to have an else on here to present an error
}
}

[drawmack]

or make that a unique field and then let the dbms handle it for you.

[kymberm]

Thanks so much, I"m new to PHP, working on getting better, this worked@!!