Trying to check if an email with the same value already exists

Question

Im trying to make so it checks if the email already exists in the database before inserting the data in it.

Here is my code:

<?php
    $servername = "servername";
    $username = "username";
    $password = "password";
    $dbname = "dbname";

    $Mail = $_POST['email'];

    // Create connection
    $conn = new mysqli($servername, $username, $password, $dbname);

    // Check connection
    if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
    }

    //Inputed in E-Mail Field
    $Mail = $_POST['email'];

    $SearchEmail = $conn->query("SELECT (Mail) FROM betakey WHERE Mail = '$Mail'");

    if ($SearchEmail->num_rows > 0) {
        print "That Email is already registered for the closed alpha";
    }
    else {
        $conn->query("INSERT INTO betakey VALUES ('$Mail')");
    }

    $conn->close();
    ?>

It doesn't give me any error when i access the page but it also doesn't echo that it exists or inserts the data when it doesn't.

Can you try this: "SELECT count(Mail) FROM betakey WHERE Mail = '$Mail'" — ahmet, Aug 27 '15 at 18:23
you might want to add an `echo 'mail sent';` statement before the last brace in your code to help you with debugging. — Mike -- No longer here, Aug 27 '15 at 18:27
enabling error_log to a single file and post it, you may have a sql or php syntax error. — Quijote Shin, Aug 27 '15 at 18:33
Pat4561 is right. I thought that the problem is detecting whether the mail exists or not. — ahmet, Aug 27 '15 at 18:42
Does this answer your question? [PHP - Check if row exists before inserting](https://stackoverflow.com/questions/31225765/php-check-if-row-exists-before-inserting) — Dharman, Dec 03 '20 at 19:37

score 0 · Answer 1 · answered Aug 27 '15 at 18:40

0

The problem is in your insert query. An insert query is always structured like shown below.

INSERT INTO table_name (column1, column2, column3,...) VALUES (value1, value2, value3,...) See: http://www.w3schools.com/php/php_mysql_insert.asp

Applying this to your code, results in the following.

<?php
$servername = "servername";
$username = "username";
$password = "password";
$dbname = "dbname";

$Mail = $_POST['email'];

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);

// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}

//Inputed in E-Mail Field
$Mail = $_POST['email'];

$SearchEmail = $conn->query("SELECT count(Mail) FROM betakey WHERE Mail = '$Mail'");

if ($SearchEmail->num_rows > 0) {
    print "That Email is already registered for the closed alpha";
}
else {
    $conn->query("INSERT INTO betakey (Mail) VALUES ('$Mail')");
}

$conn->close();
?>

And I suggest you take away one of the two '$Mail = $_POST['email'];', because it is useless to declare a variable twice in the same way.

answered Aug 27 '15 at 18:40

P.Yntema

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2
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I don't think thats the answer for it since i tried to register an email that already existed and it didn't give me the "print "(...)" statement. – KillZoneZ Aug 27 '15 at 19:37
What is the output of $SearchEmail when you try to print it? – ahmet Aug 27 '15 at 19:45
ahmet, this happens Catchable fatal error: Object of class mysqli_result could not be converted to string in /home/a9508844/public_html/BetaRegistration.php on line 31 Thats probably why it isnt working but how do i fix it – KillZoneZ Aug 27 '15 at 19:52
Sorry, I'm not good at php but maybe you can try to print as in the answer of @mailo ( http://stackoverflow.com/questions/5157905/mysql-query-result-in-php-variable ) – ahmet Aug 27 '15 at 20:09
@KillZoneZ, no, the fatal error happens because you try to print an array. If you want to see the output you should use the print_r($array) funtion. Could you provide us with the full code? At my side it is working when I try to register an email that already exists, so it should work at your side too. – P.Yntema Aug 27 '15 at 21:28

Trying to check if an email with the same value already exists

1 Answers1