Checking if email is already taken

Question

In my PHP file I want to check if the email is already taken before inserting the data into my mysql database.

I cant find anything wrong with my code but it wont work.

PHP code:

<?php


$conn = mysqli_connect("s.amazonaws.com", "username", "pass", "SellerAccounts");

//If statement echos back to the browser if we connected to the server or not.

    //Store the data from the POST (text from the user) into a variable.
    $Sellers_CompanyName = $_POST["SellersCompanyName"];
    $Sellers_CompanyWebsite = $_POST["SellersCompanyWebsite"];
    $Sellers_IndustryName = $_POST["SellersIndustryName"];
    $Sellers_SecondaryIndustryName = $_POST["SellersSecondaryIndustryName"];
    $Sellers_FirstContactFirstName = $_POST["SellersFirstContactFirstName"];
    $Sellers_FirstContactLastName = $_POST["SellersFirstContactLastName"];
    $Sellers_FirstContactNumber = $_POST["SellersFirstContactNumber"];
    $Sellers_FirstContactEmail = $_POST["SellersFirstContactEmail"];
    $Sellers_SecondContactFirstName =    $_POST["SellersSecondContactFirstName"];
    $Sellers_SecondContactLastName = $_POST["SellersSecondContactLastName"];
    $Sellers_SecondContactNumber = $_POST["SellersSecondContactNumber"];
    $Sellers_SecondContactEmail = $_POST["SellersSecondContactEmail"];
    $Sellers_Password = $_POST["Password"];


    $result = mysql_query("SELECT * FROM user_info WHERE     Sellers_FirstContactEmail = '".$Sellers_FirstContactEmail."'");

            if ( mysql_num_rows($result) > 0 ){

     echo("Email is already in use");

            }else{
 //
    $statement = mysqli_prepare($conn, "INSERT INTO user_info                (Sellers_CompanyName, Sellers_CompanyWebsite, Sellers_IndustryName,     Sellers_SecondaryIndustryName, Sellers_FirstContactFirstName, Sellers_FirstContactLastName, Sellers_FirstContactNumber, Sellers_FirstContactEmail, Sellers_SecondContactFirstName, Sellers_SecondContactLastName, Sellers_SecondContactNumber, Sellers_SecondContactEmail, Sellers_Password) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?) ");

mysqli_stmt_bind_param($statement, "sssssssssssss", $Sellers_CompanyName,    $Sellers_CompanyWebsite, $Sellers_IndustryName, $Sellers_SecondaryIndustryName, $Sellers_FirstContactFirstName, $Sellers_FirstContactLastName, $Sellers_FirstContactNumber, $Sellers_FirstContactEmail, $Sellers_SecondContactFirstName, $Sellers_SecondContactLastName, $Sellers_SecondContactNumber, $Sellers_SecondContactEmail, $Sellers_Password);
    mysqli_stmt_execute($statement);
    mysqli_stmt_close($statement);

    if(mysqli_query($connection, $sql_query)){
       // echo" Data insertion success...";
     }
            }

mysqli_close($conn);

?>

i can insert everything fine and connect to the DB

What's your table structure? Did you select the correct column? There's no problem with your SQL query statement, if you selected the correct column. — Panda, Mar 06 '16 at 02:31
The error is not in the code above. Maybe in the connection. Or the column name is not `Sellers_FirstContactEmail`. You have to provide more code. — fusion3k, Mar 06 '16 at 02:33
yea in my table user_info i have a column called Sellers_FirstContactEmail — Lazar, Mar 06 '16 at 02:34
Try echoing out `$Sellers_FirstContactEmail` to see if it holds anything — Panda, Mar 06 '16 at 02:35
We need more info or more code to help you. Also double-check that `$Sellers_FirstContactEmail` is not empty. Also check if mysql query returns errors... — fusion3k, Mar 06 '16 at 02:37

Panda · Accepted Answer · 2016-03-06T02:48:08.630

You cannot mix MySQL functions with MySQLi functions. And also your connection is $conn, so using $connection will not work.

<?php

$conn = mysqli_connect("s.amazonaws.com", "username", "pass", "SellerAccounts");

$Sellers_CompanyName = $_POST["SellersCompanyName"];
$Sellers_CompanyWebsite = $_POST["SellersCompanyWebsite"];
$Sellers_IndustryName = $_POST["SellersIndustryName"];
$Sellers_SecondaryIndustryName = $_POST["SellersSecondaryIndustryName"];
$Sellers_FirstContactFirstName = $_POST["SellersFirstContactFirstName"];
$Sellers_FirstContactLastName = $_POST["SellersFirstContactLastName"];
$Sellers_FirstContactNumber = $_POST["SellersFirstContactNumber"];
$Sellers_FirstContactEmail = $_POST["SellersFirstContactEmail"];
$Sellers_SecondContactFirstName =    $_POST["SellersSecondContactFirstName"];
$Sellers_SecondContactLastName = $_POST["SellersSecondContactLastName"];
$Sellers_SecondContactNumber = $_POST["SellersSecondContactNumber"];
$Sellers_SecondContactEmail = $_POST["SellersSecondContactEmail"];
$Sellers_Password = $_POST["Password"];


$result = mysqli_query($conn, "SELECT * FROM user_info WHERE     Sellers_FirstContactEmail = '$Sellers_FirstContactEmail'");

if ( mysqli_num_rows($result) > 0 ){

    echo("Email is already in use");

} else {

$statement = mysqli_prepare($conn, "INSERT INTO user_info                (Sellers_CompanyName, Sellers_CompanyWebsite, Sellers_IndustryName,     Sellers_SecondaryIndustryName, Sellers_FirstContactFirstName, Sellers_FirstContactLastName, Sellers_FirstContactNumber, Sellers_FirstContactEmail, Sellers_SecondContactFirstName, Sellers_SecondContactLastName, Sellers_SecondContactNumber, Sellers_SecondContactEmail, Sellers_Password) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?) ");

mysqli_stmt_bind_param($statement, "sssssssssssss", $Sellers_CompanyName,    $Sellers_CompanyWebsite, $Sellers_IndustryName, $Sellers_SecondaryIndustryName, $Sellers_FirstContactFirstName, $Sellers_FirstContactLastName, $Sellers_FirstContactNumber, $Sellers_FirstContactEmail, $Sellers_SecondContactFirstName, $Sellers_SecondContactLastName, $Sellers_SecondContactNumber, $Sellers_SecondContactEmail, $Sellers_Password);

mysqli_stmt_execute($statement);

mysqli_stmt_close($statement);

    if(mysqli_query($conn, $sql_query)){
        // echo" Data insertion success...";
    }
}

mysqli_close($conn);

?>

score -1 · Answer 2 · answered Mar 06 '16 at 02:28

-1

how about you replace your query with the following?:

$result = mysql_query("SELECT * FROM user_info WHERE Sellers_FirstContactEmail = '".$Sellers_FirstContactEmail."'");

Note the use of string concatenation in the above. In your case, your query is trying to find out emails that match the string "$Sellers_FirstContactEmail", which is not intended - you want to check for the value of that variable.

answered Mar 06 '16 at 02:28

Ashraf Iqbal

412
3
11

Actually, your solution is **identical** to the OP original query. The OP query **don't** find out emails that match the string "$Sellers_FirstContactEmail". – fusion3k Mar 06 '16 at 02:31
not really identical, see the use of dots (.) in my solution which is absent in the original query. @Lazar, do you have a php variable called $Sellers_FirstContactEmail and that contains the correct email address you are querying for? – Ashraf Iqbal Mar 06 '16 at 02:37
1

@AshrafIqbal It's the same with the use of dots and without – Panda Mar 06 '16 at 02:39

Checking if email is already taken

2 Answers2