verify if email is in db

capte

I'm trying to build a code, but i've got no ideas how to achieve such a result ...

i wanna use a <form> text box to verify if email is already in my db. And display a message according to the result
like "sorry your email is already in our db" or "please register using this link"

I'm tried to find some info about that over the web ... but no luck

I'm trying to build a php file that connects to a mysql db

any ideas ?

😕

capte

Almost got it ... but i just wanna check the email i don't need the name ... but i can't seems to be able to remove it ... geez i know it's simple ... but i dont get it ! .... please help ... any ideas ?

It also write about 20 times "nope" when he doesn't find anything ! any idea why ?

🙁

if ($submit = "Login")
{

mysql_connect ('localhost', '', '*') or

die ("Could not connect to database");

mysql_select_db("***") or
dir ("Could not select database");

$result=mysql_query("select * from nlAbonnes where email='$email'")
or die ("cant do it");

while ($row=mysql_fetch_array($result)) {
if ($row["name"]==$name )
{
printf("Successfully Logged In!");
}else{
print("Nope");

}
}
}
?>

leatherback

If you do not need the name (It is a very skiummy login though..

 	Almost got it ... but i just wanna check the email i don't need the name ... but i can't seems to be able to remove it ... geez i know it's simple ... but i dont get it ! .... please help ... any ideas ?



<?

if ($submit = "Login")
{

mysql_connect ('localhost', '***', '****') or
die ("Could not connect to database");

mysql_select_db("***") or
dir ("Could not select database");

$result=mysql_query("select * from nlAbonnes where email='$email'")
or die ("cant do it");
if(mysql_num_rows($result) >= 1)
  {
  printf("Successfully Logged In!");
  }
}

capte

ok here,s the problem i'm having ...

The echo in the top of my code is the login box ... but i'd like it to show first and if the email is entered then it shows the text "Successfully Logged In!" or "Nope".

The other thing is that when i open this code in my browser it automatically says "Succesfully Logged In!" even if nothing as been entered.

I'm pretty sure it can work with some IF or elseIF ... but i really don't know how to structure my code ...

Any help would be appreciated ... thanks !

echo "

<form method=post action=''>
Email :<b></b> <br>
<input type=text size=15 name='email'>
<br>
<input type=submit name=submit value='Login'>
<input type=reset name=reset value='Clear'>
<br>

</form> ";

if ($submit = "Login")
{

mysql_connect ('localhost', '', '') or
die ("Could not connect to database");

mysql_select_db("***") or
dir ("Could not select database");

$result=mysql_query("select * from nlAbonnes where email='$email'")
or die ("cant do it");
if(mysql_num_rows($result) >= 1)
{
printf("Successfully Logged In!");
}else{
print("Nope");

}
}

😕

mg542

$email = "";
if (!isset($POST['email'])){
?>
<form method=post action=<? '$SERVER['PHP_SELF']' ?>>
Email :<b></b> <br>
<input type=text size=15 name='email'>
<br>
<input type=submit name=submit value='Login'>
<input type=reset name=reset value='Clear'>
<br>
</form>
<?
}
$email = $_POST['email'];
else if (isset($email)){
$host = "localhost";
$user = "**";
$pass = "";
$db = "***";

$connection = mysql_connect($host, $user, $pass) or die ("Unable to connect!");


mysql_select_db($db) or die ("Unable to select database!");

$sql_email_check = mysql_query("SELECT email FROM ***** WHERE email='$email'");

$email_check = mysql_num_rows($sql_email_check);

 if($email_check >  0){

...random code....

}
else if ($email_check){
echo ' nope';
}
}
?>

or you could do the twoon separate pages by replacing $_SERVER with ***.php.
i hope i didn't make to many spelling errors