[RESOLVED] if(!result) test not working like I expected

[dmacman1962]

Hi All,

I am trying to check my database to see if a user has already registered and I am getting a match even though my table is empty?

Here is what I am doing...

$query = 'SELECT * FROM birthdayclub WHERE EmailAddress="'.$EmailAddress.'"';

echo $query . '<br>'; //Shows a correct query with a valid email address

$result = mysql_query($query) or die(mysql_error()); 

echo '<br /><br />'.$result.'<br /><br />'; // gives a Resource id #4

if(!$result){
			//saving record to MySQL database
			$sql ="insert into birthdayclub";
			$sql .="(Date, Time, EmailAddress, Birthday, FirstName, LastName)";
			$sql .="values('$Date', '$Time', '$EmailAddress', '$Birthday', '$FirstName', '$LastName')";
			if(!mysql_query($sql)){
				echo mysql_errno().'<br>'.mysql_error().'<br><br>';
									}
			die('Congrats!');
		} else {
			die('You already signed up for our Free Gift Offer. Please hit the "Back" button to return');
		} 

I did an

echo $result; 

and I get a "Resource id #4" . There was some records in the table, but I deleted them to try and test my code, hence the results.

Thanks in advance,

Don

[thorpe]

echo $query . '<br>'; //Shows a correct query with a valid email address

If you supply a valid email address to your query then surely a result resource is what you expect to be returned?

[devinemke]

$result will return TRUE on a successful query even if that query returns 0 rows. use [man]mysql_num_rows[/man] to see how many rows were returned and do your INSERT query based on that.

[dmacman1962]

Sorry Thorpe, I did not word that correctly.

What I meant by "it shows a valid email address" was that it shows the address I entered into my form, not an email address that exists in the table.

I used similar code for another email signup page...

$query = 'SELECT * FROM eletter WHERE EmailAddress="'.$EmailAddress.'"';
$result = mysql_query($query) or die(mysql_error()); 

if(!$result){
$query_insert = "insert into eletter EmailAddress, Date, Time, Confirmed values('$EmailAddress','$Date','$Time', '$Confirmed')";
$insert = mysql_query($query_insert, $conn) or die(mysql_error());
$var = "Thank you for signing up<br>for our e-Newsletter.";
}
else {
    $query_update = "update eletter set EmailAddress='$EmailAddress', Date='$Date' , Time='$Time', Confirmed='$Confirmed' WHERE `EmailAddress` = '$EmailAddress'";
    $update = mysql_query($query_update,$conn) or die(mysql_error());
$var = "Thank you for updating<br>your email address.";
} 

And that worked as I planned (inserting the email address if it did not exist in the database, or , update the email address if it did exist).

So I just modified it to my first post, and I expected it to check the db for the email address and ...

A) insert the address if it did not exist and thank you.

or

😎 check and see the address already exists and tell you that you already signed up.

What I am getting is it checks the db and does not find the address and then tells you you already signed up.

Thats my confusion.

Does that make more sense?

Thanks again,
Don

[dmacman1962]

Thanks Devinemke,

I changed my code to this...

$query = 'SELECT * FROM birthdayclub WHERE EmailAddress="'.$EmailAddress.'"';
$result = mysql_query($query) or die(mysql_error());
$num_rows = mysql_num_rows($result);
if($num_rows<1){
//insert data
} else {
//tell them they already signed up
}

and it works great!

Thank you very much!

Don