Help With Database Query

Anon

NEWBIE Question

Trying to query a sql database to check and see if an email address exists from a user input. If it does it will not allow the user to send another email.

Here is what I have from some help from another website but still doesn't work. I'm sure it must be my syntax or logic. ANY Help appreciated.

Thanks in advance
Rodney

// CODE
$Link = mysql_connect ($Host, $User, $Password);

$Query = "SELECT Email from $TableName where Email = '$Array[Email]'";
$Result = mysql_db_query ($DBName, $Query, $Link);
print "$Result "; //To see if it returns ANYTHING, only thing it returns is Resource id #2
print "$Array[Email] "; //Just to make sure the value is being returned and it is.

if ($Result) {
print "The email address $Array[Email] has already been used\n";
} else {
DO THIS
}

edwardsbc

Usually when "searching" searching for string matches in SQL you should use LIKE or CONTAINS instead of '='.

What if you try this:

$Query = "SELECT Email FROM $TableName WHERE Email LIKE '$Array[Email]'";

Anon

Thanks Craig but it still won't get me past saying that the email that the user entered already exists and I get the following results:

Resource id #2
r
The email address r has already been used

The page I'm testing it on is
http://www.islanddog.com/form2.php

everything was working until I started trying to check if the email address already exists in the database so if anyone needs the rest of the code let me know.

Thanks

bfranco

what's wrong there is that you are asking for $Result, and it will always hav a value, no matter if no records match your query.

you should use the mysql_num_rows() function:

if (mysql_num_rows($Result)) {
print "The email address $Array[Email] has already been used\n";
} else {
DO THIS
}

Anon

Works beautifully, thank you so much.

Where can I go to get more education on the mysql functions and their use inside of PHP?

Rodney

Anon

See the php manual

http://www.php.net/manual/en/ref.mysql.php

Richie

geocities.com/akindele/

Anon

Much thanks to everyone, I learned alot dealing with that little bit and still have much to learn.

Got another question while I'm here.

Trying to implode (change Array into String)

I know that it is my lack of knowing proper commands and syntax but here's what I'm doing.

I have a second table with a list of names, emails and states. What happens is based on the state the user inputs I do a search of the 2nd table to get ALL the email addresses associated with that state. No Problem here.

// Code
$Query = "SELECT * from $TableName2 where cstate = '$Array[State]'";
$Result = mysql_db_query ($DBName, $Query, $Link);

while ($Row = mysql_fetch_array ($Result)) {
print ("$Row[cemail] ");
}
//END CODE

This works just fine and shows me all the emails that it has retrieved from the database. I now want to email all of these people but can not get the $Row[cemail] to implode. My first guess is that $Row[cemail] is not an Array ... yet, OR I need to know the sperator that naturally comes back in the results

bfranco

I'm not sure what you're asking...

but if you need an array, try this:

while ($Row = mysql_fetch_array ($Result))
{
print ("$Row[cemail] ");
$Emails[]=$Row[cemail];
}

and $Emails[] is an array of email addresses.
If you want a string, try this:

$Row = mysql_fetch_array ($Result);
while ($Row)
{
print ("$Row[cemail] ");
$Email=$Row[cemail];
$Row = mysql_fetch_array ($Result);
$Emails.=$EMail.($Row?", ", "");
}

if you asked any other thing, sorry, didn't get it. 🙂