t; element names refer directly to the variables after SELECT in your SELECT
> query.
> I did the same in the Phone line.
>
> -Original Message-
> From: Jason Gerfen [mailto:[EMAIL PROTECTED]]
> Sent: Tuesday, October 23, 2001 2:23 PM
> To: [EMAIL PROTECTED]
> Subject
Let me guess, the parse error is on the last line of the file?
I get that all the time, its usually caused by a loop thats still open...
and, in your case, your "do" and "if" statements arent closed
try this.
mysql_connect ("db_hostname", "db_username", "db_password");
mysql_select_db ("
MAIL PROTECTED]]
Sent: Tuesday, October 23, 2001 2:23 PM
To: [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Re: PHP and MySQL queries...
Ok here is what I have done after your suggestions...
if ($date == "")
{$date = '%';}
$result = mysql_query ("SELECT
fname,lname,address,city,s
Ok here is the script after changes...
Name: ".$row['fname, lname']."\n";
if (mysql_num_rows($results) )
{
while( $row = mysql_fetch_array($result) )
{
}
}
else {
print "No rows found";
}
?>
And now I get a parse error on line 201 which is below the html and php
code, so there isn't anything to
First o all, GET RID of the 19 constructs using mysql_result(). You've
already captured the row with the $row=mysql_fetch_array().
mysql_result() cannot be mixed with mysql_fetch_arry().
So instead of:
echo "Address: ";
printf(mysql_result($result,$count,"address"));
echo "\n";
use:
print "Addr
