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Code problem
Posted: Thu Aug 04, 2005 5:17 pm
by bla5e
Code: Select all
if (!$name) {echo "You must enter your full name."; } else {
if (!$email) {echo "We need your Email"; } else {
if (!$phone) {echo "We must know your phone number to get back to you."; } else {
if (!$location) {echo "Please fill out the location section."; } else {
if (!$other ) {
if (!$yr) {echo "Please fill out the year of your car."; } else {
if (!$make) {echo "Please fill out the make of your car"; } else {
if (!$model) {echo "Please fill out the model of your car"; } else {
}
}
}
}
if (!$car) {echo "Please include a link to a picture of your car."; } else {
if (!$self) {echo "Please include a link to a picture of your self."; } else {
include ("join/config.php");
mysql_connect($server, $user, $password) or die(mysql_error());
mysql_select_db($database) or die(mysql_error());
$sql = mysql_query("INSERT INTO `join` (id, name, email, phone, location, yr, make, model, car, self) VALUES ('', '".$_POST['name']."','".$_POST['email']."','".$_POST['phone']."','".$_POST['location']."','".$_POST['yr']."','".$_POST['make']."','".$_POST['model']."','".$_POST['car']."','".$_POST['self']."')") or die(mysql_error());
mysql_query($sql) or die(mysql_error());
echo ("Thank you for your request to Join our Team.<br> We will get back to you as soon as possible.<br> ");
$SiteName = "Team Exile";
$SiteEmail = "hondavtecdriver@gmail.com";
$ThankYouMessage = "Thank you from Team Exile";
$SiteUserName = "Team Exile";
$AdminMessage .= "Another Request to Join the Team \n";
mail("$SiteEmail", "Team Exile", $AdminMessage, "From: $email");
}
}
}
}
}
}
The Error:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1
But the thing is, it will still submit it to the database... It also will not email me.
help please? i cant figure out whats wrong
Posted: Thu Aug 04, 2005 5:56 pm
by feyd
echo out the query string to determine where in the query it's failing, and check visually if you see anything wrong.. I'm guessing an input may contain a quote mark.
Posted: Fri Aug 05, 2005 11:01 am
by bla5e
Code: Select all
<?php
case '4b':
if (!$name) {echo "You must enter your full name."; } else {
if (!$email) {echo "We need your Email"; } else {
if (!$phone) {echo "We must know your phone number to get back to you."; } else {
if (!$location) {echo "Please fill out the location section."; } else {
if (!$other ) {
if (!$yr) {echo "Please fill out the year of your car."; } else {
if (!$make) {echo "Please fill out the make of your car"; } else {
if (!$model) {echo "Please fill out the model of your car"; } else {
}
}
}
}
if (!$car) {echo "Please include a link to a picture of your car."; } else {
if (!$self) {echo "Please include a link to a picture of your self."; } else {
echo mysql_error();
include ("join/config.php");
echo mysql_error();
mysql_connect($server, $user, $password) or die(mysql_error());
mysql_select_db($database) or die(mysql_error());
echo mysql_error();
$sql = mysql_query("INSERT INTO `join` (id, name, email, phone, location, yr, make, model, car, self) VALUES ('', '".$_POST['name']."','".$_POST['email']."','".$_POST['phone']."','".$_POST['location']."','".$_POST['yr']."','".$_POST['make']."','".$_POST['model']."','".$_POST['car']."','".$_POST['self']."')") or die(mysql_error());
mysql_query($sql) or die(mysql_error());
echo ("Thank you for your request to Join our Team.<br> We will get back to you as soon as possible.<br> ");
echo mysql_error();
$SiteName = "Team Exile";
$SiteEmail = "hondavtecdriver@gmail.com";
$ThankYouMessage = "Thank you from Team Exile";
$SiteUserName = "Team Exile";
$AdminMessage .= "Another Request to Join the Team \n";
mail("$SiteEmail", "Team Exile", $AdminMessage, "From: $email");
echo mysql_error();
}
}
}
}
}
}
break;
?>
still same error
please help!
Posted: Fri Aug 05, 2005 11:06 am
by s.dot
Usually when I get errors at line 1 I have a single quote in my SQL syntax
something like
Code: Select all
$var = "don't tell me";
$SQL = "INSERT INTO table (col) VALUES('$var')";
This will give me that kind of error.
Posted: Fri Aug 05, 2005 11:24 am
by feyd
bla5e, you're not echoing out the query string...

Posted: Fri Aug 05, 2005 12:09 pm
by bla5e
ok i echod it and this is what i get now
mysql_query(1,Resource id #3) or die(mysql_error())
Thank you for your request to Join our Team.
We will get back to you as soon as possible.
Posted: Fri Aug 05, 2005 12:19 pm
by s.dot
You're echoing out the result, not the query it's self, which is why you get "result resource"
Separate your sql from the query
Code: Select all
$SQL = "INSERT INTO foo(col) VALUES bar('$value')";
echo $SQL;
Posted: Fri Aug 05, 2005 12:27 pm
by shiznatix
you are inserting a blank value into a auto incrementing fiend (your id)
you cant do that, just take out id and take out the
'' that you have for its value other than that i dont see a
problem in your sql.
plus it is a good idea to seperate your sql
Code: Select all
$sql = '
select
*
from
table_name
where
stuff = "'.$stuff.'"
';
$do_query = mysql_query($sql) or die(mysql_error());
Posted: Fri Aug 05, 2005 1:03 pm
by bla5e
i got it to work, but thanks for the help everyone! greatly appriciated