Re: [PHP] using mysql in php problem

[raditha dissanayake]

Garth Hapgood - Strickland wrote:
I have created a registration page
http://www26a.your-server.co.za/matchm/registration4.php
On this page I have many fields, some of them pulling data out from my
database. I have a javascript validation scipt thats checks on submission of
the form to see if all required fields are filled in correctly.
IF thats the case it should submit the form. Which it isnt doing! It should
start with submission after the validation check is passed.
 

This part of the questions should be posted on a js list.
Then I check to see whether the BusinessName & BusinessOwner already exist,
if not then I am trying to to an insert (which it doesnt do either). Here is
the php code:
$errmsg = "";
$business_query = "SELECT BusinessName, BusinessOwner from business WHERE
BusinessName = '$BusinessName' AND BusinessOwner = '$BusinessOwner'" ;
$result = mysql_query($business_query);
if (mysql_num_rows($result) > 0) {$errmsg .= "You are already registered at
Match Makers";}
 

You will do yourself a great service by punctuating indenting your code 
properly. If you read the code of some open source php scripts you will 
find that they are so much easier to read.

else {
if (isset($submit) == "Register Now"){mysql_query("INSERT INTO business
(BusinessName, BusinessOwner, StreetAdress, Suburb, CityTown, ProvinceID,
StreetCode, PostalAddress, PostalCode, WebsiteAddress, IndustryID,
IndustryOther, ProductService, BusinessAge, AnnualTurnoverID,
EmpowermentProfileID, BEEScoreCardID, PreferredLocationID,
PreferredProductService, PreferredMatchPartner, PreferredIndustryID,
PreferredIndustryOther, PreferredAgeID, PreferredTurnoverID,
PreferredEmpowermentProfileID, DateRegistered, MarketSourceID)
VALUES
('" . $HTTP_POST_VARS['business_name'] . "', '" .
$HTTP_POST_VARS['street_address'] . "', '" . $HTTP_POST_VARS['suburb'] . "',
$city_town, $province_desc, $street_code, $postal_address, $postal_code,
$website_address, $industry_desc, $industry_other, $product_service,
$business_age, '', '', '', '', '', '', '', '', '', '',)");
 

See what i mean :-)
$id = mysql_insert_id();
 

..snipped ..
I am not sure where exactly I am going so wrong, and it really is starting
to drive me crazy!
 

The best thing to do is to liberally douse your code with 
error_log(mysql_error()) statement you will then start seeing error 
messages in your log (or your browser depending on your config) about 
what's going wrong.

As has often been pointed out echoing out the queries produced by your 
php script and running them through the mysql client usually helps too.

Regards
Garth
 


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Raditha Dissanayake.
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