Creating Tables with Variable Names

[Studio_Junkies]

Its my first project. I sell a book that teaches musicians how to set up a home recording studio on Trademe.co.nz.

I want to develop this site so it will strip the buyers details from the email that Trademe sends when a book is sold, then automatically email the buyer with a link to my website so they can enter their address and the reference number they used for their payment. The buyer will be able to see previous orders and print invoice/receipts.

I'm stuck on creating tables. I get a "Parse error: parse error in C:\apachefriends\xampp\htdocs\Business\process_sale.php on line 39"

This is my code:

---

//Connection to DB
include 'connectdb.php';

//Variables from form while developing code
$trademeid =$HTTP_GET_VARS['trademeid'];
$email =$HTTP_GET_VARS['email'];
$itemname=$HTTP_GET_VARS['item'];
$auctionnumber =$HTTP_GET_VARS['auctionnumber'];
$saledate =$HTTP_GET_VARS['saledate'];
$price =$HTTP_GET_VARS['price'];
$shippingcost =$HTTP_GET_VARS['shippingcost'];
$status = 1;

$connect; $selectdb;

// I only want to make a new table for a new customer!!
function tableexists ($tablename, $db){
$connect; $selectdb;
$result = mysql_list_tables ($db);
$rcount = mysql_num_rows ($result);
for ($i= 0; $i < $rcount; $i++){
if (mysql_tablename($result, $i)==$tablename) return true;
}
return false;
};

if (tableexists( $trademeid, "business") == false) {
mysql_query 'CREATE TABLE $trademeid (
transactionnumber tinyint(1) NOT NULL auto_increment PRIMARY KEY,
itemname varchar(25) NOT NULL,
auctionnumber int (10) unsigned NOT NULL,
saledate int (8) NOT NULL,
price int (8) NOT NULL,
shippingcost int (6) NOT NULL,
status tinyint (1) NOT NULL
)' // This is line 39 - the Parse error
};

---

I've tried all sorts, but it seems to be getting stuck here. Have I missed something?

Thanks.

Anthony

[halojoy]

//Connection to DB 
include 'connectdb.php'; 

//Variables from form while developing code 
$trademeid =$HTTP_GET_VARS['trademeid']; 
$email =$HTTP_GET_VARS['email']; 
$itemname=$HTTP_GET_VARS['item']; 
$auctionnumber =$HTTP_GET_VARS['auctionnumber']; 
$saledate =$HTTP_GET_VARS['saledate']; 
$price =$HTTP_GET_VARS['price']; 
$shippingcost =$HTTP_GET_VARS['shippingcost']; 
$status = 1; 


$connect; $selectdb; 

// I only want to make a new table for a new customer!! 
function tableexists ($tablename, $db){ 
$connect; $selectdb; 
$result = mysql_list_tables ($db); 
$rcount = mysql_num_rows ($result); 
for ($i= 0; $i < $rcount; $i++){ 
if (mysql_tablename($result, $i)==$tablename) return true; 
} 
return false; 
}; 

if (tableexists( $trademeid, "business") == false) { 
mysql_query 'CREATE TABLE $trademeid ( 
transactionnumber tinyint(1) NOT NULL auto_increment PRIMARY KEY, 
itemname varchar(25) NOT NULL, 
auctionnumber int (10) unsigned NOT NULL, 
saledate int (8) NOT NULL, 
price int (8) NOT NULL, 
shippingcost int (6) NOT NULL, 
status tinyint (1) NOT NULL 
)' // This is line 39 - the Parse error 
};

The first thing that look strange here is

$connect; $selectdb;

doesnt do much
does it?

halojoy
😉

[planetsim]

Please use [php][/php] tags when posting code.

Errors I can see from a quick look

mysql_query 'CREATE TABLE $trademeid (

I thought mysql_query was a function try this

mysql_query ('CREATE TABLE $trademeid (

Secondly your table you create is going to be called $trademeid everytime you try as your using single quote use double quotes.

mysql_query("CREATE TABLE $trademeid (
transactionnumber tinyint(1) NOT NULL auto_increment PRIMARY KEY,
itemname varchar(25) NOT NULL,
auctionnumber int (10) unsigned NOT NULL,
saledate int (8) NOT NULL,
price int (8) NOT NULL,
shippingcost int (6) NOT NULL,
status tinyint (1) NOT NULL
)");

Try that now, hopefully should work.

[Studio_Junkies]

Hey thanks for the help!

This PHP stuff is niggly, but fun . . . Spend hours on a problem, then finally it snaps in and you get this great feeling of accomplishment !!

OK, lets see if these PHP tags work:

<?php

function checklogin ($trademeid, $cookiecontents) {
if ($cookiecontents == '1'){
return (1);
} else {
echo "$trademeid, You need to log in to access this page. <A HREF=\"index.php\">Click Here.</A>";
exit;
mysql_close();
};};
?>

This is not a problem, just a bit of code to see if the colors and formatting work like DW.

Anthony

[rachel2004]

As far as the "line 39" error, you had some typos. I cleaned them up for you. See if this works for you. I didn't check the rest of your code, so it's possible other errors exist.

if (tableexists( $trademeid, "business") == false) {
        mysql_query('CREATE TABLE '.$trademeid.' (
                        transactionnumber tinyint(1) NOT NULL auto_increment,
                        PRIMARY KEY (transactionnumber),
                        itemname varchar(25) NOT NULL,
                        auctionnumber int (10) unsigned NOT NULL,
                        saledate int (8) NOT NULL,
                        price int (8) NOT NULL,
                        shippingcost int (6) NOT NULL,
                        status tinyint (1) NOT NULL
                     )');
}

[Studio_Junkies]

WOW! -- I have come so far in the last few days!!!

Spot the errors in my first code? - Nah! Just throw it away!!

Thanks for all your help there, I ended up just checking for the $trademeid in the details table to see if the customer already exists.

Heres an update on the code I posted above, re: check for session id. This is just to make sure the user has logged in on this session. One of the big problems I saw when I wrote this code, is they can manually enter '1' into their cookie!

What I did: create 'sessionid' column in the details table. When the user successfully logs in, they get 2 cookies:

 
// Pick a lucky number between 100 & 999 and assign it for this session. 
// These cookies expire when the browser is closed.
$loggedin = rand(100,999);
setcookie ("trademeid","$trademeid");
setcookie("loggedin","$loggedin");
// Put the lucky number in the database to compare to the cookie
$sessionid="UPDATE details SET sessionid='$loggedin' WHERE trademeid='$trademeid'";
mysql_query($sessionid);

Then I call the checklogin() function at the top of every page AFTER connecting to the DB:

 
$trademeid = $HTTP_COOKIE_VARS['trademeid'];
$loggedin = $HTTP_COOKIE_VARS['loggedin'];
include 'connectdb.php';
connect(); selectdb();
include 'loggedincheck.php';
checklogin($trademeid,$loggedin);

Heres the function definition in loggedincheck.php:

<?php

function checklogin ($trademeid, $cookiecontents) {
$query = "SELECT sessionid FROM details WHERE trademeid = '$trademeid'";
$results = mysql_query($query);
list ($sessionid) = mysql_fetch_array($results);

if (($cookiecontents) == ($sessionid)) {
return (1);
} else {
echo "$trademeid, You need to log in to access this page. <A HREF=\"index.php\">Click Here.</A>";
mysql_close();
exit;
};};
?>

This forces the user to login via index.php and get their cookies every time they visit.

Just thought I'd leave it here in case someone else finds it useful.

Thanks for all your help everyone!! - I'll be back! 😉

Anthony