[SOLVED] verifying data against one table and writing data to a new table

[lookee]

How do I verify data obtained via an HTML form with one db table, and if everything matches, write data to a new table?

 

Currently, I'm able to either validate form data in one table OR write data to ANOTHER TABLE, but when I merge all the code, I get a

 

Warning: mysql_query() [function.mysql-query]: Access denied for user 'earlier'@'localhost' (using password: NO) in /home/earlier/public_html/now/request.php on line 93

 

Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in /home/earlier/public_html/now/request.php on line 93

Could not query the database:

Access denied for user 'earlier'@'localhost' (using password: NO)

 

My code:

------------

 

<?php

session_start();

 

 

// Assign the Query

$sql="SELECT * FROM shoe WHERE username='$username'";

 

//Execute the Query

$result=mysql_query($sql);        <-------------LINE 93 in Error Message

 

if (!$result){

die ("Could not query the database: <br />". mysql_error());

 

 

 

//Saves Type Request to a MYSQL Database

//Insert a row of information into the table "shoepay" /

 

mysql_query("INSERT INTO shoepay

 

VALUES('request_number','$username','$first_name','$last_name','$email','$user_id','$shoe_size','$paid_shoes','date') ")

or die(mysql_error());

 

Header("Location: successful_payment.php");

 

}

 

mysql_close();

 

include 'close_db.php';

?>

 

----------------------------

 

It seems I'm opening one query but not finishing it then starting another, and I'm mixing up my "mysql_query" so that login information is getting crossed.  Or something.

 

Any help would be greatly appreciated.

 

[waterssaz]

Hi,

where have you set your database connection properties?. You have not passed any as the second arguement to the mysql_query(). So by default it will look for the last connection properties set in a mysql_connect(), but if it can't find any or you didn't set them anywhere then it will usef default empty parameters. In other words it will try to connect to a database using an invalid login and password. This looks like the error that you are experiencing.

Hope this makes sense and helps :-)

[lookee]

I have them in an include.

 

Both tables are within the same database, so I figured the same connection information should work. 

[fenway]

Well, it's not.

[waterssaz]

I have them in an include.

 

Both tables are within the same database, so I figured the same connection information should work.

 

Is the code above an exact copy of what you are trying to use, because forgive my ignorance but I can't see where your include file is that would make the above code work.