PHP - Php Linking Mysql Tables

Hi I got three tables (employers , company ,  and  Jobs) Employer table holds info about employer. Company hold info about the company and jobs table holds info about jobs. I was just wandering what would be a good way to link these tables in the database ?  Here is the coding for each table.


Employer Table

Code: [Select]


CREATE TABLE IF NOT EXISTS `employers` (
  `id` int(11) NOT NULL,
  `username` varchar(50) NOT NULL,
  `password` varchar(50) NOT NULL,
  `email` varchar(50) NOT NULL,
  PRIMARY KEY  (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;



Jobs Table

Code: [Select]

CREATE TABLE IF NOT EXISTS `Jobs` (
  `id` int(11) NOT NULL,
  `JobTitle` varchar(200) default NULL,
  `Company` varchar(200) default NULL,
  `Salary` varchar(30) default NULL,
  `Description` varchar(2000) default NULL,
  `CompanyURL` varchar(200) default NULL,
  `PhoneNumber` varchar(30) default NULL,
  `Requirements` varchar(2000) default NULL,
  `JobCategory` varchar(100) default NULL,
  `JobType` varchar(100) default NULL,
  `Apply_To` varchar(1000) NOT NULL,
  `Email` varchar(200) NOT NULL,
  `modified_at` datetime NOT NULL,
  `PostedOnDate` timestamp NOT NULL default CURRENT_TIMESTAMP on update CURRENT_TIMESTAMP,
  `Address` varchar(250) NOT NULL,
  `State` varchar(200) NOT NULL,
  `City` varchar(200) NOT NULL,
  `Country` varchar(100) NOT NULL,
  `Zipcode` varchar(100) NOT NULL,
  `JobID` varchar(100) NOT NULL,
  `WorkExperience` varchar(2000) NOT NULL,
  `EducationRequirement` varchar(2000) NOT NULL,
  `WebsitePostedFrom` varchar(200) NOT NULL,
  PRIMARY KEY  (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;



Company Table

Code: [Select]


CREATE TABLE IF NOT EXISTS `Company` (
  `id` int(11) unsigned NOT NULL auto_increment,
  `CompanyName` varchar(100) NOT NULL,
  `Address` varchar(100) NOT NULL,
  `Logo` varchar(100) NOT NULL,
  `PhoneNumber` varchar(25) NOT NULL,
  `ContactPerson` varchar(25) NOT NULL,
  PRIMARY KEY  (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=2 ;

<?php 

$con = database_connect();
$sql = "SELECT *
FROM anime1, episode1
WHERE animeid='$animeid'";
$result = mysql_query($sql);
while ($row = mysql_fetch_assoc($result)) {
$title = $row['title'];
$ep = $row['ep'];
}


 ?>

keep giving me back error

Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in C:\wamp\www\studying\take 2\addin12.php on line 45

So I have 3 tables: student, course_student and course. The student table is a student's registration info and course is the different courses the student can register for. The course_student table should contain the id fields of the other 2 which I should then use to link the tables and for example display all student's registered for current courses.  Student table has an id field: sno and course table has : cid. Course_student contains both. How do I go about linking the tables so the sno id field of the student table and the cid from course updates with those in the course_table?

i have two tables Loan Application table and Loans table. There is status option and date loan was applied section. After status has been chosen as Pending from the dropdown field and date chosen and loan application saved, it goes under the loans section.

Now, admin has to go and approve the loan and pick the date the loan was approved under the loans section but date can be chosen for the approved date but the status cant be changed to approve because it shows pending. Please what can i do so when i select the date for approval , the status should change to approved automatically

I have two tables.  Let's call the first one items and the second one item categories.  Now the items table would look something like this:

id(auto_increment id)
name
description
categoryid

The item categories table would look something like this:
categoryid
name
description

An entry in the items table would look something like this:
categoryid = 1,2
name = Thing
description = something
id = 1

Say I want to retrieve records that contain "1" in the categoryid column.  How would I do that?