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    <p>The problem is you are trying to insert into the salaries table, but you aren't specifying what employee the salary record is for. You would need to provide a value for the <code>emp_no</code> column that is in the <code>employees</code> table.</p> <p>ETA:</p> <p>Glad you posted more code - what you should do is modify your <code>add_emp</code> function so that it returns the id of the employee that was created (I'm assuming here that <code>emp_no</code> is and auto-increment column in your database). Like so:</p> <pre><code>function add_emp($firstname,$lastname,$gender,$date_of_birth,$jobtitle,$dept,$hiredate) { $data = array( 'first_name' =&gt; $firstname, 'last_name' =&gt; $lastname, 'gender' =&gt; $gender, 'birth_date' =&gt; $date_of_birth, 'hire_date' =&gt; $hiredate); $this-&gt;db-&gt;trans_start(); $this-&gt;db-&gt;insert('employees', $data); $this-&gt;db-&gt;trans_complete(); if ($this-&gt;db-&gt;trans_status() === FALSE) { //$msg = "Adding the new employee failed."; //return $msg; return -1; // indicates failure } else { //$msg = "Successfully Added Employee."; //return $msg; return $this-&gt;db-&gt;insert_id(); } } </code></pre> <p>Note that the code that calls <code>add_emp</code> will need to change a bit to look at the return value as an integer instead of a message now, but now you have the id of the employee and you can use that in your add_salary function instead of having to pass in all that other data, like so:</p> <pre><code>function add_salary($emp_no,$salary, $salary_from_date,$salary_to_date) { $salarystuff = array('emp_no' =&gt; $emp_no, 'salary' =&gt; $salary, 'from_date' =&gt; $salary_from_date, 'to_date' =&gt; $salary_to_date); $this-&gt;db-&gt;insert('salaries', $salarystuff); } </code></pre> <p>So the code to add an employee and their salary would be something like this:</p> <pre><code>$emp_no = add_emp([Your parameters]); if ( $emp_no &gt; 0 ) { add_salary($emp_no, [other parameters]); } else { // Show some error message that the employee creation failed. } </code></pre> <p>EDIT AGAIN:</p> <p>If you ignore my advice above, the simple answer to why you are getting an error is because <code>$selected_employee</code> is a result set and not an actual employee number. You would need to change that code like so:</p> <pre><code>$selected_employee = $this-&gt;db-&gt;get(); $result = $selected_employee-&gt;result(); $emp_no = $result[0]-&gt;emp_no; $salarystuff = array('emp_no' =&gt; $emp_no, 'salary' =&gt; $salary, 'from_date' =&gt; $salary_from_date, 'to_date' =&gt; $salary_to_date); $this-&gt;db-&gt;insert('salaries', $salarystuff); </code></pre>
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    1. POHow to add foreign key value in insert query - mysql - CodeIgniter
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    1. COyes that's what i'm trying to do, but how? can i use active records to retrieve that?
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    2. COActive Record is just a pattern, nothing less nothing more. You should have a reference to which customer you would like to insert data for, and you should use that reference to do the insert. (the reference i was talking about is the customer id most likely)
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    3. COI think I've gotten that if you would kindly look at my edit, but its not getting it. I think the problem is the $selected_employee. The error I'm getting is: Object of class CI_DB_mysql_result could not be converted to string
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