UPDATE:
How could/would i be able to display the results. My mind is blocked, i cant seem to think coz all my mind prioritizes is the design/look of how it would be displayed. If someone could show me in code how to display each of the data, i would be able to finish.
I have little to no time to make a better database structure so one table would suffice. :)
Table columns: id name meal cuisine cooking difficulty description
Search View: 1st step
<h2>Results:</h2>
<?php if($result == 0):?>
No records found.
<?php else:?>
<?php foreach($result as $r):?>
<?=anchor('recipe_controller/show_recipe_search/'.$r->id,$r->name);?><br />
<?php endforeach;?>
<?php endif;?>
<p>
<?=$links?>
</p>
Controller: 2nd step
public function show_recipe_search()
{
$to_show = $this->uri->segment(3);
$this->db->where('id', $to_Show);
$query = $this->db->get('recipe');
$data['result'] = $query->result_array();
$this->load->view('display_recipe',$data);
}
View: 3rd step
<?php foreach($result as $r):?>
<h2><?php echo '$name'; ?></h2> // Recipe name
<?=$r->echo rest of row;?><br />
<?php endforeach;?>
Since my view is incomplete as my mind is rendering me paralyzed. IDK why. I just cant think of how i would be able to display the result. Should be easy but until i can picture the design i cant echo the results yet.
What does this error imply?
a php error was encountered
severity: notice
message: trying to get property of non-object
filename: views/display_recipe.php
line number: 63
line 63 is <h2><?php echo $r->name; ?></h2>
Your error is generally thrown when an object cannot access a property, due to it being null. I'd imagine that the main idea is that you're not actually returning anything from the DB.
I suspect that your problem is on this line:
$this->db->where('id', $to_Show);
Can you try a var_dump() right after you execute your query or going through XDebug?