CODEWITHSUNDEEP
phpcodeigniter
Bill
Bill

Reputation: 5668

passing sql query results from controller into view with code igniter

So I'm having this problem, it should be pretty simple, but I don't know why I can't figure it out. I"m new to the whole idea of MVC, and I'm trying to pass a database query from my controller into a view and display the results in the view. The way I'm doing it now says "undefined variable, sql" when I load the view. This is what I have:

CONTROLLER

function make_login()
{
    //Select list of departments for dropdown
    $this->load->database();
    $sql = $this->db->query('SELECT departmentName FROM department ORDER BY departmentName ASC');



    $this->load->view('siteheader.php');
    $this->load->view('makelogin.php', $sql->result_array());
    $this->load->view('sitefooter.php');
}

VIEW

<?php
 foreach($sql->result_array() as $row)
    {
        echo $row['departmentName'];
    }
    ?>

(If I just echo it out in the controller, it displays the results)

Any help would be awesome... THANKS!

Upvotes: 5

Views: 50211

Answers (4)

Michael Ozeryansky
Michael Ozeryansky

Reputation: 8043

It seems that either CI is confusing you or you are also new to PHP. They are just functions so you can't pass variables like that.

When passing an associative array it will take the key and make that into a variable with the value in the array, using native PHP functions. What Ross said is exactly what you are supposed to do.

Model: all database stuff Controller: uses models to pass variables to views View: outputs the variables (a view should never have any sql in it)

Also note that result and result_array have the same data but result returns objects and result_array returns associative arrays.

Upvotes: 1

Ross
Ross

Reputation: 17967

few tips ~

your make_login should be in a model. your controller will look something like this:

function make_login
{
    $this->load->model('login_model'); // whatever you call it

    $data['departments'] =  $this->login_model->get_departments();

    /* note - you don't need to have the extension when it's a php file */
    $this->load->view('siteheader');
    $this->load->view('makelogin', $data);
    $this->load->view('sitefooter');
}

now in your model, have something like:

function get_departments()
{
    $sql = $this->db->query('SELECT departmentName FROM department ORDER BY departmentName ASC');
    return $sql->result();
    /* you simply return the results as an object
     * also note you can use the ActiveRecord class for this...might make it easier
     */
}

and finally, your view:

<?php
    foreach($departments as $store)
    {
        echo  $store->name . '<br />'; // your fields/whatever you want to output.
    }
?>

Upvotes: 16

Peter
Peter

Reputation: 4141

The SQL query should be done in the model.

Cheap mnemonic device: the D in model = database.

In the Controller, you assign part of the $data array to the results of the query:

$this->load->model('blog_model');
$data['posts'] = $this->blog_model->getPosts();

// Load the view with the data
$this->load->view('blog', $data);

In the Model, you do the actual query:

public function getPosts()
{
    // Method chaining supported in PHP 5
    $this->db->select('*')->from('posts');

    // Assign the query object to a variable
    $query = $this->db->get();

    // We don't want to return an empty result, so let's handle that error somehow
    if (!$query->num_rows() > 0) {
        die("There are no posts in the database.");
    }

    // Make a new array for the posts
    $posts = array();

    // For the purposes of this example, we'll only return the title
    foreach ($query->result() as $row) {
        $posts[$row->id] = $row->title;
    }

    // Pass the result back to the controller
    return $posts;
}

Now, in the view, each element of $data will be its own variable:

    <div class="post">
    <?php foreach ($posts as $id => $title) : ?>
        <h1 class="post-title"><?php echo $title; ?> (ID: <?php echo $id; ?>)</h1>
        <p class="post-content">
        .......
        </p>
        <?php endforeach; ?>
    </div>

That's it!

Upvotes: 3

iBiryukov
iBiryukov

Reputation: 1740

 foreach($array as $row)
    {
        echo $row['departmentName'];
    }
    ?>

You need to pass the $array in. I am not sure what ->load->view() does and how it treats the incoming parameters.

Upvotes: 0

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