Error in Passing Data in View in CodeIgniter

Question

I'm storing an SQL query result in a $data variable and passing it to a search_view like this from my model:

        $this->db->like('LOWER(title)', strtolower($query));
        $q = $this->db->get('questions');
        $data = $q->row_array();
        $this->load->view('search_view', array('data' => $data));

And I am trying to display all results in my view using following structure:

<?php if ($data['search_type'] == 'question') { 
        foreach ($data as $question) {
?>


<div class="question-result">
    <h2><?php echo $question['title']; ?></h2>
    <p>Posted by: <a href="#"><?php echo $question['username']; ?>username</a></p>
    <p>Date Posted: <?php echo $question['date']; ?></p>
</div>


<?php } } ?>

Where $data['search_type'] is something I need to check to see what type of search was performed, this bit is tested and works fine, no need to worry about it, however

I get an illegal string offset error for all variables like username, title and date that I use.

eddit* example dump:

array(7) { ["question_id"]=> string(1) "3" ["title"]=> string(19) "First Question Ever" ["content"]=> string(13) "Hello Content" ["date"]=> string(19) "2015-01-03 15:08:30" ["user_id"]=> string(2) "14" ["username"]=> string(4) "Ilya" ["search_type"]=> string(8) "question" }

for some reason this is not returning more than one question, although there should be more for a search performed.

Preface your snippet with var_dump($data); to see what you actually got. — mario
– mario, Commented Jan 3, 2015 at 15:26

Kevin · Accepted Answer · 2015-01-03 15:38:04Z

1

You don't need to use another foreach since its not in another dimension anymore. Once you've used ->row_array() only one row is returned.

<?php if ($data['search_type'] == 'question') { ?>

<div class="question-result">
    <h2><?php echo $data['title']; ?></h2>
    <p>Posted by: <a href="#"><?php echo $data['username']; ?>username</a></p>
    <p>Date Posted: <?php echo $data['date']; ?></p>
</div>

<?php } ?>

If you expect to have more results then use ->result_array() instead:

$data = $q->result_array();

Then on view:

<?php foreach ($data as $question): ?>
    <?php if ($question['search_type'] == 'question') { ?>
        <div class="question-result">
            <h2><?php echo $question['title']; ?></h2>
            <p>Posted by: <a href="#"><?php echo $question['username']; ?>username</a></p>
            <p>Date Posted: <?php echo $question['date']; ?></p>
        </div>
    <?php endif; ?>
<?php endforeach; ?>

Or might as well just add $this->db->where('search_type', 'question') so you wouldn't have to put an if inside the foreach in above loop.

edited Jan 3, 2015 at 15:38

answered Jan 3, 2015 at 15:32

Kevin

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2 Comments

Ilja Over a year ago

Fixed the issue where I was only getting one result now I can see both and formated correctly however after that I still get the errors :/

Kevin Over a year ago

@Ilya could you elaborate those errors, which are working and which are not? however after that I still get the errors :/ its hard to guess on my side :)

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