PHP / CakePHP - Creating a list of values from a database

Question

I created an array:

$alphabet = range('A', 'Z');

Which created a list of all of the letters of the alphabet. Then I used a for loop to print out all of the letters of the alphabet:

<ul>

<? for ($i = 0; $i < 26; $i++): ?>

    <li><span class="head-menu"><?= $alphabet[$i]; ?></span>

        <ul>

            <li><a href="#">Some Item</a></li>

            <li><a href="#">Some Item</a></li>

        </ul>

    </li>   

<? endfor ?>                

</ul>

In that code, I want to replace:

<li><a href="#">Some Item</a></li>

With a list of values from a database such that the first letter of the value in the name column in the database is the same letter as $alphabet[i].

So for example, if I have a database called "food" with just a "name" field, I want an output as such:

A
    Apple
B
    Banana
C
    Carrots
    Crackers

Where Apple, Banana, Carrots and Crackers are values in the database.

How can I go about doing this?

Machavity · Accepted Answer · 2013-11-15 02:52:53Z

What I would do is read your database out into an array and populate it that way

$sql = 'SELECT field FROM table ORDER BY field';
$res = $mysqli->query($sql);
$data = [];
$current = NULL;
while($row = $res->fetch_assoc()) {
    if($row['field']{0} != $current) {
        $current = $row['field']{0};
        $data[$current] = [];
    }
    $data[$current][] = $row['field'];
}

So now you have an array that looks like this

Array( 
    'A' => Array('Apples'),
    'B' => Array('Bananas', 'Blueberries')
    etc
}

Then you just iterate your code

<ul>
<?php for ($i = 0; $i < 26; $i++): ?>
    <li><span class="head-menu"><?= $alphabet[$i]; ?></span>
        <ul>
            <?php foreach($data[$alphabet[$i]] as $fruit): ?>
                <li><a href="#"><?php echo $fruit ?></a></li>
            <?php endforeach ?>
        </ul>
    </li>   
<? endfor ?>                
</ul>

user2964568 · Accepted Answer · 2013-11-15 02:54:30Z

0

$alphabet = range('A', 'Z');
$r = mysql_query("SELECT name FROM food");
if ($r) {
    while($row = mysql_fetch_row($r)) {
        $alphabet[$row[0][0]][] = $row[0];
    }
}

For cake

$alphabet = range('A', 'Z');
$food = $this->Food->find('list');
foreach ($food as $v)
    $alphabet[$v[0]][] = $v;

answered Nov 15, 2013 at 2:54

user2964568

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