Create a dropdown list from a multi-dimensional array

Question

I got this array:

Array
(
    [0] => Array
        (
            [id] => 1
            [name] => tomato
        )

    [1] => Array
        (
            [id] => 2
            [name] => carrot
        )

    [2] => Array
        (
            [id] => 3
            [name] => apple
        )

)

I want to print each key/value pair in an HTML form, like so:

<select>
    <option value="1">tomato</option>
    <option value="2">carrot</option>
    <option value="3">apple</option>
</select>

So, I'm using a foreach loop to iterate over the three items in the outer array and then try to print the items in the inner array on a single line. I'm stuck with the last bit. The closest I've got so far is this:

foreach ($food_opts as $key => $value) {
    foreach ($value as $k => $v) {
        echo '<pre>' . $v . '</pre>';
    }
}

This retrieves the data I need but not in a usable format:

1
tomato
2
carrot
3
apple

In short, how do you target individual items in an inner array? Something like:

foreach ($food_opts as $key => $value) {
    foreach ($value as $k => $v) {
        echo '<pre>' . $v[0] . ' - ' . $v[1] . '</pre>';
    }
}

I understand why the above code doesn't work but can't figure out how to get the data how I want it.

Amal · Accepted Answer · 2014-02-24 01:38:49Z

5

You don't need a nested foreach here. Just do:

foreach ($food_opts as $key => $arr) {
    echo '<option value="'.$arr['id'].'">'.$arr['name'].'</option>', PHP_EOL;
}

Or, you can use printf() for a more cleaner approach:

foreach ($food_opts as $key => $arr) {
    printf('<option value="%s">%s</option>', $arr['id'], $arr['name']).PHP_EOL;
}

The printf() family of functions uses % character as a placeholder. %s means "take the next argument and print it as a string". Similarly, %d means "take the next argument and print it as an int".

edited Feb 24, 2014 at 1:38

answered Feb 24, 2014 at 1:33

Amal

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1 Comment

unhack Over a year ago

Thanks so much - yes, that makes sense. I've accepted your answer as it's more complete than Fabio's.

Fabio · Accepted Answer · 2014-02-24 01:33:49Z

3

You need only one foreach loop to have all details available, you can then call them using variable assigned in the loop and keys in the scopes as follow

echo '<select>';
foreach ($food_opts as $value)
{
    echo '<option value="'.$value['id'].'">'.$value['name'].'</option>';
}
echo '</select>';

answered Feb 24, 2014 at 1:33

Fabio

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2 Comments

unhack Over a year ago

Many thanks - works a treat. I tried accepting both your answer and Amal Murali's but see I can only accept a single answer ;(.

Fabio Over a year ago

It's ok, it's a pleasure to help people

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