Cannot make PHP variable option in <select> tag html

Question

I have an array that looks like this:

Array (
    [0] => Array ( [denumireObiect] => Telefon ) 
    [1] => Array ( [denumireObiect] => Laptop ) 
    [2] => Array ( [denumireObiect] => Tableta ) 
    [3] => Array ( [denumireObiect] => Obiect ) 
)

I am trying to take all of those words and make them options for a <select> tag.

This is the code I am using for that:

foreach ($result as $i) {
            echo '<option value = ', $result[$i],'>', $result[$i], '</option>';
          }

This paragraph gives me Illegal offset type error.

This is a var_dump($result) result.

array(4) { 
[0]=> array(1) { 
["denumireObiect"]=> string(7) "Telefon" 
} 

[1]=> array(1) 
{ ["denumireObiect"]=> string(6) "Laptop" } 

[2]=> array(1) {
["denumireObiect"]=> string(7) "Tableta" 
} 

[3]=> array(1) {
["denumireObiect"]=> string(6) "Obiect" } 
} 
...

EDIT:

I have tried doing it like this:

foreach ($result as $i => $val) {
        echo '<option value = ', $i,'>', $i, '</option>';
      }

and it returns 0, 1, 2, 3

Any help is appreciated!

is $result the array you mentioned and could you please provide a var_dump($result)? In your foreach you should be abled to access $i without referencing the $result array. But without knowing $result und $i I can not tell you how exactly — j4g0
– j4g0, Commented Sep 17, 2020 at 17:46
@j4g0 I have edited the post. added a var_dump result and tried a new way to do this. Hope it helps! — Victor
– Victor, Commented Sep 17, 2020 at 17:57
I have what feels like a very dumb question, but it might be an internationalization thing... does using , as a string concatenator work? AFAIK you need to use . (php.net/manual/en/language.operators.string.php), but maybe that's because I'm in the US. echo "test","ing"; vs echo "test"."ing"; - I would expect the first to throw an error, while the second would output testing. — WOUNDEDStevenJones
– WOUNDEDStevenJones, Commented Sep 17, 2020 at 18:26
i do not think it is concatenating but rather one echo outputting several strings. It threw me off at first, too :D — j4g0
– j4g0, Commented Sep 17, 2020 at 18:28
whoa, TIL. I had no idea you could list things like that to be echoed. Sure enough, it's documented on the echo page: php.net/manual/en/function.echo.php — WOUNDEDStevenJones
– WOUNDEDStevenJones, Commented Sep 17, 2020 at 18:38

WOUNDEDStevenJones · Accepted Answer · 2020-09-17 18:24:40Z

1

@WOUNDEDStevenJones made a good point. I have updated my code to make it more readable. Note that you would have to rename the $result variable to $results.

Since every $result from your code is an associative array with the same "denumireObiect" key, you could get the values like this:

foreach ($results as $result) {
    echo "<option value=\"{$result['denumireObiect']}\">{$result['denumireObiect']}</option>";
}

If you do not like using the interpolation you could reformat the echo statement like this:

echo '<option value="' . $result['denumireObiect'] . '">' . $result['denumireObiect'] . '</option>';

or use your format from before :)

edited Sep 17, 2020 at 18:24

WOUNDEDStevenJones

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answered Sep 17, 2020 at 18:05

j4g0

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3 Comments

Victor Over a year ago

This gives me Trying to access array offset on value of type int error.

j4g0 Over a year ago

I removed the => $val from the foreach, we do not need it

WOUNDEDStevenJones Over a year ago

To be more clear, $i is usually used as the index counter in loops (i.e. for($i=0; $i<count($result); $i++)) so I would recommend keeping this style and either writing foreach ($result as $i => $val) and then using $val['denumireObiect'] in the loop (or $result[$i]['denumireObiect'], but that seems unnecessarily silly in this case). This way, $i is essentially still an index counter, which is about the only thing single character variable names should be used for.

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