PHP Display Values from Object within an Array Based on Key

Question

I have the following array which contains both objects and arrays. How do I get only the specific values (for each object) based on their keys? I've tested and the array is displaying (see below) but I cannot isolate the 'name" value as needed.

I have tried the following code to get the name value:

case 'field_prgm_housing' :
$node = 'field_color';
$tids =  field_get_items('node', $node, $key, $node->language);
$terms = taxonomy_term_load_multiple(array(), array('tid' => $tids));
$nameonly = $terms->[0]->name[0];
return = $nameonly;
break;      
Colors (Array, 2 elements)
    12 (Object) stdClass
      tid (String, 2 characters ) 12
      vid (String, 1 characters ) 3
      name (String, 9 characters ) Blue
      description (String, 0 characters )
      format (String, 13 characters ) filtered_html
      weight (String, 1 characters ) 0
      vocabulary_machine_name (String, 15 characters ) colors
      rdf_mapping (Array, 5 elements)
      path (Array, 1 element)
    13 (Object) stdClass
      tid (String, 2 characters ) 13
      vid (String, 1 characters ) 3
      name (String, 8 characters ) Green
      description (String, 0 characters )
      format (String, 13 characters ) filtered_html
      weight (String, 1 characters ) 0
      vocabulary_machine_name (String, 15 characters ) colors
      rdf_mapping (Array, 5 elements)
      path (Array, 1 element)

If the array's output is the var_dump of $terms: this one should work $terms[$i]->name. (Just replace $i with the relevant key) — Ofir Baruch
– Ofir Baruch, Commented Nov 15, 2017 at 11:15
$names = Array_column($array, "name") is that what you are looking for — Andreas
– Andreas, Commented Nov 15, 2017 at 11:19
@OfirBaruch in this case what would be the relevant key if not name? — Sam
– Sam, Commented Nov 15, 2017 at 11:21
According to your example $i should be 12 or 13. But since this data is dynamic you should find a way to get it by using a loop or a sort of a condition. — Ofir Baruch
– Ofir Baruch, Commented Nov 15, 2017 at 11:23
@Andreas In this case "$arrray" would be terms correct? '$nameonly = Array_column($terms, "name");' returns an empty array — Sam
– Sam, Commented Nov 15, 2017 at 11:23

K.B · Accepted Answer · 2017-11-15 13:14:19Z

2

Try this


    $node = 'field_color';
    $tids =  field_get_items('node', $node, $key, $node->language);
    $terms = taxonomy_term_load_multiple(array(), array('tid' => $tids));

    //loop all the values and get the require value
   $name = array();
    foreach($terms as $term){
          $name[] = $term->name;
    }
    return $name;

edited Nov 15, 2017 at 13:14

answered Nov 15, 2017 at 11:24

K.B

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4 Comments

Sam Over a year ago

I'm operating within switch statements, for each throws errors.

Sam Over a year ago

I need to use the values elsewhere hence when I them to be available as the $nameonly variable; is this possible with foreach?

Sam Over a year ago

Here is what I've tried so far but its only showing the first value: foreach($terms as $term){ $name=$term->name; } $return[$key] = $name;

K.B Over a year ago

$name = array(); foreach($terms as $term){ $name[] = $term->name; } return $name;

MilanG · Accepted Answer · 2017-11-15 12:14:12Z

0

What $terms->[0] suppose to mean?

$terms is an array, so you have to access it by specifying the index you want to get, like: $terms[12], $terms[13]...

Elements of that array are object so when you get one you have to use "->" operator to get it's field (or method).

So it has to be like:

$nameonly = $terms[12]->name;
$nameonly = $terms[13]->name;

Or you can iterate trough all the array items, as @kranthi suggested.

answered Nov 15, 2017 at 12:14

MilanG

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