PHP tips, Array Value problems!

PHP tips, Array Value problems! The number level is like this: (I picked a group of data from a large pile. In fact, I want to extract all the data and echo it to the window)
[16] => Array
(
[Option_id] => 30
[Option_name] => att_type
[Option_value] => rar, zip, gif, jpg, jpeg, png, txt, pdf, docx, doc
)


The purpose is to print the rar, zip, gif, jpg, jpeg, png, txt, pdf, docx, and doc.
Currently, we know that we can assign values using loops, but we can only use echo $ option_name to obtain the att_type value. we cannot use echo $ att_type to retrieve rar, zip, gif, jpg, jpeg, png, txt, pdf, docx, doc. Let's help me write a piece of code to realize my thoughts. let's ask for help !!!!!!

Reply to discussion (solution)

Come and save me ..........

There is something wrong with it. all arr_types are wrong. it should be att_type.

Come and save me ...........

It is difficult to help without posting code

It is difficult to help without posting code

The code is very simple, that is, first select from the database, put it in an array, and then print_r, that's it.
$ Arr = $ this-> Setting_model-> get_setting (); // $ query = $ this-> db-> get ('options '); return $ query-> result_array ();
$ This-> load-> p ($ arr); die;

The database seems to have nothing to do with this. I will not post it.

$ This-> load-> p () // public function p () {echo'

;print_r($arr);echo '

';}

However, your demand is too different and strange.

$a = array(    16 => Array        (            'option_id' => 30,            'option_name' => 'att_type',            'option_value' => 'rar,zip,gif,jpg,jpeg,png,txt,pdf,docx,doc',        ),);foreach($a as $r) {  ${$r['option_name']} = $r['option_value'];  echo $att_type;}

Rar, zip, gif, jpg, jpeg, png, txt, pdf, docx, doc

I don't know. it means no.

$arr =  Array    (            'option_id' => 30,            'option_name' => 'att_type',            'option_value' => 'rar,zip,gif,jpg,jpeg,png,txt,pdf,docx,doc'    );$a[$arr['option_name']] = $arr['option_value'];extract($a);var_dump($att_type);

However, your demand is too different and strange.

$a = array(    16 => Array        (            'option_id' => 30,            'option_name' => 'att_type',            'option_value' => 'rar,zip,gif,jpg,jpeg,png,txt,pdf,docx,doc',        ),);foreach($a as $r) {  ${$r['option_name']} = $r['option_value'];  echo $att_type;}

Rar, zip, gif, jpg, jpeg, png, txt, pdf, docx, doc

This will definitely not work. an error will be reported. The undefined att_type

I don't know. it means no.

$arr =  Array    (            'option_id' => 30,            'option_name' => 'att_type',            'option_value' => 'rar,zip,gif,jpg,jpeg,png,txt,pdf,docx,doc'    );$a[$arr['option_name']] = $arr['option_value'];extract($a);var_dump($att_type);

Error: undefined option_name, option_value... And so on. many (Undefined index: option_value)

Yes? Take it for granted ?!
I have all the running results.

I know that the extract function will certainly be used. the function is like assigning the name value to the value.

Yes? Take it for granted ?!
I have all the running results.

Capture a graph of your results. how can I run it and report an error?

However, your demand is too different and strange.

$a = array(    16 => Array        (            'option_id' => 30,            'option_name' => 'att_type',            'option_value' => 'rar,zip,gif,jpg,jpeg,png,txt,pdf,docx,doc',        ),);foreach($a as $r) {  ${$r['option_name']} = $r['option_value'];  echo $att_type;}

Rar, zip, gif, jpg, jpeg, png, txt, pdf, docx, doc

I did not understand $ {$ r ['option _ name']} = $ r ['option _ value, so we changed it to $ r ['option _ name'] = $ r ['option _ value. Sorry. This is solved, and you need to ask, $ {$ r ['option _ name']} = $ r ['option _ value']; means? I really don't understand

To be fair, I'm presenting it to you in the online editor of conengu.

Variable reading

Variable reading

What happened again !!!

To use the variable name of the array, you need to solve the ambiguity problem. That is, if you write $ a [1], the parser needs to understand whether you mean to treat $ a [1] as a variable, we still need to regard $ a as a variable. [1] refers to the index of this variable. The syntax for solving this ambiguity is: $ {$ a [1]} is used in the first case, and $ {$ a} [1] is used in the second case.

Thank you !!! Learned