Retrieve an associative array value with a variable as key

Question

Here's some code that I have trouble with. I don't know why, I feel I have used this code many times without any problems.

$people['firstname'] = "Fred";
$t = "firstname";
echo $people[$t] ;

echo returns nothing, whereas i expect it to return Fred.

Thanks for your help, Marc

There is nothing wrong with your code at all. Double check grammar, spelling and character case. — NemoStein
– NemoStein, Commented Feb 24, 2011 at 14:17
Cannot reproduce your issue. Pasted your code verbatim in a PHP 5.3 project. Works flawlessly. — mingos
– mingos, Commented Feb 24, 2011 at 14:18
Thanks for confirming the code is ok. Just have to find now why it doesn't work on my page. — Marco
– Marco, Commented Feb 24, 2011 at 14:20
@Marco I have a similar problem. Did you resolve yours? If yes, can you tell what the problem was? — Solace
– Solace, Commented Sep 14, 2014 at 21:48
I have same problem the array return noting when I use variable $CCodes2=$CCodes[$country]; but when i use echo $CCodes["LY"]; gives me correct value here is the array $CCodes=array("BH"=>"BHD","DZ"=>"DZD","EG"=>"EGP","IQ"=>"IQD","JO"=>"JOD","KW"=>"KWD","LB"=>"LBP","LY"=>"LYD", "MA"=>"MAD","OM"=>"OMR","QA"=>"QAR","SA"=>"SAR","SD"=>"SDG","TN"=>"TND","YE"=>"YER","MR"=>"MRO"); — Salem
– Salem, Commented Jan 18, 2015 at 11:41

Ian P · Accepted Answer · 2011-02-24 14:18:26Z

2

Not sure why this isn't working for you.

$people['firstname'] = 'testvalue';

$key = 'firstname';

$value = $people[$key];

echo $value;

Works as expected, echos out "testvalue"

Double check your spelling and be consistent with your ticks (purely stylistic, I'm sure.)

answered Feb 24, 2011 at 14:18

Ian P

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Comments

Wai Ha Lee · Accepted Answer · 2016-04-02 13:31:45Z

1

I think you can pass associative array value as variable . this is working for me

@$username=$_POST['username'];
@$password=$_POST['password'];

$result=array(
'username'=> "".$username."",
'password'=> "".$password.""
);

edited Apr 2, 2016 at 13:31

Wai Ha Lee

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answered Apr 2, 2016 at 12:58

Tanver

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Comments

Salem · Accepted Answer · 2021-10-14 12:35:10Z

1

OK I found solution for my code, looks like variable contains, none ASCII character so I had to remove them, to make it works.

$country = preg_replace('/[[:^print:]]/', '', $country);
$CCodes2=$CCodes[$country];

You should check your php file encoding, e.g if you are use WYSIWYG editor or formatted text, make sue remove any of these none ASCII text before paste it .

edited Oct 14, 2021 at 12:35

answered Jan 18, 2015 at 12:00

Salem

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2 Comments

Waris Ali Over a year ago

this solved my issue

Salem Over a year ago

OG, my English were horrible back than, I fix some mistakes I typed.

gadolf · Accepted Answer · 2020-02-13 08:23:56Z

$params_to_json = '{"' . str_replace(['=', '&'], ['":"', '","'], $_SERVER['QUERY_STRING']) . '"}';
$json_to_array = json_decode($params_to_json, true) ?? array(); 
extract($json_to_string); // Converts array keys into variable names and array values into variable value

As a function:

function url_params()
{
    $params_to_json = '{"' . str_replace(['=', '&'], ['":"', '","'], $_SERVER['QUERY_STRING']) . '"}';
    return json_decode($params_to_json, true) ?? array();
}
extract(url_params());

For example URL: example.com?id=3

echo $id //3

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