I'm sorry if this is not what you're talking about...
/**
* assume your table structure just like this.
* table: users
* | id | country | name | population |
* ------------------------------------------------------------------
* | 1 | Texas | Fort Worth | 400000 |
*
*/
function getPopulationById($id) {
// echo 'SELECT country, name, population FROM users WHERE id = ' . $id;die;
$result = mysql_query('SELECT country, name, population FROM users
WHERE id = ' . $id);
if(mysql_num_rows($result) != 1) return false;
$row = mysql_fetch_assoc($result);
$pop[$id][$row['country']][$row['name']] = $row['population'];
return $pop;
}
$array = getPopulationById(1);
print_r($array);
// print_r($array) should be like this:
// Array ( [1] => Array ( [Texas] => Array ( [Fort Worth] => 400000 ) ) )
On Fri, Mar 6, 2009 at 3:41 PM, Anton Heuschen <antonfh@xxxxxxxxx> wrote:
>
> This might sound trivial, but for the live of me cant seem to get it to
> work, and I am not familiar with such a thing.
>
> What I have is a query lets say :
>
> select country,name,population from USERS where id= 'some_id' ";
>
>
> Now I want to assign the result to one set (The above example might have 3+
> entries per telephone, thus what would be nice is something like an array of
> :
>
>
> [id][country][name] = population .......... or to be exact if I echo the
> value for each id and country and name to get the population value
>
>
> Like :
>
> Array {
>
> [id] array {
> [country] array {
> [0] = USA
> }
> [name] array {
> [0] = test
>
> }
>
> }
>
>
>
>
> I dont know something like that, maybe Im over comlicating the question now
> even, the main thing is wondered was in the first place was with a standard
> query and variable assign, from the query like:
>
> select country,name,population from USERS where id= 'some_id' ";
>
> normally you would just assign each field to one variable. like
>
> $country = result["country"]
> $name = result["name"]
>
>
> But now I want all 3 fields as one variable (which would be an array) ..but
> how ?
>
> I hope I have not totally confused the question now.
>
> Thanks in advance
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