Re: multiple function returns

At 9:08 PM -0800 1/24/01, jeremy brand wrote:

$array[0] will be the lang_pref and $array[1] will be the currency_pref. Jeremy

An easy way to read the returned array values into scalar variables is: list($lang_pref, $currency_pref) = PrefChange($language, $currency, $state); You can only do this if your PrefChange function _always_ returns the same number of elements (two here; it could be any number). If you have a function that - for example - returns a 2-element array or FALSE if there's an error, then you have to do something a little more complex: $Stuff = PrefChange($language, $currency, $state); if (is_array($Stuff)) { # ...do stuff with $Stuff ! } else { # ...show error message } - steve

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Get your own Free, Private email at http://www.smackdown.com/ On Thu, 25 Jan 2001, Jamie wrote:

Date: Thu, 25 Jan 2001 13:04:27 +0800 From: Jamie <jssmith@echidna.stu.COWAN.EDU.AU> To: PHP <php-general@lists.php.net> Subject: [PHP] multiple function returns Can anyone help me I'm trying to read two results from a function by passing them out via an array but I'm not sure how to access them once I have done that. In the PHP manual it shows a list call which I can't seem to even find what the hell it does the code below runs without any errors but does not seem to work. here the function gest passed the language and currency preference and a variable (1,0) to decide what it will do with them - update the preferences of the user or read out and define the preferences. (once updated I want it to read out the prefs aswell) function PrefChange($lang,$curry,$state){ global $DB_Server, $HTTP_Host, $DB_Login, $DB_Password, $DB_Name, $DocRoot ; if ($state){//read preferences from page if (!($result = mysql_db_query($DB_Name,"SELECT language_pref, currency_pref FROM users WHERE user_id ='$cookie_user'"))){

    DisplayErrMessage(sprintf("internal error %s %s %s %d:%s\n",$DB_Server,

$DB_Login, $DB_Password,

          mysql_errno(), mysql_error()));
    while ($row = mysql_fetch_array($results)){
          $lang_pref = $row["language_pref"];
          $currancy_pref = $row["currency_pref"];
     return array($lang_pref,$currancy_pref);
     }
    }

} else {///update preferences from page if (!($result = mysql_db_query($DB_Name,"UPDATE users SET language_pref='$lang' AND currency_pref='$curry' WHERE user_id='$cookie_user'"))){

    DisplayErrMessage(sprintf("internal error %s %s %s %d:%s\n",$DB_Server,

$DB_Login, $DB_Password,

          mysql_errno(), mysql_error()));
    PrefChange(0,0,1) ;
    }

} } an example of what I'd like to do when calling the function (here I'm only reading the prefrences from the DB ), a select box that is showing the preference that is chosen with the preference read from the above function and the list created by the function below <select name="form_lang_pref" style="font-family:verdana, Arial;font-size:8pt" CLASS:menu"> <? list ($form_language_pref, $form_currency_pref) = PrefChange(0,0,1);

           $select_id = $form_language_pref;
           $db = mysql_connect("$DB_Server", "$DB_Login, $DB_Password");
           mysql_select_db("$DB_Name",$db);
           $results = mysql_query("SELECT * FROM language ORDER BY language

",$db);

           while ($row = mysql_fetch_array($results)){
               $rowid = $row["language_id"];
               $name = $row["language"];
               if ($rowid == $select_id) {

echo("<option SELECTED value=\"$id\">$name</option>\n");

} else { echo("<option value=\"$id\">$name</option>\n"); } // endwhile; } echo("</select>"); ?>

-- +--- "They've got a cherry pie there, that'll kill ya" ------------------+ | Steve Edberg University of California, Davis | | sbedberg@ucdavis.edu Computer Consultant | | http://aesric.ucdavis.edu/ http://pgfsun.ucdavis.edu/ | +-------------------------------------- FBI Special Agent Dale Cooper ---+