Im 试图创建一个函数,该函数接受输入,确定其值,并从脚本中包含的现有数组中输出一个单词。问题是输出是空白的,我相信该函数会忽略脚本中已有的变量,有没有办法更改它,以便函数不会忽略现有变量?
这是函数:
由于多语言要求,单词需要来自数组。
function get_genre($id)
{
if($id == "1"){
$genre = $lang['277'];
}
if($id == "2"){
$genre = $lang['278'];
}
if($id == "3"){
$genre = $lang['279'];
}
if($id == "4"){
$genre = $lang['280'];
}
if($id == "5"){
$genre = $lang['281'];
}
if($id == "6"){
$genre = $lang['282'];
}
if($id == "7"){
$genre = $lang['283'];
}
if($id == "8"){
$genre = $lang['284'];
}
if($id == "9"){
$genre = $lang['285'];
}
if($id == "10"){
$genre = $lang['286'];
}
if($id == "11"){
$genre = $lang['287'];
}
if($id == "12"){
$genre = $lang['288'];
}
if($id == "13"){
$genre = $lang['289'];
}
if($id == "14"){
$genre = $lang['290'];
}
if($id == "15"){
$genre = $lang['374'];
}
return $genre;
}
function get_genre($id)
{
global $lang;
....
}
或
function get_genre($id, $lang) //Must pass $lang array to function here
{
}
function get_genre($id)
{
global $lang;
if($id == "1"){
$genre = $lang['277'];
}
if($id == "2"){
$genre = $lang['278'];
}
if($id == "3"){
$genre = $lang['279'];
}
if($id == "4"){
$genre = $lang['280'];
}
if($id == "5"){
$genre = $lang['281'];
}
if($id == "6"){
$genre = $lang['282'];
}
if($id == "7"){
$genre = $lang['283'];
}
if($id == "8"){
$genre = $lang['284'];
}
if($id == "9"){
$genre = $lang['285'];
}
if($id == "10"){
$genre = $lang['286'];
}
if($id == "11"){
$genre = $lang['287'];
}
if($id == "12"){
$genre = $lang['288'];
}
if($id == "13"){
$genre = $lang['289'];
}
if($id == "14"){
$genre = $lang['290'];
}
if($id == "15"){
$genre = $lang['374'];
}
return $genre;
}
虽然这并不理想,但您是否考虑过使用关联数组?
var $lookupArray = array();
$lookupArray["1"] = $lang['274'];
....
然后你可以这样称呼它:
function get_genre($id)
{
return(array_key_exists($id,$lookupArray)) ? $lookupArray[$id] : null;
}