我正在使用CodeIgniter 3,我想扩展默认语言助手。
默认情况下,CodeIgniter允许函数lang(language_key)
从语言文件中获取值,但我的语言文件不同。
默认语言文件:
$lang = [
'MONTH_JANUARY' => 'January',
'MONTH_FEBRUARY' => 'February',
];
我的档案:
$lang = [
'MONTH' => [
'JANUARY' => 'January',
'FEBRUARY' => 'February',
],
];
我想用爆炸创建一个辅助函数。
例如,使用lang('MONTH.JANUARY')
它会查找$ lang ['MONTH'] ['JANUARY']并返回'January'。
最好的方法是什么?
答案 0 :(得分:0)
使用引用遍历数组
function lang($string) {
$lang = [
'MONTH' => [
'JANUARY' => 'January',
'FEBRUARY' => 'February',
],
];
$p = &$lang;
foreach(explode('.', $string) as $x) {
if(! isset($p[$x])) {
// key does not exist - do something
}
else {
$p = &$p[$x];
}
}
return $p;
}
echo lang('MONTH.JANUARY'); // 'January'