求一個高效簡單的多維陣列字元編碼轉換函式

來源:互聯網
上載者:User
function arrayCv($data) {if (is_array($data)) {foreach ($data as $key => $val) {if (!is_array($val)) {$arr[$key] = iconv('UTF-8', 'GBK',  $val);} else {$arr[$key] = arrayCv($val);}}} else {return iconv('UTF-8', 'GBK',  $data);}return $arr;}

現在是這個樣的感覺不優雅,有用array_map, array_walk來實現的嗎

回複內容:

function arrayCv($data) {if (is_array($data)) {foreach ($data as $key => $val) {if (!is_array($val)) {$arr[$key] = iconv('UTF-8', 'GBK',  $val);} else {$arr[$key] = arrayCv($val);}}} else {return iconv('UTF-8', 'GBK',  $data);}return $arr;}

現在是這個樣的感覺不優雅,有用array_map, array_walk來實現的嗎

試試 array_walk_recursive 。

function arrayCv($data) {      if (is_array($data)) {            foreach ($data as $key => $val) {                 $arr[$key]=arrayCv($val);           }      } else {             return iconv('UTF-8', 'GBK',  $data);      }}

==================================================

function myConv(&$value,$key) {    $value=iconv('UTF-8', 'GBK',$value);}function arrayCv($data) {    array_walk($data,"myConv");}

function ac($input = array()){    return json_decode(iconv('UTF-8', 'GBK//IGNORE',  json_encode($input,JSON_UNESCAPED_UNICODE)),true);}

不知道這個行不行。有人用php5.4麼。測測。

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