tp5.1 獲取原始上傳文件名

打印一下上傳的文件對象可以的得出，大致對象結構如下

object(think\File)#77 (13) {
  ["error":"think\File":private] => string(0) ""
  ["filename":protected] => string(14) "/tmp/phpkIAvIy"
  ["saveName":protected] => NULL
  ["rule":protected] => string(4) "date"
  ["validate":protected] => array(0) {
  }
  ["isTest":protected] => NULL
  ["info":protected] => array(5) {
    ["name"] => string(10) "ebrima.ttf"
    ["type"] => string(24) "application/octet-stream"
    ["tmp_name"] => string(14) "/tmp/phpkIAvIy"
    ["error"] => int(0)
    ["size"] => int(907232)
  }
  ["hash":protected] => array(0) {
  }
  ["pathName":"SplFileInfo":private] => string(14) "/tmp/phpkIAvIy"
  ["fileName":"SplFileInfo":private] => string(9) "phpkIAvIy"
  ["openMode":"SplFileObject":private] => string(1) "r"
  ["delimiter":"SplFileObject":private] => string(1) ","
  ["enclosure":"SplFileObject":private] => string(1) """
}

看這結構，訪問權限是protected，如何獲取呢?

獲取方法:

對象->getInfo() 獲取info屬性

//對象->getInfo() 獲取info屬性,['name']  獲取name的值
$file->getInfo()['name']