上传多个文件时，您应该如何使用“CakePhpRequest”数组？
我有这个结果
CakeRequest Object
(
[params] => Array
(
[plugin] =>
[controller] => users
[action] => upload
[named] => Array
(
)
[pass] => Array
(
)
[form] => Array
(
[files] => Array
(
[name] => Array
(
[0] => php.exe
[1] => php.gif
[2] => php.ini
[3] => php.ini-development
[4] => php.ini-production
)
[type] => Array
(
[0] => application/x-msdownload
[1] => image/gif
[2] => application/octet-stream
[3] => application/octet-stream
[4] => application/octet-stream
)
[tmp_name] => Array
(
[0] => C:\xampp\tmp\php455E.tmp
[1] => C:\xampp\tmp\php456E.tmp
[2] => C:\xampp\tmp\php456F.tmp
[3] => C:\xampp\tmp\php4570.tmp
[4] => C:\xampp\tmp\php4571.tmp
)
[error] => Array
(
[0] => 0
[1] => 0
[2] => 0
[3] => 0
[4] => 0
)
[size] => Array
(
[0] => 73728
[1] => 2523
[2] => 78907
[3] => 72908
[4] => 72941
)
)
)
)
[...] other data
我是否必须计算已上传的文件数量，然后使用数字迭代所有文件？
如果输出是这样的，不是更容易吗？
CakeRequest Object
(
[params] => Array
(
[plugin] =>
[controller] => users
[action] => upload
[named] => Array
(
)
[pass] => Array
(
)
[form] => Array
(
[files] => Array
(
[0] => Array(
[name] => php.exe
[type] => application/x-msdownload
[tmp_name] => C:\xampp\tmp\php455E.tmp
[error] => 0
[size] => 73728
)
[1] => Array(
[name] => php.gif
[type] => image/gif
[tmp_name] => C:\xampp\tmp\php456E.tmp
[error] => 0
[size] => 2523
)
[...] more data
)
)
)
[...] other data
这样我只需要使用 foreach ($files as $file)
对不起，我的英语不好