我有一个上传的文件,当试图将其保存到文件系统时,它只是失败了
我的var_export($_FILES['bom_file'])
产生
array (
'name' => '_bomUpload.tester.csv',
'type' => 'application/octet-stream',
'tmp_name' => 'C:''wamp''tmp''php837A.tmp',
'error' => 0,
'size' => 265,
)
这是我的move_uploaded_file()
尝试:(我已经尝试了几个不同的文件路径)
$file_path = 'C:/wamp/NetBeansProjects/BomProject/application/uploads/boms/testfile.csv'; // For Testing ---->
$testVar = move_uploaded_file($_FILES['bom_file']['name'], $file_path);
echo ($testVar)? 'The file has been uploaded': 'There was an error uploading the file';
echo '<pre>UPLOAD?: ', var_export($testVar), '</pre>'; // For Testing ---->
我一直只收到错误消息,而$testVar
为FALSE。
我错过了什么??
更改
$testVar = move_uploaded_file($_FILES['bom_file']['name'], $file_path);
至
$testVar = move_uploaded_file($_FILES['bom_file']['tmp_name'], $file_path);
name
只是文件的"名称"。实际上传的文件引用在tmp_name
数组密钥
尝试$testVar = move_uploaded_file($_FILES['bom_file']['tmp_name'], $file_path);
-文件应该仍在tmp_name中作为点