Question

我想在这里做什么，获取当前文件然后上传它，找到文件的扩展名并重命名它！并回应结果!!但它似乎错了，我不知道哪个部分！：（（

    $fieldname = $_REQUEST['fieldname'];
    $uploaddir = 'uploads/';
    $uploadfile = $uploaddir . basename($_FILES[$fieldname]['name']);


    if (move_uploaded_file($_FILES[$fieldname]['tmp_name'], $uploadfile)) {

    //find the extension
     $extension= pathinfo($uploadfile);

    //rename the file
      rename ($uploadfile, "newfile.".$extenion['extension']."");
 echo "uploads/newfile.'".$extension['extension']."'"; // "success"

}

Answer 1

我认为你应该这样做：

$fieldname = $_POST['fieldname']; // don't use $_REQUEST
$extension = pathinfo($_FILES[$fieldname]['name'], PATHINFO_EXTENSION);

$uploaddir = 'uploads/';
$uploadfile = $uploaddir . 'newfile.'.$extension;

if (move_uploaded_file($_FILES[$fieldname]['tmp_name'], $uploadfile)) {
    // success
}

move_uploaded_file已经“重命名”了该文件，无需手动调用rename。只需在一次操作中完成。

您可能还注意到我已将PATHINFO_EXTENSION传递给pathinfo，因为您只需要扩展名而不是完整路径信息。

最后，我使用$_POST代替$_REQUEST。除非你真的知道自己在做什么，否则不应该使用$_REQUEST。例如，这可能会导致意外变量篡改cookie或会话。

Answer 2

您应该使用新文件名作为move_uploaded_file的第二个参数：

$fieldname = $_REQUEST['fieldname'];
$tmpfile = $_FILES[$fieldname]['tmp_name'];

$info = pathinfo($tmpfile);
$filename = 'newfile.' . $info['extension'];

$uploaddir = 'uploads/';
$uploadfile = $uploaddir . $filename;

if(move_uploaded_file($tmpfile, $uploadfile)) {
    echo $uploadfile;
}

Answer 3

$fieldname = $_POST['fieldname']; #its never going to be a GET! ..not sure y u need this though, the file field should hold the array key, but anyway...
#get the extension by removing everything before the last dot
$extension = preg_replace('@.+\.@', '', $_FILES[$fieldname]['name']);
$newname = "newfile.".$extension;

$uploaddir = 'uploads/';
$uploadfile = $uploaddir . $newname;

if (move_uploaded_file($_FILES[$fieldname]['tmp_name'], $uploadfile)) {

 echo "$uploadfile"; // "success"

}

未经测试，但逻辑直截了当。

在上传时找到扩展名的文件？

3 个答案: