<html><head><title>File Upload</title></head>
<body bgcolor = "lavender"><div align = "center">
<?php
if (isset($_FILES['file']) && move_uploaded_file(
$_FILES['file']['name']))
{
echo "<font color = 'green'>The file has been uploaded.</font>";
}
else echo "<font color = 'red'>There was an error uploading the file.</font>";
?>
</div></body>
</html>
这是我的代码。
我正在尝试使用“get”方法通过单独网页中的单独表单上传文件。
所示表单的代码如下:
<form enctype="multipart/form-data" action = "fileupload.php" method = "get">
<input type = "file" name = "file"><br>
<input type = "submit" value = "Upload">
</form>
由于某种原因,我不断收到错误消息 - 尽管我很确定我做得正确。这是我第一次这样做,建议将不胜感激。
它应该是 tmp_name
if (isset($_FILES['file']) && move_uploaded_file($_FILES['file']['tmp_name'],'ftp/' . $_FILES['file']['name']))
并且不要将其作为 GET 发送
<form enctype="multipart/form-data" action = "fileupload.php" method = "post">
(更改为发布)
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