Question

我正在尝试将图像存储在FTP服务器上以便在其他页面上使用，但是我在尝试使用此工作时遇到了多个错误。在XAMPP上运行一切。首先，我在html页面上使用输入：

<input type="file" name="image" required>

然后我将它翻过来并尝试上传：

$image = $_POST["image"];
$ftpCon = ftp_connect("127.0.0.1", "21") or die("Could not connect to FTP");
ftp_fput($ftpCon, "image.png", $image, FTP_BINARY);
ftp_close($ftpCon);

使用此代码我收到此错误：“ftp_fput（）期望参数3为资源，字符串为”

Answer 1

更改行

$image = $_POST["image"];

到

$image = $_FILES['image']['tmp_name'];

Answer 2

当您通过表单上传项目（文件）时，它会填充在$ _FILES超全局中。

通过的上传到当前脚本的关联项目数组 HTTP POST方法。

http://se2.php.net/manual/en/reserved.variables.files.php

确保您还使用enctype='multipart/form-data'

设置表单

所以PHP的第一行必须改为：

$image = $_FILES['image']['tmp_name'];

$ _FILES是关联的，包含以下数据：

（userfile = image，在您的情况下）

$_FILES['userfile']['name']
The original name of the file on the client machine.
$_FILES['userfile']['type']
The mime type of the file, if the browser provided this information. An example would be "image/gif". This mime type is however not checked on the PHP side and therefore don't take its value for granted.
$_FILES['userfile']['size']
The size, in bytes, of the uploaded file.
$_FILES['userfile']['tmp_name']
The temporary filename of the file in which the uploaded file was stored on the server.
$_FILES['userfile']['error']
The error code associated with this file upload. This element was added in PHP 4.2.0

从HTML输入上载图像到ftp

2 个答案: