从本地服务器检索PHP中的图像内容并保存在远程服务器中

Question

$imgData = base64_encode(file_get_contents('F:\images\home\pic.jpg')); 

The above is my local host path. For local host, it is working.

But, how can I get it in my remote server?

I have not worked on FTP. Please guide me.

I need to get the content of the image alone. I don't need to upload the file to remote server.

Answer 1

You seem to be developing on a Windows machine because of the \\ directory separator. It is best practice to use forward-slash / as it will work on both Windows and Linux, provided you use relative paths (the path with reference to root directory is called absolute. The path with reference to current directory is called relative). You should aim to use relative paths whenever possible.

Example:

• F:/images/home/pic.jpg is an absolute path. Notice I replaced your backslash \\ with a forwardslash / .
• http://www.example.com/images/pic.png is an absolute path.
• ./pic.jpg is a relative path and is equivalent to pic.jpg
• ../images/pic.jpg ia a relative path ( .. means you've exited the current folder, entering the working directory where the images folder is located too).

On your remote (Linux) server, you will have to know the relative path

Let's assume your remote server is a linux/apache server. Maybe you should also tag your question with apache instead of javascript/angular ?

You should determine the relative path to your image and use the examples above to rewrite your code to something like this:

$imgData = base64_encode(file_get_contents('../images/home/pic.jpg'));

You can also use an absolute path ( http://www.example.com/images/pic.png ) but that is not always possible or recommended.

How to find the relative path to your image? There are multiple ways to do that, using PHP, SSH or FTP. Your FTP software will show you the path to your image if you navigate to it, but you need to adapt it to your working directory path. This image is from Filezilla, an open source FTP client:

In this case, in /home/bg29ll/public_html/food2/images that you see in Filezilla, food2 is your working directory, so you would only need to use:

$imgData = base64_encode(file_get_contents('images/pic.jpg'));

You might also include this line in your php code and run it on your server:

print_r( dirname(__FILE__) );

This will let you know what your absolute path is.

EDIT: to save the image in a database (although I would recommend to upload images to the server and only save the location/name into the DB) you would have to use a blob column type (one of its variations, TINYBLOB, BLOB, MEDIUMBLOB, LONGBLOB ). Mediumblob would allow for file sizes up to 16 MB.

CREATE TABLE images (
    id   INT           AUTO_INCREMENT PRIMARY KEY,
    mime VARCHAR       (255) NOT NULL,
    data MEDIUMBLOB          NOT NULL
);

You would then use something like this to insert an image in the DB (using PDO ):

    $blob = fopen($filePath, 'rb');

    $sql = "INSERT INTO files(mime,data) VALUES(:mime,:data)";
    $stmt = $this->pdo->prepare($sql);

    $stmt->bindParam(':mime', $mime);
    $stmt->bindParam(':data', $blob, PDO::PARAM_LOB);

    $stmt->execute();