我使用VichUploaderBundle上传图片到AmazonS3 symfony中-2。

I am using VichUploaderBundle to upload images to AmazonS3 in symfony-2.

我按照本文档的 https://github.com /dustin10/VichUploaderBundle/blob/master/Resources/doc/index.md 并建立了我的实体类。

I have followed this documentation https://github.com/dustin10/VichUploaderBundle/blob/master/Resources/doc/index.md and created my entity class.

所以我在实体有一个setter方法​​

So I have a setter method in the entity

/**
 * @param UploadedFile $file
 */
public function setFile(File $file = null)
{
    $this->file = $file;
    $this->updated = new \DateTime();
}

客户端（这是一个Web应用程序）将发送的图像base_64格式。 所以，我不知道我将如何得到一个文件对象从该字符串的？ （由于setter方法​​的参数是FILE）

The client (This is a web application) will be sending the image in base_64 format. So I dont understand how will i get a FILE object out of that string? (Since the parameter of setter method is FILE)

推荐答案

好了，所以你首先需要将图像保存到一个临时文件 - 这里是一个解决方案：How保存PNG图像服务器端，从一个base64数据串

Ok so first you need to save the image to a temporary file - here is a solution for that: How to save a PNG image server-side, from a base64 data string

然后，你需要使用该文件创建UploadedFile的实例，有点像这个例子： https://github.com/dustin10/VichUploaderBundle/issues/203

Then you need to create an instance of an UploadedFile using that file, kind of like this example: https://github.com/dustin10/VichUploaderBundle/issues/203

然后发送这UploadedFile的给您setFile（）方法。

And then send this UploadedFile to your setFile() method.