如何使用 cloudinary + Laravel 上传图片？答案

【问题标题】：How Upload images with cloudinary + Laravel?如何使用 cloudinary + Laravel 上传图片？
【发布时间】：2021-09-29 07:38:11
【问题描述】：

好吧，我想通过 Laravel 将图像上传到 Cloudinary，我按照文档中所述的步骤进行操作：

https://github.com/cloudinary-labs/cloudinary-laravel 在这里：https://cloudinary.com/blog/laravel_file_upload_to_a_local_server_or_to_the_cloud

我正在使用 Laravel 8，这是我的代码：

 public function store(Request $request){

    $request->validate([
        'name' => 'required|string|unique:festivals|max:255', //unique:table
        'description' => 'required|string|max:255',
        'image' => 'required|image|dimensions:min_width=200,min_height=200',
    ], self::$messages);

    //Here I create an instance and upload file to server 
    $festival = new Festival($request->all());
    $path = cloudinary()->upload($request->file('image')->getRealPath())->getSecurePath();
    dd($path);

    //Then at field image of festivals que save the path and that goes to the database
    $festival->image = 'festivals/' . basename($path);
    $festival->save();


    return response() -> json($festival, 201); //code 201 created

}

当我尝试通过 Postman 创建新记录时，会发生这种情况：

Error after register new festival

但是，图片已上传到 Cloudinary：

image uploaded

然后我尝试检查记录是否已创建，但没有创建。

我能做什么，有人知道吗？

谢谢

【问题讨论】：

标签： php laravel cloudinary

【解决方案1】：

好的，我解决了:)

    //Here I create an instance and upload file to server
    $festival = new Festival($request->all());
    $result = $request->image->storeOnCloudinary();
     
    //Here I get the url
    $path = $result->getPath();

    //Then at field image of festivals que save the path and that goes to the database

    //$festival->image = $path; output: "https://res.cloudinary.com/test/image/upload/v1226931211/zuiyb17vfs8dre6gvuov.jpg"
    //$festival->image = dirname($path); output: "https://res.cloudinary.com/test/image/upload/v1526711711"
    //$festival->image = basename($path); output: "zuiy31lvfs4dre5gvuov.jpg"
    $base =  basename(dirname($path).basename($path)); //output: "v1636941221zuiyb14vfsmdre6gvuov.jpg"
    $folder =  substr($base, 0, -24); //output: "v1626915261"
    //So:
    $img_path = $folder.'/'.basename($path);

    $festival->image = 'festivals/'.$img_path;  //imageFolder/imageName.jpg;

但是如果有人有最好的方法，请告诉我

【讨论】：