创建页面改变形象名称的问题Laravel

0

的问题

这是在创建网页,并希望改变形象名称遵循的视频id在Laravel项目。 但是,当我上传和保存图像在创建页面,保存在数据库中的名称成为/tmp/phpRXDHFh,我想像的姓名保存在数据库作为公共/视频/themnull/视频-id.png。 如何解决这个?

这里是代码的控制器

public function store(Request $request)
{

    $data = $request->validate([
        'video_path'   => 'nullable|mimes:mp4,avi,mpeg,ogx,oga,ogv,ogg,webm| max:25600',
        'user_id'      => 'required | integer',
        'category_id'  => 'required | integer',
        'video_section_id'  => 'required | string',
        'title'         => 'required | string',
        'post_date'    => 'nullable | date',
        'description'   => 'nullable | string',
        'youtube_link'  => 'nullable | string',
        'image_path'    => 'nullable | image |  mimes:jpeg,png,jpg,gif | max:2140',
    ]);


    if ($request->check_type == 1) {

        $data['type'] = 'link';
    }

    $data['post_date'] = $request->post_date ?? date('Y-m-d');

    if ($request->hasFile('video_path') != '') {

        $art_video = $request->file('video_path');
        $ad_video_name = uniqid('video_') . Str::random('10') . '.' . $art_video->getClientOriginalExtension();
        $created_id = 0;

        $video_image_path = $request->file('image_path');
        $video_image_name =  $art_video->id. '.' . $video_image_path->getClientOriginalExtension();
        $video_image_path_resize = Image::make($video_image_path->getRealPath());
         $video_image_path_resize->resize(400, 200);
        if ($video_image_path->isValid()) {

            $video_image_path_resize->save(public_path('video/themnull/' . $video_image_name));
            $video_image_path = 'public/video/themnull/' . $video_image_name;
            $data['image_path'] = $video_image_path;
        }

        if ($art_video->isValid()) {
            $fileName = $ad_video_name;
            $foldername = '/video';
            $name = $fileName;
            $video_path = $art_video->storeAs($foldername, $name, 'public');

            $data['youtube_link'] = null;
            $data['type'] = 'directly';
            $data['video_path'] = 'public/' . $video_path;
            $created_id = Video::create($data);
        }
    } else {
        $created_id = Video::create($data);
    }
    if ($request->filled('tag_name')) {
        if ($created_id) {
            $last_inserted_id = $created_id->id;
            foreach ($request->tag_name as $tag_name) {
                $data2 = ([
                    'video_id' => $last_inserted_id,
                    'tag_name' => $tag_name
                ]);
                $created_id = VideoTag::create($data2);
            }
        }
    }
    try {
        $this->successfullymessage('Video  Added successfully ');
        return redirect()->back();
    } catch (\Exception $e) {
        $this->failmessage($e->getMessage());
        return redirect()->back();
    }
}
file-upload image image-upload laravel
2021-11-24 04:55:44
1

最好的答案

0
$extension = $request->file('image_path')->getClientOriginalName();
$fileName = time().'-'.$art_video->id..'.'.$extension; // you can use time with name if not, then use only $art_video->id..'.'.$extension;.
       
// $file->move('video/thumbnail',$fileName); to store in public folder

//  If you want to keep files in storage folder, you can use following : -

Storage::disk('public')->put('video/thumbnail/'.$fileName,File::get($request->file('image_path')));

// Dont't forget to run 'php artisan storage:link'
// It will store into your storage folder and you can access it by Storage::url('file_path)
2021-11-24 05:41:42

嗨,我需要放或更换?
Sam

从那里你是修改的图象的姓名和储存你的形象。 $video_image_path = $request->file('image_path'); $video_image_name = $art_video->id. '.' . $video_image_path->getClientOriginalExtension(); 把这个不是 $video_image_path$video_image_name 在顶 $extension = $request->file('image_path')->getClientOriginalName(); $fileName = time().'-'.$art_video->id..'.'.$extension;
Masudul Hasan Shawon

可能你想要储存你的图片在这里 $video_image_path_resize->save(public_path('video/themnull/' . $video_image_name)); 所以替换以下有- Storage::disk('public')->put('video/thumbnail/'.$fileName,File::get($request->file('image_path'))); 和检查你的储存文件。
Masudul Hasan Shawon

嗨,仍然同样的图像的名字存储在数据库仍/tmp/phpPfJm8Q
Sam

我需要的名字图像存储在数据库(image_path)公共/视频/themnull/视频-id.png
Sam

因为你储存 $data 而不是修改的数据。 这就是为什么该系统得到一个临时的名称。 商店 'video/thumbnail'."/".$fileName 你的数据库。 喜欢,你可以储存 Video::create(['file_path' => 'video/thumbnail'."/".$fileName, 'foo' => 'bar', '...' => '...']).
Masudul Hasan Shawon

所以意思是,我需要改变$data['youtube_link']=null;$data['type']='直接';$data['video_path']='公共/'. $video_path;$created_id=视频:创建($data);
Sam

嗨Masudul,这样如何?
Sam

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