Question

我从用户表单中获取图像。我需要调整图像大小并将其保存到public/images。图片的链接必须保存到数据库中。现在，我已经临时存储了数据库中存储的/tmp/phpNgculU文件。因此，我无法在浏览器中显示图像。帮助我更改代码。我真的不明白我的错误在哪里。这需要在没有php artisan storage:link

的情况下完成

这是我控制器的“更新”方法的一部分：

if(!empty($request->file())) {
    $task->images = $request->file('file');
    $img = Image::make($task->images)->resize(320, 240, function ($constraint) {
        $constraint->aspectRatio();
    });
    $imgName = rand(11111, 99999).'.'.$task->images- >getClientOriginalExtension();
    $img->storeAs('images/', $imgName);
}

我也收到错误消息：

命令（StoreAs）对驱动程序（Gd）不可用。

Answer 1

您可以这样做：

如果要存储上载图像的路径，则需要一个数据库表。

Schema::create('images', function (Blueprint $table) {
            $table->increments('id');
            $table->string('name');
            $table->string('path');
            $table->timestamps();
        });

还有Image模型：

<?php
namespace App;
use Illuminate\Database\Eloquent\Model;
class Image extends Model
{
    protected $table = 'images';
    protected $fillable = [
      'path',
      'name',
    ];
}

在repository目录中创建一个app\Repositories（可能您也需要创建目录）：

namespace App\Repositories;

use Illuminate\Support\Facades\Storage;
use Symfony\Component\HttpFoundation\Response;

class ImageRepository
{
    /**
     * Upload the image
     * @param $image
     * @return \Illuminate\Contracts\Routing\ResponseFactory|\Illuminate\Http\Response
     */
    public function uploadImage($image, $name=null)
    {
        return $this->upload($image);
    }
    /**
     * Upload the image
     *
     * @return \Illuminate\Contracts\Routing\ResponseFactory|\Illuminate\Http\Response
     */
    private function upload($image, $name=null)
    {
        try{
            $name == null ? $name = uniqid() : $name = $name;
            $path = Storage::disk('public')->put('images', $image);
            $uploadedImage = Image::create([
                'path' => $path,
                'name' => $name,
            ]);
            return $uploadedImage;
        }catch (\Exception $exception){
            return response('Internal Server Error', Response::HTTP_INTERNAL_SERVER_ERROR);
        }
    }
}

您的控制器代码：

//Create a variable to the image repository:
private $imageRepository;

public function __construct(ImageRepository $imageRepository)
{
    $this->imageRepository = $imageRepository;
}

//In your update function:
public function update(Request $request)
{
    $image = Input::file('file'); //i think you call your input on the html form as 'file'
    $img = Image::make($image)->resize(320, 240, function ($constraint) {
        $constraint->aspectRatio();
    });
    $imgName = rand(11111, 99999).'.'.$image->getClientOriginalExtension();
    $this->imageRepository->uploadImage($img, $imgName);

    //Continue your logic here
}

记住：您需要将enctype添加到HTML表单：

enctype =“ multipart / form-data”

希望有帮助。

Answer 2

您可以使用Laravel File Storage

HTML表格

 <form method="post" action="/upload" enctype="multipart/form-data">
        {{ csrf_field() }}
        <input type="file" name="image">
        <button type="submit">Submit</button>
    </form>

图像方法助手

public static function saveImage($image, $path, $option){
    try {
        if($image != null) {
            $image_path = $image->store($path, $option);
            return $image_path;
        } else {
            return null;
        }
    } catch (\Exception $e) {
        echo 'Image Helper saveImage ' .$e->getMessage();
        }
    }

控制器

public function store(Request $request){
        $category = new Category;
        $category->name = $request->name;
            if($request->has('image')){
                $category->image_path = saveImage($request->file('image'),'category', 'public');
            }
        $category->save();
        return $category
    }

如何在Laravel中保存图像？

2 个答案: