我認爲,正確的做法是讓一個Image
模型一個對應的表格,那麼你可以設置它與其他模型的關係。例如:
public function store(Request $request)
{
$model = new RelatedModel(); // This is a related model example
$images = $request->file("image.*");
foreach($images as $uploadedImage)
{
$path = $uploadedImage->store('path/images', 'local'); // disk can be null, it will then use the default disk in filesystems.php
$image = new Image();
// A way you want to use to give the image a name
$image->name = $this->generateName();
$image->path = $path;
// Where relatedModel is the method on Image model defining a belongsTo relation for example with RelatedModel
$image->relatedModel()->associate($model);
$image->save();
}
}
我不知道你爲什麼要按照問題中指定的方式保存圖片。但是,如果你堅持,你必須在你的代碼添加新字段
id | image1 | image1_name | image2 | image2_name ...
然後:
public function store(Request $request)
{
$model=new Model();
// This is a function you would make to generate a different name than the path
$model->image1_name = $this->generateName();
$model->image1 = $request->file("image.0");->store('path/images', 'local');
$model->image2_name = $this->generateName();
$model->image2 = $request->file("image.1");->store('path/images', 'local');
// ...etc.
$model->save();
}
什麼是你需要幫助的具體問題?是的,你可以上傳5張圖片並將他們的名字保存在數據庫中。 –
如果你用不同的id將它們分成不同的記錄會更好嗎?像ID,文件名,圖像(BLOB?),創建時間,更新時間 – Jigs