【发布时间】:2022-01-15 18:03:21
【问题描述】:
我尝试更新包含文件(图像)的表单并删除旧图像。更新工作正常,但旧图像无法删除。我试过这段代码,但图像没有被删除。请帮我。提前致谢。
public function update(Request $request, $id)
{
$slug = SlugService::createSlug(Category::class, 'slug', $request->title);
$request->validate([
'title'=>'required',
'category_image'=>'image'
]);
if ($request->hasFile('category_image')) {
$image = $request->file('category_image');
$newImageName = uniqid().'-'.$request->title.'.'.$image->getClientOriginalExtension();
$location = public_path('/categoryImage');
$OldImage = public_path('categoryImage/'.$request->category_image);
$image->move($location, $newImageName);
Storage::delete($OldImage);
}else {
$newImageName = $request->category_image;
}
Category::where('id', $id)->update([
'slug'=>$slug,
'title'=>$request->input('title'),
'details'=>$request->input('details'),
'category_image'=>$newImageName
]);
return redirect('category')->with('success', 'Category Successfully Updated');
}
【问题讨论】:
-
当我在 if 语句中使用 dd('$request->category_image') 时不会产生图像的名称,而在 else 语句中使用时会显示图像名称。我无法弄清楚我做错了什么。
标签: php sql laravel image slug