类App\Models\KamarModel的对象无法转换为代码点火器中的字符串

Question

我试图编辑现有数据，但我们想更新它，出现了错误。如何将此图像对象转换为字符串，以便将其插入phpmyadmin

我的控制器

public function update($id){
$gambarLama = new KamarModel();
$kamarGambar = $gambarLama->find($id);
$file = $this->request->getFile('gambar');
if($file->isValid() && !$file->hasMoved()){
$gambarLama = $kamarGambar['gambar'];
if(file_exists("images/".$gambarLama)){
unlink("images/" . $gambarLama);
}
$imageName = $file->getRandomName();
$file->move("images/",$imageName);
}else{
$imageName = $gambarLama;
}
$data = [
'nama_kamar'        => $this->request->getPost('nama_kamar'),
'deskripsi'         => $this->request->getPost('deskripsi'),
'tipe_kamar'        => $this->request->getPost('tipe_kamar'),
'harga_kamar'       => $this->request->getPost('harga_kamar'),
'status'            => $this->request->getPost('status'),
'fasilitas'         => $this->request->getPost('fasilitas'),
'gambar'            => $imageName
];
$gambarLama->update($id,$data);
return redirect()->to('/dataHotel');
}

我的路线$routes->post('/editKamar/(：num('，'Admin:：update/$1'(；

Answer 1

您首先需要它的文件名，然后移动到上传文件夹

if (isset($_FILES['image'])) {

$uploadFolder=ROOTPATH."upload";

$avatar = $this->request->getFile('image');
$avatar->move($uploadFolder, time() . '.' . $avatar->getClientExtension());
$pathStoreToDatabase= $avatar->getName();

}
```
and remember use post method