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如何从Laravel上的数组图像中获取getClientOriginalName()?_Laravel - Fatal编程技术网

如何从Laravel上的数组图像中获取getClientOriginalName()?

如何从Laravel上的数组图像中获取getClientOriginalName()?,laravel,Laravel,我有一个来自请求的数组,如下。。。。我有order_details表,我想更新列类别、备注和foto数据,其中有我从请求中获得的orderid和skuid array:7 [▼ "orderid" => "191121120129KUR" "_token" => "1ov7aCP8JzzokWNfMpwntTH0Dv084HqwIqMUfLd1" "orderdet" => array:5 [▼ 0 => array:4 [▼ "skuid

我有一个来自请求的数组,如下。。。。我有order_details表,我想更新列类别、备注和foto数据,其中有我从请求中获得的orderid和skuid

array:7 [▼
  "orderid" => "191121120129KUR"
  "_token" => "1ov7aCP8JzzokWNfMpwntTH0Dv084HqwIqMUfLd1"
  "orderdet" => array:5 [▼
    0 => array:4 [▼
      "skuid" => "10006"
      "category" => array:2 [▼
        0 => "1"
        1 => "2"
      ]
      "remark" => "remarks 1"
      "foto" => array:2 [▼ 
        0 => Illuminate\Http\UploadedFile {#348 ▶}
        1 => Illuminate\Http\UploadedFile {#349 ▶}
      ]
    ]
    1 => array:4 [▼
      "skuid" => "10007"
      "category" => array:1 [▼
        0 => "1"
      ]
      "remark" => "remarks 2"
      "foto" => array:4 [▼
        0 => Illuminate\Http\UploadedFile {#350 ▶}
        1 => Illuminate\Http\UploadedFile {#351 ▶}
        2 => Illuminate\Http\UploadedFile {#352 ▶}
        3 => Illuminate\Http\UploadedFile {#353 ▶}
      ]
    ]
    2 => array:2 [▼
      "skuid" => "10008"
      "remark" => null
    ]
    3 => array:2 [▼
      "skuid" => "10138"
      "remark" => null
    ]
    4 => array:2 [▼
      "skuid" => "10078"
      "remark" => null
    ]
  ]
  "actioncomplaint" => "1"
  "salahsiapa" => "3"
  "remarkcust" => null
  "remarkfresh" => "Dani"
]
我想使用getClientOriginalName()存储图像的名称,。。。 如何获得图像的名称,这是我的控制器

foreach($request->orderdet as  $row)
        {
            $data = OrderDetail::where('data_order_id', $request->orderid)->where('skuid', $row['skuid'])->first();
            if(isset($row['category'])) {

                $data->remarkscomplaint = $row['remark'];
                $data->categorycomplaint = implode(',',  $row['category']);
                $data->fotocomplaint = $row['foto'];
                $data->save();
            }
        }
尝试以下方法:

$fotos = $request->only(['foto']);
$fotoNames = []
foreach($fotos as $foto) {
 $fotoNames[] = $foto->getClientOriginalName();
}

$implodedFotos = implode(',' $fotoNames);
然后

然后为了保存你的图像会帮助你

$foto->storeAs('path', 'filename');
尝试以下方法:

$fotos = $request->only(['foto']);
$fotoNames = []
foreach($fotos as $foto) {
 $fotoNames[] = $foto->getClientOriginalName();
}

$implodedFotos = implode(',' $fotoNames);
然后

然后为了保存你的图像会帮助你

$foto->storeAs('path', 'filename');

使用html表单提交文件时,只能使用
getClientOriginalName()
。保存文件后,以后将无法获取原始文件名。在数据库中存储文件时,请尝试保存文件名。使用html表单提交文件时,只能使用
getClientOriginalName()
。保存文件后,以后将无法获取原始文件名。在DB.foreach($fotos as$foto)中存储时尝试保存文件名{$fotoNames[]=$foto->getClientOriginalName();//$foto->move(public_path()。/images/complaint/,$fotoNames);}是否保存?文件图像?sirforeach($fotos as$foto){$fotoNames[]=$foto->getClientOriginalName();//$foto->move(public_path()。/images/complaint/,$fotoNames);}是否保存?文件图像?先生