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