Laravel 如何从雄辩的选择中生成变量
我有来自表petugas的selectid的代码,还有来自auth的where语句Laravel 如何从雄辩的选择中生成变量,laravel,variables,eloquent,laravel-authentication,Laravel,Variables,Eloquent,Laravel Authentication,我有来自表petugas的selectid的代码,还有来自auth的where语句 $petugas = petugas::where('ID_PETUGAS','=',auth()->user()->id); 我想把这个id放到这个密码里 Peminjaman::create([ 'ID_ANGGOTA' => $request->ID_ANGGOTA, 'ID_BUKU' => $request->ID_B
$petugas = petugas::where('ID_PETUGAS','=',auth()->user()->id);
我想把这个id放到这个密码里
Peminjaman::create([
'ID_ANGGOTA' => $request->ID_ANGGOTA,
'ID_BUKU' => $request->ID_BUKU,
'ID_PETUGAS' => $petugas->ID_PETUGAS, --> to this
'TANGGAL_PINJAM' => $request->TANGGAL_PINJAM,
'TANGGAL_KEMBALI' => $request->TANGGAL_KEMBALI
]);
如何输入这些变量?使用first()方法:
然后
我收到一条消息,“试图获取非对象的属性'ID_PETUGAS'”表中是否存在字段ID_PETUGAS?您必须将“id_petugas”替换为所需的字段。很抱歉,我在阅读右栏时出错,您的代码工作正常,谢谢。。。
$petugas = petugas::where('ID_PETUGAS','=',auth()->user()->id)->first();
echo $petugas->ID_PETUGAS;