Question

我有一个模型，打算将输入保存到两个不同的表中，但是有错误

Symfony \ Component \ Debug \ Exception \ FatalThrowableError参数1 传递给Illuminate \ Database \ Eloquent \ Builder :: create（）的类型数组，给定的对象，在 C：\ xampp \ htdocs \ blog \供应商\ laravel \ framework \ src \ Illuminate \ Support \ Traits \ ForwardsCalls.php 在第23行

第一个表是Transaction_in，第二个表是Transaction_in_detail。我计划将idTransaction_in用作2个表之间的连接器，并且该列不是primaryKey（这是一种好习惯）吗？我可以将输入保存到Transcation_in表中，但是仍然无法将输入保存到第二表中，而且我不知道如何将idTransaction_in列用作连接器。

public function store(Request $request)
{
    $request->validate([
        'supplier_name' => 'required',
        'transaction_in_date' => 'required|before_or_equal:today',
        'device_type_name' => 'required',
        'device_brand_name' => 'required',
        'device_spec' => 'required|max:255',
        'price' => 'required',
        'amount' => 'required',
        'total_price' => 'required',
        'keterangan' => 'Nullable',

        ]);


    $transaction_in = new Transaction_in();
    $transaction_in->idTransaction_in = "0";
    $transaction_in->Supplier_id = $request->input('supplier_name');
    $transaction_in->tanggal_transaksi = $request->input('transaction_in_date');
    $transaction_in->save();
    $transaction_in->update(['idTransaction_in' => sprintf('TIN-%04d', $transaction_in->id)]);

    $lastid=Transaction_in::create($transaction_in)->idTransaction_in;
    if(count($request->device_type_name)>0){
        foreach ($request->device_type_name as $item => $v) {
            $data2=array(
                'Transaction_in_id' => $lastid,
                'DeviceType_id' => $request->device_type_name[$item],
                'DeviceBrand_id' => $request->device_brand_name[$item],
                'spek_device' => $request->device_spec[$item],
                'harga_device' => $request->price[$item],
                'jumlah_device' => $request->amount[$item],
                'total_harga_device' => $request->total_price[$item]
            );
            Transaction_in_detail::insert($data2);
        }
    }

    return redirect('/transactionsin')->with('success', 'Transaction success');
}

DeviceType_id和Devicebrand_id是外键。

Answer 1

据我所知，您在此行中遇到了问题

$lastid=Transaction_in::create($transaction_in)->idTransaction_in;

$transaction_in是对象，并且create方法要求传入array。

如果其他所有方法都正确，则应该可以：

$lastid=Transaction_in::create($transaction_in->toArray())->idTransaction_in;

Answer 2

您已经save了数据，但也使用了create。它将在您的数据库中创建重复的记录。

我会给你一个解决方案：

$transactionIn = Transaction_in::create([
    'idTransaction_id' => '0',
    'supplier_id' => $request->input('supplier_name') // why supplier name assigned supplier_id?
    'tanggal_transaksi' => $request->input('transaction_in_date'),
]);

$transactionIn = tap($transaction_in)->update([
    'idTransaction_in' => sprintf('TIN-%04d', $transactionIn->id)
]);

$lastId = $transactionIn->idTransaction_in;

传递给Illuminate \ Database \ Eloquent \ Builder :: create（）的参数1必须为数组类型

2 个答案: