Question

我想不出如何在Eloquent中处理这个问题。我有一个多对多的关系，需要一对一的关系分配给它。

这是我以最简单的形式设计的基本数据库结构：

using System;
using System.IO;
using System.Security.Cryptography;
using System.Text;

public class Program
{
    public static void Main()
    {

        var key = Encoding.ASCII.GetBytes("0123456789abcdef");//Change the key.

        var data = Encrypt("Hello World",key);
        var str = Decrypt(data, key);
        Console.WriteLine(str);
    }

    public static byte[] Encrypt(string plaintext, byte[] key)
    {

        using(var desObj = Rijndael.Create())
        {
            desObj.Key = key;
            desObj.Mode = CipherMode.CFB;
            desObj.Padding = PaddingMode.PKCS7;
            using(var ms = new MemoryStream())
            {
                //Append the random IV that was generated to the front of the stream.
                ms.Write(desObj.IV, 0, desObj.IV.Length);

                //Write the bytes to be encrypted.
                using(CryptoStream cs = new CryptoStream(ms, desObj.CreateEncryptor(), CryptoStreamMode.Write))
                {
                    var plainTextBytes = Encoding.UTF8.GetBytes(plaintext);
                    cs.Write(plainTextBytes, 0, plainTextBytes.Length);
                }
                return ms.ToArray();
            }
        }
    }

    public static string Decrypt(byte[] cyphertext, byte[] key)
    {

        using(MemoryStream ms = new MemoryStream(cyphertext))
        using(var desObj = Rijndael.Create())
        {
            desObj.Key = key;
            desObj.Mode = CipherMode.CFB;       
            desObj.Padding = PaddingMode.PKCS7;

            //Read the IV from the front of the stream and assign it to our object.
            var iv = new byte[16];
            var offset = 0;
            while(offset < iv.Length)
            {
                offset += ms.Read(iv, offset, iv.Length - offset);
            }
            desObj.IV = iv;

            //Read the bytes to be decrypted
            using(var cs = new CryptoStream(ms, desObj.CreateDecryptor(), CryptoStreamMode.Read))
            using(var sr = new StreamReader(cs, Encoding.UTF8))
            {
                return sr.ReadToEnd();
            }
        }
    }
}

每个帐户属于多个代理商。

每个代理商都属于多个帐户。

每个“Account_Agent”（多对多数据透视表）属于收费。

如何在Eloquent模型中定义第三个关系？

我希望能够像这样（或类似）访问“account_agent”费用：

ACCOUNTS: id

AGENTS: id

FEES: id

ACCOUNT_AGENT: account_id, agent_id, fee_id

谢谢，希望我的问题很明确。

Answer 1

见这里：

https://laravel.com/docs/5.4/eloquent-relationships#many-to-many

向下滚动到定义自定义中间表模型

基本上你要做的就是定义一个这样的类：

class AccountAgent extends Illuminate\Database\Eloquent\Relations\Pivot
{
    /**
     * Many:1 relationship with Fee model
     *
     * @return \Illuminate\Database\Eloquent\Relations\BelongsTo
     */
    public function fee()
    {
        return $this->belongsTo(Fee::class);
    }
}

然后，您在Account类中定义多个：多个关系，如下所示：

/**
 * The agents that belong to the account.
 */
public function agents()
{
    return $this->belongsToMany(Agent::class)->using(AccountAgent::class);
}

然后在你的代码中，而不是这个：

foreach ($account->agents as $agent) {
  echo $agent->fee;
}

......这样做：

foreach ($account->agents as $agent) {
  /** @var Fee $fee_object */
  $fee_object = $agent->pivot->fee;
}

在该循环中，$ fee_object属于费用类（涵盖费用表的模型类），因此您可以echo $fee_object->fee_amount或其他任何需要使用的列。

LARAVEL：数据透视表上的一对一关系？

1 个答案: