如何使用 laravel 8 同时保存一对多关系数据
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【中文标题】如何使用 laravel 8 同时保存一对多关系数据【英文标题】：How to save One to Many relationqhip data at same time using laravel 8 【发布时间】：2021-08-29 01:55:10 【问题描述】：我正在尝试将数据存储到数据库中，但遇到了一种情况，我不知道这样做是否可行。
所以我在数据库中有两个下表
public function up()
Schema::create('castings', function (Blueprint $table)
$table->increments('id_casting');
$table->string('nom');
$table->string('prenom');
$table->string('cine');
$table->string('date_naissance');
$table->integer('lieu_naissance');
$table->string('mineur');
$table->integer('id_representant')->unsigned();
$table->timestamps();
$table->foreign('id_representant')->references('id_representant')->on('representants');
);
还有表格代表：
public function up()
Schema::create('representants', function (Blueprint $table)
$table->increments('id_representant');
$table->string('nom_prenom');
$table->string('cine');
$table->string('lien_casting');
$table->integer('actif');
$table->timestamps();
);
我的模特：
class Casting Model
use HasFactory;
extends protected $fillable = ["id_casting",
"nom", "prenom" , "cine" , "date_naissance","lieu_naissance" ,"mineur"
];
public function representants()
return $this->belongsTo('App\Representant');
还有模特代表：
class Representant extends Model
use HasFactory;
protected $fillable = ["id_representant",
"nom_prenom", "cine" , "lien_casting","actif"
];
function castings()
return $this->hasMany('Casting');
我有以下表格：
<form id="castingform" method="post" action="castingss" enctype="multipart/form-data">
csrf_field()
<input type="hidden" id="id_hidden" name="id" />
<div class="form-row">
<div class="form-group col-md-6">
<label for="casting_name">Nom</label>
<input type="text" class="form-control" id="nom" name="nom" placeholder="Nom" >
</div>
<div class="form-group col-md-6">
<label for="casting_name">Prénom</label>
<input type="text" class="form-control" id="prenom" name="prenom" placeholder="Prenom" >
</div>
</div>
<div class="form-row">
<div class="form-group col-md-6">
<label for="casting_cin">CIN</label>
<input type="text" class="form-control" id="cine" name="cine" placeholder="Cin">
</div>
</div>
<div class="form-row">
<div class="form-group col-md-6">
<label for="casting_date_naissancee">Date de naissance</label>
<div class="input-group date">
<input type="text" class="form-control" id="date_naissance" name="date_naissance">
<span class="input-group-text input-group-append input-group-addon" id="spanEstPaymentDate">
<i class="simple-icon-calendar"></i>
</span>
</div>
</div>
<div class="form-group col-md-6">
<label for="casting_lieu_naissance">Lieu de naissance</label>
<input type="text" class="form-control" id="lieu_naissance" name="lien_naissance" placeholder="Lieu de naissance">
</div>
</div>
<div class="form-group col-md-6 js-guarantor-container" hidden>
<label for="guarantor">Nom et prénom du représentatnt légal </label>
<input type="text" class="form-control" id="nom_prenom" name="nom_prenom" placeholder="Représentant">
</div>
<div class="form-group col-md-6 js-guarantor_identification_number-container" hidden>
<label for="guarantor_identification_number">CIN du représentatnt légal</label>
<input type="number" class="form-control" id="cine" name="cine" placeholder="CINE">
</div>
<div class="form-group col-md-6 js-guarantor_lien-container" hidden>
<label for="guarantor_identification_number">Lien</label>
<input type="text" class="form-control" id="lien_casting" name="lien_casting" placeholder="Lien">
</div>
<div class="form-group" align="center">
<input type="hidden" name="action" id="action" />
<input type="hidden" name="hidden_id" id="hidden_id" />
<input type="submit" name="action_button" id="action_button" class="btn btn-warning" value="ADD" />
</div>
<div class=" col-md-6">
<span id="form_result"></span>
</div>
</form>
我设法将数据插入到单个表中，但我发现这个级别的复杂性：
在我的表格中插入时，当我从出生日期计算年龄并且发现年龄小于 18 岁时，我必须在铸件表中插入代表的 id 并插入表中代表信息名称，代表电影和链接，同时代表。
有可能做到吗？如果您有任何可以帮助我的想法，我将不胜感激。
更新
我在控制器中做了以下代码：
public function store(Request $request)
$representants = new Representant();
$castings = new Casting();
$date = $request['date_naissance']; // 24/8/1995
$age = Carbon::now()->format('Y')- Carbon::createFromFormat('dd/mm/YYYY', $date)->format('Y');
if ($age < 18)
$representants->nom_prenom = $request['nom_prenom'];
$representants->cine = $request['cine'];
$representants->lien_casting = $request['lien_casting'];
// assign your data to the models needed
$representants->save();
$castings->nom = $request['nom'];
$castings->prenom = $request['prenom'];
$castings->cine = $request['cine'];
$castings->date_naissance = $request['date_naissance'];
$castings->lieu_naissance = $request['lieu_naissance'];
// $representant->lien_casting = $request['lien_casting'];
$castings->representant_id = $representants->representant_id;
$castings->save();
else
$castings->nom = $request['nom'];
$castings->prenom = $request['prenom'];
$castings->cine = $request['cine'];
$castings->date_naissance = $request['date_naissance'];
$castings->lieu_naissance = $request['lieu_naissance'];
但我收到以下错误：
message: "Unexpected data found. ↵Data missing",…
exception: "Carbon\\Exceptions\\InvalidFormatException"
file: "D:\\Projet_Cast_Infl\\vendor\\nesbot\\carbon\\src\\Carbon\\Traits\\Creator.php"
line: 643
message: "Unexpected data found.\r\nData missing"
trace: [file: "D:\Projet_Cast_Infl\vendor\nesbot\carbon\src\Carbon\Traits\Creator.php", line: 665,…,…]
【问题讨论】：
【参考方案1】：我想你可以！ :) 但是你想要的大部分应该在控制器中完成。
yourController.php（您要发布到的控制器）
public function store(Request $request)
$representant = new Representant();
$castings = new Casting();
$date = $request['casting_date_naissance'] // 24/8/1995
$age = Carbon::now()->format('Y') - Carbon::createFromFormat('d/m/Y', $date)->format('Y');
if ($age < 18)
$representant->nom = $casting->nom = $request['casting_name'];
// assign your data to the models needed
$representant->save();
$casting->representant_id = $representant->id;
$casting->save()
【讨论】：
感谢您的帮助，请查看 mu 更新Carbon::createFromFormat('d/m/Y') 尝试将您的d/m/Y 更改为您的日期输入方式，例外是输入日期和日期格式之间存在差异。以上是关于如何使用 laravel 8 同时保存一对多关系数据的主要内容，如果未能解决你的问题，请参考以下文章
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