我已经在模型和Match
模型TokenMatch
的两个模型之间创建了"一对一"的雄辩关系,但是当我尝试将TokenMatch与匹配相关联时,我遇到了一个错误:
"Field 'tokenmatch_id' doesn't have a default value (SQL: insert into `matches` (`id`, `updated_at`, `created_at`) values (, 2019-04-22 08:55:51, 2019-04-22 08:55:51))
"
我的代码问题:
$match = new Match();
$tokenmatch = TokenMatch::find(1);
$match->token()->associate($tokenmatch)->save();
类匹配
class Match extends Model
{
public function token()
{
return $this->belongsTo('AppTokenMatch' , 'id', 'tokenmatch_id');
}
}
类令牌匹配
class TokenMatch extends Model
{
protected $table = 'tokensmatch';
public function match()
{
return $this->hasOne('AppMatch','tokenmatch_id');
}
}
匹配表
Schema::create('matches', function (Blueprint $table) {
$table->bigIncrements('id');
$table->boolean('isFinished')->nullable();
$table->boolean('isWon')->nullable();
$table->unsignedBigInteger('tokenmatch_id');
$table->foreign('tokenmatch_id')->references('id')->on('tokensmatch');
$table->timestamp('created_at')->default(IlluminateSupportFacadesDB::raw('CURRENT_TIMESTAMP'));
$table->timestamp('updated_at')->default(IlluminateSupportFacadesDB::raw('CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP'));
});
令牌匹配表
Schema::create('tokensmatch', function (Blueprint $table) {
$table->bigIncrements('id');
$table->boolean('isUsed')->default(false);
$table->string('token', 15)->unique();
$table->dateTimeTz('expiryDate')->nullable();
$table->boolean('isValid')->default(true);
});
所以我期望当我保存新的模型匹配时,在表的"tokenmatch_id"字段上匹配不为空
...
$match->token()->associate($tokenmatch)->save();
...
我认为您在模型中定义的关系存在问题Match
。它应该是:
public function token()
{
return $this->belongsTo('AppTokenMatch' ,'tokenmatch_id', 'id');
}
请参考:
https://laravel.com/docs/5.4/eloquent-relationships#one-to-many-inverse
更改
$table->unsignedBigInteger('tokenmatch_id');
自
$table->unsignedBigInteger('tokenmatch_id')->nullable();