本文介绍了如何从laravel雄辩地从多个表中检索数据的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
Model relation
---------------------
language.php
----
public function attributeDetail()
{
return $this->hasMany(AttributeDetail::class, 'language_id');
}
attribute.php
----
public function attributeDetail()
{
return $this->hasMany(AttributeDetail::class, 'attribute_id');
}
attributeDetail.php
----
public function language()
{
return $this->belongsTo(Language::class);
}
public function attribute()
{
return $this->belongsTo(Attribute::class);
}
我想显示这样的json对象
I want to show the json object like this
{
'attribute_id' => 101,
'available_language' => [
{'id' => 1,'language_name' => 'English'},
{'id' => 2,'language_name' => 'French'}
],
}
表结构:
languages(`id`, `language_name`, `translate_version`, `is_default`, `status`);
attributes(`id`, `required`, `type`, `status`);
attributedetails(id`, `attribute_id`, `language_id`, `attribute_name`, `status`);
推荐答案
尝试类似的东西
$results = Attribute::select('id')->with(['attributeDetail.language' => function ($query) {
$query->select('id', 'language_name');
}])->get();
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