php 向 Laravel 中的选择查询添加一个带有值的新列
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Add a new column with a value to the select query in Laravel
提问by John Doeherskij
I would like to know how to create a default variable called "type" and set a value to "car" while doing a select join in Laravel.
我想知道如何在 Laravel 中进行选择连接时创建一个名为“type”的默认变量并将值设置为“car”。
Here is my code so far:
到目前为止，这是我的代码：
$items = DB::table('items')->orderBy('created_at', 'desc')
->join('items_categories', 'items.item_category_id', '=', 'items_categories.category_id')
->select(
'items.id as items___item_id',
'items.item_title as items___item_title',
'items_categories.id as items_categories___category_id',
'items_categories.title as items_categories___category_title',
)
->take(20);
This works nice. However, I need to get/add a custom key and value for each record of this select so I can use it later in the template to filter stuff further.
这很好用。但是，我需要为此选择的每条记录获取/添加自定义键和值，以便稍后在模板中使用它来进一步过滤内容。
So, I need to add a key called typewith a value of carso in the print_r I will see type => car for every record and I can use this in my code.
所以，我需要在 print_r 中添加一个type值为carso的键，我会看到每条记录的 type => car ，我可以在我的代码中使用它。
How to do that?
怎么做？
Can I put that somhow in the select?
我可以把那个somhow放在选择中吗？
Like:
喜欢：
->select(
'items.id as items___item_id',
'items.item_title as items___item_title',
'items_categories.id as items_categories___category_id',
'items_categories.title as items_categories___category_title',
//something like this?
'type' = 'car'
)
Because right now I am getting this:
因为现在我得到了这个：
Array
(
[0] => stdClass Object
(
[items___item_id] => 10
[items___item_user_id] => 2
[items___item_title] => A new blue truck
[items_categories___category_id] => 1
[items_categories___category_title] => Truck
)
[1] => stdClass Object
(
[items___item_id] => 11
[items___item_user_id] => 2
[items___item_title] => VW Tiguan
[items_categories___category_id] => 1
[items_categories___category_title] => SUV
)
And I want to get this:
我想得到这个：
Array
(
[0] => stdClass Object
(
[items___item_id] => 10
[items___item_user_id] => 2
[items___item_title] => A new blue truck
[items_categories___category_id] => 1
[items_categories___category_title] => Truck
[type] => car
)
[1] => stdClass Object
(
[items___item_id] => 11
[items___item_user_id] => 2
[items___item_title] => VW Tiguan
[items_categories___category_id] => 1
[items_categories___category_title] => SUV
[type] => car
)
If possible, not in the model file, but during the one query, because it's only one time when I need this modification to be done.
如果可能，不在模型文件中，而是在一次查询中，因为只有一次我需要完成此修改。
回答by W. Dan
You can use this method:
您可以使用此方法：
$data = DB::table('items')
->Select('items.id as items___item_id',
'items.item_title as items___item_title');
# Add fake column you want by this command
$data = $data->addSelect(DB::raw("'fakeValue' as fakeColumn"));
$data = $data->orderBy('items.id')->get();
Enjoy it!
好好享受！
回答by Rob Fonseca
You will want to create a model for your items table and query it that way. Using eloquent, you can create columns on the fly by adding column names to the $appends property and then defining a model attribute.
您将希望为您的 items 表创建一个模型并以这种方式查询它。使用 eloquent，您可以通过将列名添加到 $appends 属性然后定义模型属性来动态创建列。
php artisan make:model Item
Any model automatically looks for a table that is the plural of the model name (Item looks for 'items'). In the Item model, add the following lines
任何模型都会自动查找作为模型名称复数形式的表（Item 查找“items”）。在 Item 模型中，添加以下几行
/**
* Append custom columns to the model
*
* @var array
*/
protected $appends = ['type'];
/**
* Define the type column to every Item object instance
*
* @return string
*/
public function getTypeAttribute()
{
return 'car';
}
Now update your query to use the model instead of DB::select. Make sure to use the model at the top of your controller
现在更新您的查询以使用模型而不是 DB::select。确保使用控制器顶部的模型
use App\Item;
....
$items = Item::orderBy('created_at', 'desc')
->join('items_categories', 'items.item_category_id', '=', 'items_categories.category_id')
->select(
'items.id as items___item_id',
'items.item_title as items___item_title',
'items_categories.id as items_categories___category_id',
'items_categories.title as items_categories___category_title',
)
->take(20)->get();
You need to add get() as the final method when using a Model for it to return a collection vs. DB::select.
当使用模型返回集合与 DB::select 时，您需要添加 get() 作为最终方法。
回答by John Doeherskij
Here is the solution for this problem, hope it helps somebody else in the future.
这是这个问题的解决方案，希望它可以在未来帮助其他人。
$items = DB::table('items')->orderBy('created_at', 'desc')
->join('items_categories', 'items.item_category_id', '=', 'items_categories.category_id')
->select(DB::raw('"car" AS type,
items.id as items___item_id,
items.item_title as items___item_title,
items_categories.id as items_categories___category_id,
items_categories.title as items_categories___category_title
')
)
->take(20);
回答by saber tabatabaee yazdi
i implement it sth like this sample
我像这个示例一样实现它
$first = Message::where('from_id', '=', Auth::user()->id)
->where('to_id', '=', $id)->get();
foreach ($first as $f)
$f->is_sent = true;
$second = Message::where('from_id', '=', $id)
->where('to_id', '=', Auth::user()->id)->get();
foreach ($second as $s)
$s->is_sent = false;
$chats = $first->merge($second)
->sortBy('created_at');
return $chats->toJson();
you can do it by for eachafter get() method
你可以通过 for eachafter get() 方法来完成
