标签: laravel-query-builder

给出以下代码:

DB::table('users')->get();

我想获取上面的数据库查询生成器将生成的原始SQL查询字符串.在这个例子中,它将是SELECT * FROM users.

我该怎么做呢？

我正在使用Laravel Eloquent查询构建器,我有一个查询,我希望WHERE在多个条件下有一个子句.它有效,但它并不优雅.

例:

$results = User::where('this', '=', 1)
    ->where('that', '=', 1)
    ->where('this_too', '=', 1)
    ->where('that_too', '=', 1)
    ->where('this_as_well', '=', 1)
    ->where('that_as_well', '=', 1)
    ->where('this_one_too', '=', 1)
    ->where('that_one_too', '=', 1)
    ->where('this_one_as_well', '=', 1)
    ->where('that_one_as_well', '=', 1)
    ->get();

有没有更好的方法来做到这一点,还是我应该坚持这种方法？

我怎么说 WHERE (a = 1 OR b =1 ) AND (c = 1 OR d = 1)

对于更复杂的查询,我应该使用原始SQL吗？

我想通过使用orderBy()Laravel Eloquent中的方法对Laravel 4中的多个列进行排序.查询将使用Eloquent生成,如下所示:

SELECT *
FROM mytable
ORDER BY
  coloumn1 DESC, coloumn2 ASC

我怎样才能做到这一点？

如何使用Laravel Query Builder或Eloquent ORM在Laravel 3/4中检索原始执行的SQL查询？

例如,像这样:

DB::table('users')->where_status(1)->get();

要么:

(posts (id, user_id, ...))

User::find(1)->posts->get();

否则,至少我如何保存执行到laravel.log的所有查询？

I have two tables, User and Post. One User can have many posts and one post belongs to only one user.

In my User model I have a hasMany relation...

public function post(){
    return $this->hasmany('post');
}

And in my post model I have a belongsTo relation...

public function user(){
    return $this->belongsTo('user');
}

Now I want to join these two tables using Eloquent with() but want specific columns from the second table. I know I can use the Query …

如果新记录不存在,插入新记录或更新的简写是什么？

<?php

$shopOwner = ShopMeta::where('shopId', '=', $theID)
    ->where('metadataKey', '=', 2001)->first();

if ($shopOwner == null) {
    // Insert new record into database
} else {
    // Update the existing record
}

我正在尝试使用Laravel查询生成器的JOIN查询添加条件.

<?php

$results = DB::select('
       SELECT DISTINCT 
          *
          FROM 
             rooms 
                LEFT JOIN bookings  
                   ON rooms.id = bookings.room_type_id
                  AND (  bookings.arrival between ? and ?
                      OR bookings.departure between ? and ? )
          WHERE
                bookings.room_type_id IS NULL
          LIMIT 20',
    array('2012-05-01', '2012-05-10', '2012-05-01', '2012-05-10')
);

我知道我可以使用Raw表达式,但之后会有SQL注入点.我已经尝试了以下查询生成器,但生成的查询(显然,查询结果)不是我想要的:

$results = DB::table('rooms')
    ->distinct()
    ->leftJoin('bookings', function ($join) {
        $join->on('rooms.id', '=', 'bookings.room_type_id');
    })
    ->whereBetween('arrival', array('2012-05-01', '2012-05-10'))
    ->whereBetween('departure', array('2012-05-01', '2012-05-10'))
    ->where('bookings.room_type_id', '=', null)
    ->get();

这是Laravel生成的查询:

select distinct * from `room_type_info`
    left join `bookings` 
on `room_type_info`.`id` = `bookings`.`room_type_id` …

$ book = array('book1','book2'); $ book数组元素数是可变的.它可能有2个元素或20个元素
我需要进行这样的查询:

select * from book where bookname like %book1% or bookname like %book2%

要在laravel 5中进行此查询,可以选择:

$name = DB::Table('bookinfo')
          ->select('*')
          ->wherein('bookname',$book)
          ->get();

但它使用=运算符我需要使用like运算符

我正面临多形关系中的问题,我无法在哪里工作.基本上我有一个"where"条件,我想申请.关系代码正常工作以返回相关模型,但一旦应用whereHas它就会返回错误.

下面是代码

订单类:

class Order extends Model
 {

  // function to return orders 
  public static function getAllOrders()
  { 


    return $orders = Order::with('part.pcategory')->whereHas('part', function ($query) 
         {
                  $query->where('cat_id',4);
         })->get();
  }

  // the relation 
  public function part()
  { 

  return $this->morphTo(null,'department_short_code','part_stock_number','stock_number', 'dep_short_code');
  }

 }

SFD零件类:

class sfd_part extends Model
{

  public function orders()
    {   

    return  $this->morphMany('App\Order','part','department_short_code','part_stock_number');
   }

   public function pcategory()
    {

    return $this->belongsTo('App\Pcategories','cat_id', 'category_id');
    }

}

当我调用getAllOrders()时,它会给出以下错误

SQLSTATE [42S22]:未找到列:1054'where子句'中的未知列'cat_id'(SQL:选择count(*)作为orders存在的聚合(select*from ordersas laravel_reserved_0where laravel_reserved_0.id= laravel_reserved_0.part_stock_number和cat_id …