基于条件的方法链接 [英] Method Chaining based on condition

查看:59
本文介绍了基于条件的方法链接的处理方法,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

如何根据laravel 4中的条件进行方法链接?如果一个值不为假,则内部方法将被链接到if语句之前调用的方法.

How can you do method chaining based on condition in laravel 4 ? Say if one value is not false then the method inside will be chained to the method called before the if statement.

laravel有可能吗?

Is it possible in laravel?

$data = User::where('username', $somevariable );

if(isset( $somevar_again ))
{
  $data->where('age', 21);
}
$data->orderBy('reg_date', 'DESC')->get();
return $data->first();

//上面的代码已经试过,它给了我错误的结果 在codeigniter中,我可以做到

// tried code above and its giving me wrong result in codeigniter I can do this

$this->db->select('e.*, v.name_en as v_name_en')
    ->from($this->table_name . ' e, ' . $this->ptc_venues . ' v');
  $this->db->where('e.venue_id_en = v.id'); 

  if(isset($search)){
   $this->db->where('(v.name_en LIKE "%'.$search.'%")');
  }

  $this->db->limit($limit, $start);
  $this->db->order_by('e.added_date_en', 'DESC');

推荐答案

我相信您的问题发生了,因为您没有在每次查询生成器方法调用后都将结果查询存储回去.

I believe your problem happened because you didn't store back the resulting query after each query builder method call.

$query = User::query();

// Checking for username if exists
if (!empty($username)) {
    $query = $query->where('username', $username);
}

// Check for age if exists
if (isset($age)) {
    $query = $query->where('age', $age);
}

// Ordering
$query = $query->orderBy('reg_date', 'DESC');

// Get the first result
// After this call, it is now an Eloquent model
$user = $query->first();

var_dump($user);

这篇关于基于条件的方法链接的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆