我使用Laravel雄辩的查询构建器，我有一个查询，我想在多个条件上有一个where子句。它能起作用，但并不优雅。
例子:
$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，或者我应该坚持这个方法?
当前回答
使用这个
$users = DB::table('users')
->where('votes', '>', 100)
->orWhere('name', 'John')
->get();
其他回答
$variable = array('this' => 1,
'that' => 1
'that' => 1,
'this_too' => 1,
'that_too' => 1,
'this_as_well' => 1,
'that_as_well' => 1,
'this_one_too' => 1,
'that_one_too' => 1,
'this_one_as_well' => 1,
'that_one_as_well' => 1);
foreach ($variable as $key => $value) {
User::where($key, '=', $value);
}
我们使用这条指令根据用户类型分类和用户名两个条件来获取用户。
在这里，除了从profiles表中获取用户信息外，我们还使用两个条件在输入时进行过滤，以减少查询次数。
$users = $this->user->where([
['name','LIKE','%'.$request->name.'%'],
['trainers_id','=',$request->trainers_id]
])->with('profiles')->paginate(10);
如果你的条件是这样的(匹配一个值)，一个简单的更优雅的方式是:
$results = User::where([
'this' => value,
'that' => value,
'this_too' => value,
...
])
->get();
但如果你需要OR子句，那么确保你对每个OR where()子句重复必须满足条件。
$player = Player::where([
'name' => $name,
'team_id' => $team_id
])
->orWhere([
['nickname', $nickname],
['team_id', $team_id]
])
$projects = DB::table('projects')->where([['title','like','%'.$input.'%'],
['status','<>','Pending'],
['status','<>','Not Available']])
->orwhere([['owner', 'like', '%'.$input.'%'],
['status','<>','Pending'],
['status','<>','Not Available']])->get();
在Eloquent中，你可以这样做:
$results = User::where('this', '=', 1)
->orWhere('that', '=', 1)
->orWhere('this_too', '=', 1)
->orWhere('that_too', '=', 1)
->orWhere('this_as_well', '=', 1)
->orWhere('that_as_well', '=', 1)
->orWhere('this_one_too', '=', 1)
->orWhere('that_one_too', '=', 1)
->orWhere('this_one_as_well', '=', 1)
->orWhere('that_one_as_well', '=', 1)
->get();
