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Array的条件:
$users = User::where([
'column1' => value1,
'column2' => value2,
'column3' => value3
])->get();
将产生如下查询:
SELECT * FROM TABLE WHERE column1 = value1 and column2 = value2 and column3 = value3
使用匿名函数的条件:
$users = User::where('column1', '=', value1)
->where(function($query) use ($variable1,$variable2){
$query->where('column2','=',$variable1)
->orWhere('column3','=',$variable2);
})
->where(function($query2) use ($variable1,$variable2){
$query2->where('column4','=',$variable1)
->where('column5','=',$variable2);
})->get();
将产生如下查询:
SELECT * FROM TABLE WHERE column1 = value1 and (column2 = value2 or column3 = value3) and (column4 = value4 and column5 = value5)
其他回答
多个where子句
$query=DB::table('users')
->whereRaw("users.id BETWEEN 1003 AND 1004")
->whereNotIn('users.id', [1005,1006,1007])
->whereIn('users.id', [1008,1009,1010]);
$query->where(function($query2) use ($value)
{
$query2->where('user_type', 2)
->orWhere('value', $value);
});
if ($user == 'admin'){
$query->where('users.user_name', $user);
}
终于得到了结果
$result = $query->get();
使用Eloquent很容易创建多个where检查:
首先:(使用简单的where)
$users = User::where('name', $request['name'])
->where('surname', $request['surname'])
->where('address', $request['address'])
...
->get();
第二个:(在数组中分组where)
$users = User::where([
['name', $request['name']],
['surname', $request['surname']],
['address', $request['address']],
...
])->get();
你也可以在里面使用条件(=，<>，等等)，就像这样:
$users = User::where('name', '=', $request['name'])
->where('surname', '=', $request['surname'])
->where('address', '<>', $request['address'])
...
->get();
一定要对子查询应用任何其他过滤器，否则or可能会收集所有记录。
$query = Activity::whereNotNull('id');
$count = 0;
foreach ($this->Reporter()->get() as $service) {
$condition = ($count == 0) ? "where" : "orWhere";
$query->$condition(function ($query) use ($service) {
$query->where('branch_id', '=', $service->branch_id)
->where('activity_type_id', '=', $service->activity_type_id)
->whereBetween('activity_date_time', [$this->start_date, $this->end_date]);
});
$count++;
}
return $query->get();
使用纯Eloquent，像这样实现它。这段代码返回所有帐户处于活动状态的已登录用户。 $用户= \ App \用户::(“状态”,“积极”)——> (logged_in,真的)- > ();
public function search()
{
if (isset($_GET) && !empty($_GET))
{
$prepareQuery = '';
foreach ($_GET as $key => $data)
{
if ($data)
{
$prepareQuery.=$key . ' = "' . $data . '" OR ';
}
}
$query = substr($prepareQuery, 0, -3);
if ($query)
$model = Businesses::whereRaw($query)->get();
else
$model = Businesses::get();
return view('pages.search', compact('model', 'model'));
}
}
