我想要所有价格在给定价格范围内的旅馆,从我旅馆的桌子。价格存储在四个不同的行中,如价格一股,价格两股,价格三股,价格更多的股票,
public function actionNearbyHostel($min,$max,$types,$lat,$long){
$latitude = $lat;
$longitude = $long;
$type = $types;
$max_price = (int)$max;
$min_price = (int)$min;
$hostels = Hostels::find ()->select("*,(6371 *
acos(cos(radians({$latitude}))*
cos(radians(`lat`)) * cos(radians(`log`) - radians({$longitude})) +
sin(radians({$latitude})) * sin(radians(`lat`)))) AS distance")
->having("distance<:distance")
->addParams([
':distance' => 5
])->where([
'status' => Hostels::ACTIVE,
'type' => $type
])->andWhere(['between','price_one_share',$min_price,$max_price])
->all();
return $this->render('filterhostel',[
'hostels' => $hostels,
]);
} 这是我已经写好的动作
andWhere(['between','price_one_share',$min_price,$max_price])像这样,我想包括剩下的三行并得到我的结果。
这是我的表行图像http://www.clipular.com/posts/4991398571671552?k=Lud1W4LGJe5hSxOsl2ao7VOGguI
发布于 2019-04-23 21:57:59
在做了一些作业之后,我找到了答案
public function actionNearbyHostel($min,$max,$types,$lat,$long){
$latitude = $lat;
$longitude = $long;
$type = $types;
$max_price = (int)$max;
$min_price = (int)$min;
$hostels = Hostels::find ()->select("*,(6371 * acos(cos(radians({$latitude}))*
cos(radians(`lat`)) * cos(radians(`log`) - radians({$longitude})) +
sin(radians({$latitude})) * sin(radians(`lat`)))) AS distance")
->having("distance<:distance")
->addParams([
':distance' => 5
])->where([
'status' => Hostels::ACTIVE,
'type' => $type
])->andWhere(['or',
['between','price_one_share',$min_price,$max_price],
['between','price_two_share',$min_price,$max_price],
['between','price_three_share',$min_price,$max_price],
['between','price_more_share',$min_price,$max_price]
])
->all();
return $this->render('filterhostel',[
'hostels' => $hostels,
]);
} https://stackoverflow.com/questions/55786014
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