我只想取出service_id,service_name,service_price,ipad_img,cart_id这几个key和他们对应的value，我该怎么办？
回复内容：
我只想取出service_id,service_name,service_price,ipad_img,cart_id这几个key和他们对应的value，我该怎么办？
简单粗暴点，参考如下
$newarray = [];foreach($array as $key=>$val){ $newarray[$key]['service_id'] = $val['id']['service_id']; $newarray[$key]['service_name'] = $val['detail'][0]['service_name']; $newarray[$key]['service_price'] = $val['detail'][0]['service_price']; $newarray[$key]['ipad_img'] = $val['detail'][0]['ipad_img']; $newarray[$key]['cart_id'] = $val['cart_id']['cart_id'];}var_dump($newarray);
能不能贴下这个图片上的代码，好跑一下。。。
$new_arr = [];array_walk_recursive($arr, function($item, $key) use (&$new_arr) { $new_arr[$key] = $item;});print_r($new_arr);
请问你一下从数据库中取出来的数据就1条还是多条？
多条可以是下面：
$arr=[];
foreach($fruits as $k=>$v){
$arr[$k]['service_id'] = $v['id']['service_id'];$arr[$k]['service_name'] = $v['detail'][0]['service_name'];$arr[$k]['service_price'] = $v['detail'][0]['service_price'];$arr[$k]['ipad_img'] = $v['detail'][0]['ipad_img'];$arr[$k]['cart_id'] = $v['cart_id']['cart_id'];
}
print_r($arr);
一条的话可以使用楼上的
可以用left join ;
你的表结构大该是：
yld_cart:
cart_idservice_idassociator_id
yld_service:
service_idservice_nameservice_priceipag_img
sql语句可以这么写：
select cart_id , cart.`service_id` ,service.`service_name`,service.`service_price`,service.`ipad_img`from `yjd_cart` as cart left join `yjd_service` service on cart.service_id = service.service_id where cart.associator_id=1
yii2可以这么写：
$query = new query(); $associator_id = 1; $result = $query->select(['cart_id','yjd_cart.service_id','yjd_service.service_name','yjd_service.service_price','yjd_service.ipad_img']) ->from('yjd_cart') ->leftjoin('yjd_service','yjd_service.service_id = yjd_cart.service_id') ->where(['yjd_cart.associator_id'=>$associator_id]) ->all(); var_dump($result);