求大神指导数组-tutoriel php-php.cn

求大神指导数组

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：

求大神指导数组

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);

Copier après la connexion

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);

Copier après la connexion

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'];

Copier après la connexion

}
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_id	service_id	associator_id

yld_service:

service_id	service_name	service_price	ipag_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

Copier après la connexion

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);

Copier après la connexion