PHP select 不返回值

Question

我有此代码，我想获取药物表中的信息并将其显示在帐户表中的acc_id是 = 到药物表中的acc_id，其中med_timeoftheday='morning'

$postdata = file_get_contents("php://input");
if (isset($postdata)) {
     $request = json_decode($postdata);
     $User_ID = $request->acccid;
     $sql = sprintf("SELECT * FROM account_info
     join medication on account_info.acc_id = medication.acc_id 
     where account_info.acc_id='%s'",
       mysqli_real_escape_string($conn,$User_ID));
    $result=$conn->query($sql);
    if ($result->num_rows>0)
    {   
        while($row=$result->fetch_assoc()) 
        {$data[]=$row;
        }

         echo json_encode($data);
    }

}

这是我的ts:

我该怎么做？

提前谢谢您!

Answer 1

尝试这样的事情:

SELECT * FROM medication 
  INNER JOIN account_info ON account_info.acc_id = medication.acc_id
WHERE medication.med_timeoftheday='morning'