dongmao3131 2017-12-11 17:52
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PHP select不返回值

I have this code and I want to get the info in medication table and display it where acc_id in account table is = to acc_id in medication table and where 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);
    }

}

this is my ts :

how can I do that ?

Thank you in advance!

  • 写回答

2条回答 默认 最新

  • dpp66953 2017-12-11 18:06
    关注

    Try somehting like this:

    SELECT * FROM medication 
      INNER JOIN account_info ON account_info.acc_id = medication.acc_id
    WHERE medication.med_timeoftheday='morning'
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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