PHP select不会返回值

Question

我有这段代码，我希望在medication table中获取相关信息并将acc_id account table中的= acc_id medication table显示在med_timeoftheday='morning' {} 1}}和$postdata = file_get_contents("php://input"); if (isset($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); } }

{{1}}

这是我的意思：

我该怎么做？

提前谢谢！

Answer 1

试试这样的话：

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

Answer 2

首先，如果您从药物表中选择数据，请选择第一个药物表，然后使用帐户表连接。

$sql = "SELECT * FROM medication JOIN account_info ON account_info.acc_id = medication.acc_id WHERE medication.med_timeoftheday='morning'";