Question

我正在尝试从另一个选择（A）中的mysql中选择（B）数据，但是我从选择（B）中得到错误。我不知道如何修复它并发现它的问题。

// the server connection code id omitted.

//以下代码已被编辑。 $ bookId = $ _POST [“bookId”];

$sql = "SELECT * FROM feedback WHERE bookId= '".$bookId."'";// selecting(A)
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {

            // output data of each row
    while($row = mysqli_fetch_assoc($result)) {
        $sql = "SELECT * FROM userInfo WHERE userId= '". $row["userId"]."'"; //selecting(B)
        $data = mysqli_query($conn, $sql);
        $userData = mysql_fetch_array($data);

        $Arr = array("userId" => $userData ["userId"], "userName" => $userData ["userName"],
               "feedback" => $row["feedback"], "date" => $row["date"]);

         $feedbackArray[] = array("feedbackDetail"=> $Arr);
    }
} 

mysqli_close($conn);
$json= array("feedback" => $feedbackArray);
echo  json_encode($json);

我得到的结果：错误+有效的json（但是有null :(）

{"feedback":[{"feedbackDetail":{"userId":null,"userName":null,"feedback":"I love you ba~~","date":"2016-03-19 21:48:46"}},{"feedbackDetail":{"userId":null,"userName":null,"feedback":"well!!!","date":"2016-03-19 21:46:52"}}]}

任何想法？

php选择选择

0 个答案: