使用表连接选择特定的数据

Question

我试图从用户表中输入一些用户用户名，方法是将它从问题表中加入，意图是我可以显示哪个用户发布了这个特定问题。

用户id，username

discussion_q id，question_text，user_id

这是我在这里：

$sql = "SELECT q.id AS questionId, q.question_text AS questionText, q.user_id AS questionUserId, q.published AS questionPub, users.id AS userId 
        FROM discussion_q 
        JOIN users 
        ON questionUserId = userId 
        WHERE project_id = '$projectId' 
        ORDER BY published"; 

我得到0的结果返回给我，当然。我确定我已经完成了这项工作或错过了一些简单的事情？

这里是我的PHP返回结果：

$result = $conn->query($sql); 
    if (mysqli_num_rows($result) > 0) { 
     while($row = mysqli_fetch_assoc($result)) { 
      echo '<div class="twelve columns"> 
        <p><a href=".php?project_id=' . $row['id'] .'">' . $row['question_text'] . '</a></p> 
        <p>' . $row['published'] . ' by ' . $row['username'] . '</p> 
        </div>'; 
     } 
    } else { 
     echo "0 results"; 
    } 

所以最终目标是为输出与谁张贴的用户的用户名QUESTION_TEXT。

$sql = "SELECT q.id AS questionId, q.question_text AS questionText, q.user_id AS questionUserId, q.published AS questionPub, users.id AS userId 
        FROM discussion AS q 
        JOIN users 
        ON (q.user_id = users.id) 
        WHERE project_id = '$projectId' 
        ORDER BY published"; 

Answer 1

没有得到任何结果回来了PHP的让我通过我的DBM