php - 如何列出在线论坛中所有没有答案的问题

Question

我正在用 PHP 创建在线讨论论坛作为我最后一个学期的项目，我想列出所有在未回答的页面中没有答案的问题，就像 Stack Overflow 中一样。

但我不知道如何实现；我在互联网上搜索了很多，但仍然找不到如何列出所有没有答案的问题。

下面是我的代码:

<?php
@session_start();
include("include/user_header.php");
include("include/connection.php");


$query="SELECT * FROM question WHERE NOT EXISTS (SELECT * FROM answer WHERE answer.question_id = question.question_id);"
$result = mysql_query($query) or die(mysql_error());
?>
                                    <script type="text/javascript">
                                            if(action=="pass_id")
                                            {
                                                document.frm_cms.action="answer.php";
                                                <!--document.frm_que.myaction.value=action;-->
                                                document.frm_cms.id.value=id;   
                                                document.frm_cms.submit();
                                            }


                                        </script>
<div class="main-container">
     <div class="another-main-container row">        

    <div class="ask-question-container">

        <div class="ask-question-container-inner row">
            <form name="frm_cms" method="post" class="printable">
                    <div class="question-usr-prfl row">
                    <h1 class="usr-hdng"><span class="qustn-fgr-usr-prfl"></span> Questions</h1>
                    <ul class="row list-of-questions">
                        <?php
                            while($row = mysql_fetch_array($result))
                            {   
                        ?>
                                    <li class="row">
                                    <a href="javascript:setAction('pass_id','<?php echo $row['question_id']; ?>');" >
                                        <?php echo $row['title']; ?>
                                    </a>
                                    <div class="quest-tag-cntr row">
                                    <?php 
                                        $tgNm = explode(",",$row['tag_name']);
                                        for($k=0;$k<count($tgNm);$k++)
                                        {
                                            $k+1;
                                    ?>
                                            <a href='question.php?id=<?php echo $tgNm[$k]; ?>'>
                                    <?php
                                                echo getFromId("tag", "tag_name", "id", $tgNm[$k]);
                                    ?>
                                            </a>
                                    <?php
                                        }                                                       
                                    ?>
                                    </div><!--/quest-tag-cntr-->  
                                </li>
                        <?php
                            }                                                       
                        ?>                                                                                 
                    </ul>


           </div><!--/question-usr-prfl-->
          </form>
        </div>
    </div>
  </div>
</div>

<?php
include("include/user_footer.php");
?>

Answer 1

我认为这样的事情值得一试:

SELECT DISTINCT question.id
FROM question
LEFT JOIN answer ON (question.id = answer.question_id)
WHERE answer.id IS NULL

我已左连接到答案表(因为我们希望该连接失败)，然后将非空列与 NULL 进行比较。因此，返回的行是没有答案的问题。