Question

我的查询和其中一个连接有问题，初始查询工作正常但是当我尝试加入链接表时，每个文章有4个匹配的链接，查询返回每个文章的4条记录，从而显示我的网站上每篇文章4次？

所以我的表是：

tbl_news_articles => **newsID**, newsTitle, newsBody, newsDate
tbl_images => **imageID**, filename
tbl_news_images => **imageID**, **newsID**
tbl_links => **linkID**, linkTitle, linkURL
tbl_news_links => **newsID**, **linkID**

我正在使用的查询是

SELECT n.newsID, n.type, n.title, n.body, n.date, ni.imageID, i.imageID, i.filename, l.linkID, l.linkTitle, l.linkURL 
FROM tbl_news_articles AS n 
LEFT JOIN tbl_news_images AS ni ON ni.newsID = n.newsID 
LEFT JOIN tbl_images AS i ON i.imageID = ni.imageID 
LEFT JOIN tbl_news_links AS nlink ON nlink.newsID = n.newsID 
LEFT JOIN tbl_links AS l ON l.linkID = nlink.linkID

返回类似下面的内容

newsID, newsTite, newsDesc, newsDate, imageID, filename, linkID, linkTitle, linkURL
65,Article 1 Title,Article 1 description,2016-05-21,1,group.jpg,1,link1 title,linkURL
65,Article 1 Title,Article 1 description,2016-05-21,1,group.jpg,2,link2 title,linkURL
65,Article 1 Title,Article 1 description,2016-05-21,1,group.jpg,3,link3 title,linkURL
65,Article 1 Title,Article 1 description,2016-05-21,1,group.jpg,4,link4 title,linkURL

所以每篇文章每个链接返回4次。

解决此问题的正确方法是什么？我应该在我的PHP代码中运行一个单独的查询来单独获取链接吗？

感谢您的帮助

我的PHP代码是

<?php
//call the database to retreive the records
$sql = "SELECT n.newsID, n.type, n.title, n.body, n.date, ni.imageID,      i.imageID, i.filename, l.linkID, l.linkTitle, l.linkURL FROM     tbl_news_articles AS n 
LEFT JOIN tbl_news_images AS ni ON ni.newsID = n.newsID 
LEFT JOIN tbl_images AS i ON i.imageID = ni.imageID 
LEFT JOIN tbl_news_links AS nlink ON nlink.newsID = n.newsID 
LEFT JOIN tbl_links AS l ON l.linkID = nlink.linkID 
ORDER BY date DESC LIMIT $offset, $rowsperpage";
$result = $conn->query($sql);
$i = 0;
if ($result->num_rows > 0)
{
    // output data of each row
    while($row = $result->fetch_assoc())
    {
        $id = $row['newsID'];
        $link = "article.php?id=".$id;
        $type = $row['type'];
        $title = $row['title'];
        $urltitle = strtolower($title);
        $urltitle = str_replace(" ","-",$urltitle);
        $body = $row['body'];
        #$bodytext = (strlen($body) > 130) ? substr($body,0,130).'...' : $bodytext;
        $pos= strpos($body, ' ', 140);
        $bodytext = substr($body,0,$pos); 
        $bodytext .= "... <a href='$link' title='$title'>read more</a>";
        $date = $row['date'];
        $formated_date = date("d-M-Y", strtotime($date));
        $imgID = $row['imageID'];
        $filename = $row['filename'];
        if($filename != "")
        {
             $imgLink = "images/news-articles/".$id."/".$filename;
        }
        else
        {
             $imgLink = "images/news-item-placeholder.jpg";
        }
        $i++;
        if($i == 1)
        {
            echo "<div class='news-spotlight'>";
        }
        elseif($i >=2 && $i <=3)
        {
            if($i == "2")
            {
                $class = "first";
            }
            else
            {
                $class = "";
            }
            echo "<div class='news-highlight $class'>";
        }
        else
        {
            echo "<div class='news-item'>";
        }?>
        <a itemprop="url" href="<?php echo $link?>"><img itemprop="image" src="<?php echo $imgLink?>" alt="<?php echo $title?>" title="<?php echo $title?>"></a>
        <div class='data'>

        <h3><a href="<?php echo $link?>"><span itemprop="name"><?php echo $title?></span></a></h3>
        <div class="article-date"><?php echo $formated_date?></div>
        <span itemprop="startDate" content="<?php echo $date?>"></span>
        <div itemprop="location" itemscope itemtype="http://schema.org/Place">          
        <span itemprop="address" content="England"></span>
        <span itemprop="name" content="MayoShinDo Association"></span>
        </div>
        <p itemprop="description"><?php echo $bodytext?></p>

    <?php if($i == 1)
    {?>
        <ul>
             <li><a href="<?php echo $link?>"><span class="glyphicon glyphicon-ok" aria-hidden="true"></span> link 1</a></li>
             <li><a href="<?php echo $link?>"><span class="glyphicon glyphicon-ok" aria-hidden="true"></span> link 2</a></li>
             <li><a href="<?php echo $link?>"><span class="glyphicon glyphicon-ok" aria-hidden="true"></span> link 3</a></li>
             <li><a href="<?php echo $link?>"><span class="glyphicon glyphicon-ok" aria-hidden="true"></span> link 4</a></li>
        </ul><?php
    }?>
    </div><?php
    echo "</div>";
    }
}
else
{
    echo "0 Results";   
}?>

我还没有在php代码中设置链接，正如您从上面的代码中看到的那样，只是想先查询正确的

我看过你可以使用GROUP_CONCAT吗？我尝试了以下

SELECT n.newsID, n.type, n.title, n.body, n.date, ni.imageID, i.imageID, i.filename, l.linkID, l.linkTitle, l.linkURL
GROUP_CONCAT(("|", l.linkID,l.linkTitle,l.linkURL)) as "fieldvalue"
FROM tbl_news_articles AS n 
        LEFT JOIN tbl_news_images AS ni ON ni.newsID = n.newsID 
        LEFT JOIN tbl_images AS i ON i.imageID = ni.imageID 
        LEFT JOIN tbl_news_links AS nlink ON nlink.newsID = n.newsID 
        LEFT JOIN tbl_links AS l ON l.linkID = nlink.linkID

但是我说我的sql语法有错误？任何想法

谢谢你！

当JOIN返回多个结果时，减少mysql查询中返回的重复记录？

0 个答案: