本文介绍了随机播放MYSQL结果的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
以下代码显示随机数据库图像以及一个特定的图像(这是我想要的).在数据库查询之后,由于始终首先显示图像ID 11,我该如何洗牌这些结果?我希望图像ID 11随机显示在其他图像之中.
The following code displays random database images as well as one specific image (which is what I want). How can I shuffle these results after database query as image ID 11 is always displayed first? I want image ID 11 displayed randomly amongst the others.
我可以使用SHUFFLE()吗?如果是这样,那么我现在该把它放在哪里?
Can I use SHUFFLE(). If so where exactly do I put it as at the moment?
任何帮助仍然可以学习PHP
Any help would be appreciated as still learning PHP
谢谢.
<?php mysql_connect("", "", "") or die(mysql_error()) ; mysql_select_db("images") or die(mysql_error()) ; $photo=mysql_query("SELECT * FROM `profile_images` ORDER BY (ID = 11) DESC, RAND() LIMIT 7"); while($get_photo=mysql_fetch_array($photo)){ ?> <div style="width:300px;"> <img src="<? echo $get_photo['url']; ?>"> </div> <? } ?>推荐答案
在将它们检索到php后,您可以对其进行洗牌.
You can shuffle them after they are retrieved to php.
$photos = array(); while ($get_photo = mysql_fetch_array($photo)) $photos[] = $get_photo; shuffle($photos);或者您可以通过子查询来做到这一点:
Or you can do it with subqueries:
SELECT A.* FROM ( SELECT * FROM `profile_images` ORDER BY (ID = 11) DESC, RAND() LIMIT 7 ) as A ORDER BY RAND()更多推荐
随机播放MYSQL结果
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