【发布时间】:2016-05-08 22:15:47
【问题描述】:
我正忙于制作一个小型评论系统。我希望用户能够给出评分,除了给出 cmets,现在我将这些评分值(来自单选按钮)存储在数据库中。但是,当我尝试使用数据库中每个评论的评分值填充单选按钮时,以下脚本仅检查与数据库中最后一个值对应的单选按钮。
<?php
$sql = mysql_query("SELECT * FROM comments WHERE id_post = '$id_post'") or die(mysql_error());;
while($affcom = mysql_fetch_assoc($sql)){
$name = $affcom['name'];
$email = $affcom['email'];
$comment = $affcom['comment'];
$rating = $affcom['rating'];
$date = $affcom['date'];
// Get gravatar Image
// https://fr.gravatar.com/site/implement/images/php/
$default = "mm";
$size = 35;
$grav_url = "http://www.gravatar.com/avatar/".md5(strtolower(trim($email)))."?d=".$default."&s=".$size;
?>
<div class="cmt-cnt">
<img src="<?php echo $grav_url; ?>" />
<div class="thecom">
<h5><?php echo $name; ?></h5><span data-utime="1371248446" class="com-dt"><?php echo $date; ?></span>
<br/>
<p>
<?php echo $comment; ?>
</p>
<p>
<input type="radio" name="hoi" value="1" <?php echo ($rating=='1')?'checked':'' ?> />
<input type="radio" name="hoi" value="2" <?php echo ($rating=='2')?'checked':'' ?> />
<input type="radio" name="hoi" value="3" <?php echo ($rating=='3')?'checked':'' ?> />
<input type="radio" name="hoi" value="4" <?php echo ($rating=='4')?'checked':'' ?> />
<input type="radio" name="hoi" value="5" <?php echo ($rating=='5')?'checked':'' ?> />
</p>
</div>
</div><!-- end "cmt-cnt" -->
<?php } ?>
这是它现在的样子:
谁能告诉我如何检查所有的复选框(基于数据库值)?
【问题讨论】:
标签: mysql radio-button