-1

好的，我有一个脚本可以在您每次评分时更改图像，并且效果很好，但问题是人们可以向评分系统发送垃圾邮件，所以我认为可以在包含 div 之前进行某种延迟设置新图像后显示评分

应该延迟的评级栏代码：

<div id="button" onclick="changeSrc2()">
<div class="rate_widget" id="<? echo $id;?>">
<div class="star_1 ratings_stars"></div>
<div class="star_2 ratings_stars"></div>
<div class="star_3 ratings_stars"></div>
<div class="star_4 ratings_stars"></div>
<div class="star_5 ratings_stars"></div>
<div class="total_votes">vote data</div>
</div>
</div>

在评分栏上单击更改图像的代码：

$number="1";
$wrongnumber="2";
$random = mysql_query("SELECT * FROM images ORDER BY RAND()");
$place="upload/";
echo '<script type="text/javascript"> ';
while($wor = mysql_fetch_array($random))
  {
  $ids=$wor['id'];
  $name = $wor['name'];
  $images = $place . $wor['name'];
  $number=$number + 1;
  $wrongnumber=$wrongnumber + 1;
echo 'function ' . 'changeSrc' . $number . '() '; ?>
{
document.getElementById("rand").src="<? echo $images;?>";
document.getElementById("button").onclick=changeSrc<? echo $wrongnumber;?>;
document.getElementsByClassName('rate_widget')[0].id = <? echo $ids;?>;
}
<?
  }
?> 
</script>

和显示第一张图片的代码：

    /*Display images*/
$r = mysql_query("SELECT * FROM images ORDER BY RAND() LIMIT 1");

while($wor = mysql_fetch_array($r))
{  
    $place="upload/";
    $id=$wor['id'];
    $name = $wor['name'];
    $image = $place . $wor['name'];
    echo '<img id="rand" src="'.$image.'" style="max-height:330px;">';  
}
4

1 回答 1

0

尝试使用sleep()函数。它看起来像这样：

<?php
sleep(2);
echo '
 <div id="button" onclick="changeSrc2()">
 <div class="rate_widget" id=' . $id . '">
 <div class="star_1 ratings_stars"></div>
 <div class="star_2 ratings_stars"></div>
 <div class="star_3 ratings_stars"></div>
 <div class="star_4 ratings_stars"></div>
 <div class="star_5 ratings_stars"></div>
 <div class="total_votes">vote data</div>
 </div>
 </div>
';
?>
于 2013-01-04T22:59:52.250 回答