Php 新图片后div上的延迟
好的,我有一个脚本,每次你给它评分时都会更改图像,它工作得很好,但问题是人们可能会向评分系统发送垃圾邮件,所以我认为在新图像设置后包含评分的div出现之前,可能会有一些延迟 应延迟的比率条代码:Php 新图片后div上的延迟,php,javascript,html,delay,Php,Javascript,Html,Delay,好的,我有一个脚本,每次你给它评分时都会更改图像,它工作得很好,但问题是人们可能会向评分系统发送垃圾邮件,所以我认为在新图像设置后包含评分的div出现之前,可能会有一些延迟 应延迟的比率条代码: <div id="button" onclick="changeSrc2()"> <div class="rate_widget" id="<? echo $id;?>"> <div class="star_1 ratings_stars"></di
<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>
单击评级栏上的更改图像的代码:
$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>
和代码以显示第一个图像:
/*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;">';
}
尝试使用睡眠功能。它看起来是这样的:
<?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>
';
?>