新图片后div上的延迟

-1

好的我有一个脚本，每次你评价它都会改变图像，而且它工作的很好，但问题是人们可以垃圾评级系统，所以我认为它可以做一些排序包含评级股利之前延迟的显示出来后的新图像设置新图片后div上的延迟

代码应推迟的RatingBar：

<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>

代码的onclick改变评价栏上的图像：

$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，以显示所述第一图像：

/*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;">'; 
}

来源

2013-01-04 ak_37

尝试使用sleep()功能。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> 
'; 
?>

来源

2013-01-04 22:59:52 Ivan

新图片后div上的延迟

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