php、
if(){}
else
{
echo "<script><br /> $(document).ready(function() {$('.captcha').hide(); });<br /></script>";
......php代码,在此省略
}
我想要的结果是当执行else的时候页面中的captcha class的css给隐藏掉。但如上写的代码貌似不对,该怎么在php的判断语句中加jquery代码啊?
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如果代码没有问题的话
- PHP code
echo (function($){ $('.captcha').show(); })(jQuery); HTML; $error_msg = elgg_echo('loginerror'); <br><font color="#e78608">------解决方案--------------------</font><br>
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此易事尔
if(){
}else
?>
<script>//jquery</script>
}
?>
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";多了这个,加分吧,,我分少呢
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这段JS代码要放在引入jquery.js文件之后
问题关键在于,最终输出到客户端的html代码是怎样的
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我已经告诉过你了
不过记得,这个输出必须在你的captcha已经正常输出之后进行控制,否则无效。
所以其实你应该这么做 php输出一个隐藏的值例如
echo '';
在整个页面载入之后再根据这个值来执行相应的操作。
- JScript code
$(document).ready(function() { if ($('#fool').val()=='false'){ //执行代码 $('.captcha').show(); } }); <br><font color="#e78608">------解决方案--------------------</font><br>
------解决方案--------------------
你要明白一点 php是运行在服务器端而js和jquery是运行在客户端的,所以在你需要操作的html标签没有显示到客户端的时候,jquery是无法操作这些代码的。理解这点你就知道剩下的怎么做了。
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算了,告诉你一招秘技,一般人我不告诉他,记得给分。。
- PHP code
<script src="jquery-1.4.2.js"></script> <div id="dv" style="display:none;width:100px;height:100px;border:1px solid red"></div> <?php $k='<script> $(function(){ $("#dv").show(3000); }) '; $do=DOMDocument::loadHTML($k); echo $do->saveHTML(); ?> <br><font color="#e78608">------解决方案--------------------</font><br>
------解决方案--------------------
document.cookie的用法不是 名=值 一一对应的么? document.cookie="Id=111";
就算你的cookie 有内容吧
if (cookies == 'value') 这里怎么是这样判断呢.
getCookie('Id')//获取cookie的值
...
一塌糊涂啊....
http://www.satya-weblog.com/2007/05/php-and-javascript-cookie.html
先看看基本的东西.先仔细的学习一下cookie 在php和js里的用法.
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另外firebug是可以断点调试的,多多用这个功能+alter 慢慢找代码不对的地方吧
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你的描述 有问题啊

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