小编use*_*027的帖子

我有一个问题.我有一个下拉菜单,但是下拉列表总是在身体后面一层,尽管菜单的z-index设置为999且body的z索引设置为-999

请chceck http://www.w3dominik.com/x/finemoney/(右上角的菜单,它表示下拉列表,应该有2个选项,现在只有1个可见)

感谢帮助

如果我有这个:

<span data-helo="1">something</span>
<span data-helo="2">something different</span>

我想针对第一个跨度,我应该怎么做？

我尝试过类似的东西:

$('span').data('helo', '1')

但它返回了两个跨度.

我有这个代码

        <ul>
            <li><a href="index.php">Web Development</a></li>
                <span data-landing_count="1">
                    <li class="subli"><a href="1.html">E-shop</a></li>
                    <li class="subli"><a href="2.html">Prezentations</a></li>
                    <li class="subli"><a href="3.html">Custom CMS</a></li>
                    <li class="subli"><a href="4.html">WordPress & Joomla!</a></li>
                    <li class="subli"><a href="5.html">Web apps</a></li>
                    <li class="subli lastsl"><a href="index-b.html">Design</a></li>
                </span>
            <li><a href="6.html">E-courses</a></li>
                <span data-landing_count="2">
                    <li class="subli"><a href="7.html">Design & Graphics</a></li>
                    <li class="subli lastsl"><a href="8.html">Web Development</a></li>
                </span>
            <li><a href="9.html">SEO</a></li>
                <span data-landing_count="3">
                    <li class="subli"><a href="91.html">On-page SEO</a></li>
                    <li class="subli"><a href="92.html">W3C validation</a></li>
                    <li class="subli"><a href="93.html">Linking</a></li>
                    <li class="subli lastsl"><a href="94.html"><img src="img/star.png" alt="star" width="12" class="star">Copywriting</a></li>
                </span>
            <li data-landing_count="4"><a href="95.html">Web hosting</a></li>
            <li data-landing_count="5"><a href="96.html"><img src="img/star.png" alt="star" width="12" …

我想做的事情:

选择date = 2012-10-14的行,然后显示该行之后的4行.所以从这个列表中

2012-10-12 column #2
2012-10-13 column #2
2012-10-14 was very sunny.
2012-10-15 rained all day.
2012-10-16 whatever.
2012-10-17 column #2
2012-10-18 column #2
2012-10-19 rained all day.
2012-10-20 whatever.
2012-10-21 column #2
2012-10-22 column #2

它会返回这个:

2012-10-14 was very sunny.
2012-10-15 rained all day.
2012-10-16 whatever.
2012-10-17 column #2
2012-10-18 column #2

感谢帮助.

PS:在数据库中周末没有数据,所以有些日期会丢失.

我现在不知道我的错误,我的代码如下:

<?php

mysql_set_charset('utf8');
$result = mysql_query("SELECT * FROM obedy ORDER BY datum DESC LIMIT 30");
while ($row = mysql_fetch_array($result, MYSQL_NUM)) 
{
    $time = strtotime( $row[0] );
    $myDate = date( 'd.m.', $time );
    $w_day = date( 'N', $time );
    $ww_day = $w_day-1;
    $the_row= $row[0];

    echo "<p>" . $myDate . "&emsp;";
    echo "<input type='hidden' value='$row[0]' name='tdel'>";
    echo "<input type='text' name='menu1' id='menu1' class='input' value='". $row[1] ."' size='37'/>";
    echo "<input type='text' name='menu2' id='menu2' class='input' value='". $row[2] ."' size='37'/>";
    echo "<input type='text' name='menu3' id='menu3' class='input' …

我的代码有效,我只是想知道它是不是很糟糕的做法,因为我想是这样.我尝试了所有preg_replace但它似乎没有工作.所以我就是这样写的.

作为一个输入,我期待网址

google.com www.google.com http://google.com

要么

http://www.google.com

因此我需要

google.com

我的代码:

 $website = trim($website); //removes space characters
                        $website = trim($website, '/');
                        $website = trim($website, 'http://');
                        $website = trim($website, 'www.');