我有一个响应式网站,当网站位于其他导航链接旁边的"桌面"视图(屏幕可用宽度> 768)时,它具有简单的下拉登录菜单.当屏幕宽度低于768时,导航链接最终会出现在选择选项中.问题是下拉登录菜单在选择选项中不起作用.
<a href>当屏幕宽度小于768时,我想使用PHP将下拉登录菜单更改为一个简单的链接.
现在我在我的页面中<head>:
<?
$screenWidth = '<script type="text/javascript">document.write(screen.availWidth);</script>';
?>
在<body>:
<?
if($screenWidth <= "768") {
echo '<li><a href="login.php">Log in</a></li>';
} else {
?>
<div id="fancy">
<li id="login">
<a id="login-trigger" href="#">Log in <span>▼</span></a>
<div id="login-content">
<form>
<fieldset id="inputs">
<input id="username" type="email" name="Email" placeholder="Your email address" required>
<input id="password" type="password" name="Password" placeholder="Password" required>
</fieldset>
<fieldset id="actions">
<input type="submit" id="submit" value="Log in">
<label><input type="checkbox" checked="checked"> Keep me signed in</label>
</fieldset>
</form>
</div>
</li>
<? } ?>
在我的桌面上,我回显了$ screenWidth,它给出了1920.因此我希望显示"花式"下拉登录菜单.(确实如此).
在我的手机上,$ screenWidth echo给出320.然后我希望<a href>显示链接.(它没有 - 而是显示"花式"菜单).
看起来奇怪的是,在体内回显时的变量将给出不同的数字,但是当在if语句中进行比较时,它不会改变输出.
有没有更好的方法来改变输出？
编辑:jquery响应式菜单代码
jquery.responsivemenu.js:
(function($) {
$.fn.responsiveMenu = function(options) {
var defaults = {autoArrows: false}
var options = $.extend(defaults, options);
return this.each(function() {
var $this = $(this);
var $window = $(window);
var setClass = function() {
if ($window.width() > 768) {$this.addClass('dropdown').removeClass('accordion').find('li:has(ul)').removeClass('accorChild');}
else {$this.addClass('accordion').find('li:has(ul)').addClass('accorChild').parent().removeClass('dropdown');}
}
$window.resize(function() {
setClass();
$this.find('ul').css('display', 'none');
});
setClass();
$this
.addClass('responsive-menu')
.find('li.current a')
.live('click', function(e) {
var $a = $(this);
var container = $a.next('ul,div');
if ($this.hasClass('accordion') && container.length > 0) {
container.slideToggle();
return false;
}
})
.stop()
.siblings('ul').parent('li').addClass('hasChild');
if (options.autoArrows) {
$('.hasChild > a', $this)
.find('strong').append('<span class="arrow"> </span>');
}
});
}
})(jQuery);
PHP是服务器端执行的,这意味着var screenWidth包含:"document.write(screen.availWidth);" 当你比较它.您的输出在客户端执行..
Your simplest option might be to populate both options in the DOM, then use CSS3 Media queries to hide/show the proper element based on screen size.
So your HTML might look like:
<li class="login-link"><a href="login.php">Log in</a></li>
<div id="fancy">
<li id="login">
<a id="login-trigger" href="#">Log in <span>▼</span></a>
<div id="login-content">
<form>
<fieldset id="inputs">
<input id="username" type="email" name="Email" placeholder="Your email address" required>
<input id="password" type="password" name="Password" placeholder="Password" required>
</fieldset>
<fieldset id="actions">
<input type="submit" id="submit" value="Log in">
<label><input type="checkbox" checked="checked"> Keep me signed in</label>
</fieldset>
</form>
</div>
</li>
And your CSS could look like:
.login-link, #login{
display: none;
}
@media screen and (max-width: 767px){
.login-link {
display: block;
}
#login{
display: none;
}
}
@media screen and (min-width: 768px) {
#login{
display: block;
}
.login-link{
display: none;
}
}
Edit: Fixed #login reference. Edit 2: Adding JSFiddle Example JSFiddle Example