Question

我正在尝试让jQuery根据两个值之间的属性值显示/隐藏元素。

我的标记看起来像这样：

<a class="page_link" href="javascript:go_to_page(0)" longdesc="0">1</a>
<a class="page_link" href="javascript:go_to_page(1)" longdesc="1">2</a>
<a class="page_link" href="javascript:go_to_page(2)" longdesc="2">3</a>
<a class="page_link" href="javascript:go_to_page(3)" longdesc="3">4</a>
<a class="page_link active_page" href="javascript:go_to_page(4)" longdesc="4">5</a>
<a class="page_link" href="javascript:go_to_page(5)" longdesc="5">6</a>
<a class="page_link" href="javascript:go_to_page(6)" longdesc="6">7</a>
<a class="page_link" href="javascript:go_to_page(7)" longdesc="7">8</a>
<a class="page_link" href="javascript:go_to_page(8)" longdesc="8">9</a>

我有一些看起来像这样的jQuery：

//Now, we need to paginate the pagination
var page_link_count = $('.page_link').length(); 

//Hide all the .page_links
$('.page_link').css('display','none'); 

//Show active_page
$('.active_page').css('display','block'); 

//Show up to four .page_link on either side of active_page
var active_page_value = $('.active_page').attr('longdesc'); 

$('.page_link').each(function(){

    var longdesc = $(this).attr('longdesc');

    // This next part might be what's getting me...
    if( longdesc <= active_page_value+4 && longdesc >= active_page_value-4){

        $(this).css('display','block');

    }

});

我想说，如果具有page_link类的任何元素的longdesc属性值大于active_page longdesc - 4之和的值或者小于active_page longdesc + 4的总和

Answer 1

在添加之前，您需要解析数字。将替换值

if( (longdesc <= parseInt(active_page_value ,10) +4)
     &&( longdesc >= parseInt(active_page_value,10) -4))

<强> JS

//Now, we need to paginate the pagination
var page_link_count = $('.page_link').length; 

//Hide all the .page_links
$('.page_link').css('display','none'); 

//Show active_page
$('.active_page').css('display','block'); 

//Show up to four .page_link on either side of active_page
var active_page_value = $('.active_page').attr('longdesc'); 

var min = parseInt(active_page_value,10) - 4;
var max = parseInt(active_page_value,10) + 4;

$('.page_link').each(function(){
    var longdesc = $(this).attr('longdesc');

    if( longdesc <= max && longdesc >= min){
        $(this).css('display','block');
    }

});

Answer 2

要改变两件事：

更改

var page_link_count = $('.page_link').length();

要

var page_link_count = $('.page_link').length;

和

var active_page_value = $('.active_page').attr('longdesc');

要

var active_page_value = +$('.active_page').attr('longdesc');

Answer 3

过度复杂化......这就是你所需要的。

//Hide all the .page_links
$('.page_link').hide(); 

//Show active_page
$('.active_page').show(); 

//Show all links before active
$('a.active_page').show();
$('a.active_page').prevAll().show();

*注意：最后两个可以合并，但我不想让你迷惑。

jQuery显示具有一系列值的元素

3 个答案: