基于三个下拉菜单的值的Jquery选择器

Question

我正在处理三个下拉菜单，每个下拉菜单都有三个选项。这三个下拉菜单一起创建了一个系统，它可以处于二十七个状态中的任何一个状态。对于每个下拉列表，都有选项1、选项2和全部(选项1和选项2的并集)。

根据这三个下拉菜单所处的状态，我希望显示或隐藏不同的div。这些div具有class="distribution-(select-1)-(select-2)-(select-3)"形式的类

我正在试图找到一种干式的方法来编写代码，而不是依赖于27个不同的条件。

以下是下拉菜单：

<div  style = "margin-top:1cm; font-size:125%">
    <strong>Clickout Type&nbsp;</strong>
    <select class = "clickouts" >
        <option value = "all" selected="selected">All</option>
        <option value = "clickouts" >Clickouts</option>
        <option value = "preferred" >Preferred Clickouts</option>
    </select>
    <strong>Graph Type&nbsp;</strong>
    <select class = "graph-type" >
        <option value = "all" selected="selected">All</option>
        <option value = "hist" >Histograms</option>
        <option value = "percent" >Percent</option>
    </select>
    <strong>Roll-up Type&nbsp;</strong>
    <select class = "roll-up" >
        <option value = "all" selected="selected">All</option>
        <option value = "aggregate" >Aggregate</option>
        <option value = "granular" >Granular</option>
    </select>
</div>

下面是一些div类的示例：distribution-hist-clickouts-aggregate, distribution-percent-clickouts-aggregate, class="distribution-hist-clickouts-granular", distribution-percent-clickouts-granular

下面是前七个州的jquery：

$('.clickouts, .graph-type, .roll-up').change(function(){
    var graph_type = $('.graph-type').val(); 
    var roll_up = $('.roll-up').val();
    var clickouts = $('.clickouts').val(); 

    if (graph_type == 'all' & roll_up == 'all' & clickouts == 'all'){
        $('div[class*="distribution"]').css('display', 'block');
    }
    if (graph_type == 'all' & roll_up == 'all' & clickouts == 'clickouts'){
        $('div[class*="clickout"]').css('display', 'block');
        $('div[class*="preferred"]').css('display', 'none');
    }

    else if (graph_type == 'all' & roll_up == 'all' & clickouts == 'preferred'){
        $('div[class*="clickout"]').css('display', 'none');
        $('div[class*="preferred"]').css('display', 'block');
    }

    else if (graph_type == 'hist' & roll_up == 'all' & clickouts == 'all'){
        $('div[class*="hist"]').css('display', 'block');
        $('div[class*="percent"]').css('display', 'none');
    }

    else if (graph_type == 'percent' & roll_up == 'all' & clickouts == 'all'){
        $('div[class*="hist"]').css('display', 'none');
        $('div[class*="percent"]').css('display', 'block');
    }

    else if (graph_type == 'all' & roll_up == 'aggregate' & clickouts == 'all'){
        $('div[class*="aggregate"]').css('display', 'block');
        $('div[class*="granular"]').css('display', 'none');
    }

    else if (graph_type == 'all' & roll_up == 'granular' & clickouts == 'all'){
        $('div[class*="aggregate"]').css('display', 'none');
        $('div[class*="granular"]').css('display', 'block');
    }
});

Answer 1

如果要将复杂的类class="distribution-(select-1)-(select-2)-(select-3)更改为一组类class="distribution (select-1) (select-2) (select-3)，则可以使用以下脚本：

$('.clickouts, .graph-type, .roll-up').change(function(){
    var graph_type = $('.graph-type').val(); 
    var roll_up = $('.roll-up').val();
    var clickouts = $('.clickouts').val(); 

    graph_type = (graph_type == "all" ? "" : "." + graph_type);
    roll_up = (roll_up == "all" ? "" : "." + roll_up);
    clickouts = (clickouts == "all" ? "" : "." + clickouts);

    var selector = '#container div' + graph_type + roll_up + clickouts;

    $('#container div:visible').hide();
    $(selector).show();
});

代码：http://jsfiddle.net/9R5zj/9/

Answer 2

好的，有几件事可以帮助你重构它。

您应该记住的一件简单的事情是元素可以有多个类。这意味着，您可以为distribution-hist-clickouts-aggregate类提供四个类，而不是使用带有div的div，如下所示：

<div class="distribution hist clickouts aggregate"></div>

此外，不是将选择选项的value设置为all，而是将选择器放入以下值中，从而使程序更容易理解：

<option value = ".aggregate, .granular" selected="selected">All</option>    
<option value = ".aggregate" >Aggregate</option>    
<option value = ".granular" >Granular</option>    

我不确定我是否完全理解所有if语句背后的逻辑，但您可以这样做：

$("div.distribution").hide();
$(graph_type).show();
$(roll_up).show();
$(clickouts).show();

你可以在this demo上看到我的建议。