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如何根据其他下拉选择的值显示下拉选择?

[英]How to display a dropdown select based on the value of other dropdown select?

Suppose I have a dropdown select option like below: 假设我有一个下拉选择选项,如下所示:

 <form>
    <select name="option" >
                <option value="">Select a Value:</option>
                <option value="1">Option 1</option>
                <option value="2">Option 2</option>

    </select>
</form>

When any user Selects the Option 1 I want to display the following dropdown select beside it instantly 当任何用户选择选项1时,我想显示以下下拉列表,然后立即选择它

 <select name="name" >
                <option value="">Select a person:</option>
                <option value="john">john</option>
                <option value="Micheal">Micheal</option>

    </select>

Or, if any user Selects the Option 2 I want to display the following dropdown select beside it instantly 或者,如果任何用户选择选项2我想显示以下下拉列表,请立即选中它

 <select name="class" >
                <option value="">Select a person:</option>
                <option value="Class 1">Class 1</option>
                <option value="Class 2">Class 2</option>

    </select>

Would you please kindly show me how to do this? 请你告诉我怎么做? I am using Codeigniter. 我正在使用Codeigniter。

$('select[name="option"]').change(function(){
    $('.hidden').hide();
    if(this.value == 1){
        $('select[name="name"]').toggle();
    }
    else if(this.value == 2){
        $('select[name="class"]').toggle();
    }
});

Fiddle: http://jsfiddle.net/maniator/Z6Upj/ 小提琴: http//jsfiddle.net/maniator/Z6Upj/

Give this a shot: 试一试:

$(function(){
    $("select[name='option']").change(function(e){
        if ($(this).val() == "1")
            $("select[name='name']").show();
    });
});

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