Jquery $(this).parent().next()；

Question

我将跳转到包含"secound"类的<fieldset>，但跳转不起作用。我的代码出了什么问题？对不起，我的英语，我希望你能帮助我？

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$(".next").click(function(){
    textnext = "next1";
    current_fs = $(this).parent();
    next_fs = $(this).parent().next();

    if( textnext == $(".next").val()){
        next_fs = $(this).parent(".secound").next();
    }
    next_fs.show();
}

<script src="https://code.jquery.com/jquery-2.1.4.min.js"></script>

<fieldset style="height:100vh">
    <input type="button" name="next" class="next action-button" value="next1" />
</fieldset>
<fieldset style="height:100vh" class="first">
    <input type="button" name="next" class="next action-button" value="Next" />
</fieldset>
<fieldset style="height:100vh" class="secound">
    <input type="button" name="next" class="next action-button" value="Next" />
</fieldset>

​

Answer 1

如果要比较单击的按钮的值

textnext

，你需要做的是

if( textnext == $(this).val()){
    next_fs = $(this).parent(".secound").next();
}

Answer 2

.secound字段集不是您正在单击的元素的父级。

Answer 3

这就是你想要做的？

​

$(".next").click(function(){
   
    var textnext = "next1";
  
    var current_fs = $(this).parent();
  
    var next_fs = $(this).parent().nextAll('.secound');
  
   $('body').scrollTop( next_fs.offset().top);
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.8.3/jquery.min.js"></script>
<div id='test'>
  
<fieldset style="height:50vh">
    <input type="button" name="next" class="next action-button" value="next1" />
</fieldset>
<fieldset style="height:50vh" class="first">
    <input type="button" name="next" class="next action-button" value="Next" />
</fieldset>
<fieldset style="height:50vh" class="secound"  >
    <input type="button" name="next" class="next action-button" value="Next" />
</fieldset>
</div>

​