答案 0 :(得分:3)
您需要通过类名组合$.fn.filter
div:
$('#filternumbers').on('change', function () {
var number = $(this).val();
var classNames = '.' + number.join('.');
$(".numbers").hide().filter(classNames).show();
});

.numbers {display: none;}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<select id="filternumbers" multiple="multiple">
<option value="one">One</option>
<option value="two">Two</option>
<option value="three">Three</option>
<option value="four">Four</option>
<option value="five">Five</option>
<option value="six">Six</option>
</select>
<div class="numbers one two three">one two three</div>
<div class="numbers one three five">one three five</div>
<div class="numbers two four six">two four six</div>
<div class="numbers one two four">one two four</div>
&#13;
答案 1 :(得分:1)
您可以检查首先检查的所有值,然后运行每个()检查已检查的值,然后使用过滤器显示所需的值
@OneToMany(mappedBy = "evaluationType", cascade = CascadeType.ALL)
private List<Member> memberList;
参考jsfiddle http://jsfiddle.net/gsmd2jd3/