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jquery - 我想遍历 age>23 和 <27 的关联数组

问题描述

我正在使用 If 条件循环，但它只显示第一个结果，我必须在条件为 23>age<27 的关联数组中获取所有员工的名字和姓氏

     var employee=[{"firstName":"Zahir","lastName":"Alam","Age":25,"Company":"Switchme","Role":"Developer","Department":"Tech","Head":{"Id":3,"Name":"Sourasis Roy"}},{"firstName":"Amith","lastName":"Manniken","Age":25,"Company":"Switchme","Role":"Developer","Department":"Tech","Head":{"Id":3,"Name":"Sourasis Roy"}},{"firstName":"Sourasis","lastName":"Roy","Age":28,"Company":"Switchme","Role":"CTO"},{"firstName":"Aditya","lastName":"Mishra","Age":29,"Company":"Switchme","Department":"Tech","Role":"CEO"},{"firstName":"Priti","lastName":"Lata","Age":24,"Company":"Switchme","Role":"HR"},{"firstName":"Sumita","lastName":"Nath","Age":24,"Company":"Switchme","Role":"HLA Head","Department":"Crm"},{"firstName":"Tarini","lastName":"Khanna","Age":22,"Company":"Switchme","Role":"Content Writer"},{"firstName":"Abhisek","lastName":"Soni","Age":23,"Company":"Switchme","Role":"HLA","Department":"Crm","Head":{"Id":5,"Name":"Sumita Nath"}},{"firstName":"Ankit","lastName":"Pump","Age":23,"Company":"Switchme","Role":"HLA","Department":"Crm","Head":{"Id":5,"Name":"Sumita Nath"}},{"firstName":"Pogo","lastName":"Laal","Age":23,"Company":"Switchme","Role":"Designer"},{"firstName":"Sabina","lastName":"Sekh","Age":28,"Company":"Switchme","Role":"HLA Head","Department":"Crm"},{"firstName":"Sanjay","lastName":"Poudal","Age":24,"Company":"Switchme","Role":"HLA Head","Department":"Crm","Head":{"Id":10,"Name":"Sabina Sekh"}}];
    $.each( employee, function( key, value ) {
            if(value["Age"] >23 || value["Age"] <27) {
                $("div.age").text(""+ (value["firstName"])+" "+(value["lastName"]));
                $("div.age").join(",");
            }
        });
     <script src="jquery.min.js"></script>
    <h3>4. List out all employee whose age&gt;23 and age&lt;27 ( comma separated)</h3>  
<div class="age"></div>

所有的名字和姓氏

标签： jquery

解决方案

与其尝试用 DOM 做很多事情，不如直接生成所需的输出并将其放入 DOM 中？喜欢

const employee =[{"firstName":"Zahir","lastName":"Alam","Age":25,"Company":"Switchme","Role":"Developer","Department":"Tech","Head":{"Id":3,"Name":"Sourasis Roy"}},{"firstName":"Amith","lastName":"Manniken","Age":25,"Company":"Switchme","Role":"Developer","Department":"Tech","Head":{"Id":3,"Name":"Sourasis Roy"}},{"firstName":"Sourasis","lastName":"Roy","Age":28,"Company":"Switchme","Role":"CTO"},{"firstName":"Aditya","lastName":"Mishra","Age":29,"Company":"Switchme","Department":"Tech","Role":"CEO"},{"firstName":"Priti","lastName":"Lata","Age":24,"Company":"Switchme","Role":"HR"},{"firstName":"Sumita","lastName":"Nath","Age":24,"Company":"Switchme","Role":"HLA Head","Department":"Crm"},{"firstName":"Tarini","lastName":"Khanna","Age":22,"Company":"Switchme","Role":"Content Writer"},{"firstName":"Abhisek","lastName":"Soni","Age":23,"Company":"Switchme","Role":"HLA","Department":"Crm","Head":{"Id":5,"Name":"Sumita Nath"}},{"firstName":"Ankit","lastName":"Pump","Age":23,"Company":"Switchme","Role":"HLA","Department":"Crm","Head":{"Id":5,"Name":"Sumita Nath"}},{"firstName":"Pogo","lastName":"Laal","Age":23,"Company":"Switchme","Role":"Designer"},{"firstName":"Sabina","lastName":"Sekh","Age":28,"Company":"Switchme","Role":"HLA Head","Department":"Crm"},{"firstName":"Sanjay","lastName":"Poudal","Age":24,"Company":"Switchme","Role":"HLA Head","Department":"Crm","Head":{"Id":10,"Name":"Sabina Sekh"}}];

const filtered = employee.filter(e => e.Age > 23 && e.Age < 27).map(e => `${e.firstName} ${e.lastName}`);

$("div.age").text(filtered.join(", "));
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<h3>4. List out all employee whose age&gt;23 and age&lt;27 ( comma separated)</h3>
<div class="age"></div>

div我已经过滤掉了所需的员工，将它们映射到所需的字符串，然后通过加入将它们放入。

此外，您当前代码不起作用的原因是您不关心以前的值div.age是什么，您只是每次都覆盖它。

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