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Javascript jQuery单击事件以显示JSON数组中未正常工作的特定数据_Javascript_Jquery_Arrays_Json - Fatal编程技术网

Javascript jQuery单击事件以显示JSON数组中未正常工作的特定数据

Javascript jQuery单击事件以显示JSON数组中未正常工作的特定数据,javascript,jquery,arrays,json,Javascript,Jquery,Arrays,Json,我用JSON数组创建了3个虚拟用户,如下所示: var userArray = {"users":[ {"Name":"User1","Email":"user1@something.com"}, {"Name":"User2","Email":"user2@something.com"}, {"Name":"User3","Email":"user3@something.com"} ]}; $(document).ready(function(){ $(documen

我用JSON数组创建了3个虚拟用户,如下所示:

var userArray = {"users":[
    {"Name":"User1","Email":"user1@something.com"},
    {"Name":"User2","Email":"user2@something.com"},
    {"Name":"User3","Email":"user3@something.com"}
]};
$(document).ready(function(){
$(document).on("click",".allUser",function() {
    $(this),find('.emails').show(1000);
});

});
现在,我只想显示存储在具有类的div中的每个用户的名称

var i = 0
document.writeln("<div id=tableThis>");
for(i=0;i<userArray.users.length;i++)
{
    document.writeln("<div class=allUser>");
    document.writeln("<b>User :: </b>"
            + userArray.users[i].Name);
    document.writeln("<div class=emails><b>Email :: </b>"
            + userArray.users[i].Email  +"</div>");
    document.writeln("</div>");
}
}))

我更愿意切换每个名称,以便在单击每个用户的名称时显示、隐藏每个用户的电子邮件

当他们不允许
document.writeIn

是否有一种更整洁/更干净的方法来完成这个小任务?

您需要在DOM加载后动态生成html,并使用类获取子元素。email

这样做:

var userArray = {"users":[
    {"Name":"User1","Email":"user1@something.com"},
    {"Name":"User2","Email":"user2@something.com"},
    {"Name":"User3","Email":"user3@something.com"}
]};
$(document).ready(function(){
$(document).on("click",".allUser",function() {
    $(this),find('.emails').show(1000);
});

});

正是我所需要的。非常感谢。