Question

我需要用下面的代码帮助检索数据：

<div id="divResult"></div> 

$(document).ready(function() { 
    var jsonString = '{"Fakir":{"firstName":"Bharat","lastName":"Tiwari","gender":"Male","salary":50000},"Pagal":{"firstName":"Nanu","lastName":"Pagal","gender":"Male","salary":90000}}'; 
    var employeeJSON = JSON.parse(jsonString); 
    var result = ''; 

    $.each(employeeJSON.Fakir,function(i, item){ 
     result += item['firstName']+"<br>"; 
     /*result += item.lastName + "<br>"; 
     result += item.gender + "<br>"; 
     result += item.salary + "<br> <br>";*/ 
    }); 
    $("#divResult").html(result) 
}) 
$("#divResult").html(result) 

我得到的输出是undefined。请帮我找一个解决方案？为什么我没有获取数据？

Answer 1

环employeeJSON而不是employeeJSON.Fakir

$(document).ready(function() { 
 
    var jsonString = '{"Fakir":{"firstName":"Bharat","lastName":"Tiwari","gender":"Male","salary":50000},"Pagal":{"firstName":"Nanu","lastName":"Pagal","gender":"Male","salary":90000}}'; 
 

 
var employeeJSON = JSON.parse(jsonString); 
 
var result = ''; 
 
$.each(employeeJSON,function(i,item){ 
 
    result += item['firstName'] + "<br>"; 
 
    result += item.lastName + "<br>"; 
 
    result += item.gender + "<br>"; 
 
    result += item.salary + "<br> <br>"; 
 
    }); 
 
    $("#divResult").html(result) 
 
    })

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> 
 
<div id="divResult"> </div>

Answer 2

由于对象的Fakir属性本身就是一个对象，传递给each()处理函数的参数将成为对象中属性的键和值。因此，你可以将它们直接串联字符串，像这样：

$.each(employeeJSON.Fakir, function(k, v) { 
    result += v + "<br>"; 
}); 

Working example

Answer 3

。在你的jsonString没有数组变量。你为什么试图迭代这个对象？

您可以这样做，如下所示;

$(document).ready(function() { 
    var jsonString = '{"Fakir":{"firstName":"Bharat","lastName":"Tiwari","gender":"Male","salary":50000},"Pagal":{"firstName":"Nanu","lastName":"Pagal","gender":"Male","salary":90000}}'; 

    var employeeJSON = JSON.parse(jsonString); 
    var result = ''; 
    var item = employeeJSON.Fakir; 
    result += item['firstName']+"<br>"; 
    result += item.lastName + "<br>"; 
    result += item.gender + "<br>"; 
    result += item.salary + "<br> <br>"; 

    $("#divResult").html(result) 
});