现在有一组JSONArray数据，我想以其中一种元素来排序(我这里这个元素是int)，这在实际工作中还是经常遇到的，主要是通过重写Comparator。
数据

[{
    
    "repayTerm":"3","test":"2018-12-09"},
{
    
    "repayTerm":"1","test":"2018-12-19"},
{
    
    "repayTerm":"2","test":"2018-12-13"}]  

第一版

jta.sort(new Comparator<Object>() {
    
    
            @Override
            public int compare(Object o1, Object o2) {
    
    
                return ((JSONObject)o1).getIntValue("repayTerm")-((JSONObject)o2).getIntValue("repayTerm");
            }
        });

befor:
[{
    
    "repayTerm":"3","test":"2018-12-09"},
{
    
    "repayTerm":"1","test":"2018-12-19"},
{
    
    "repayTerm":"2","test":"2018-12-13"}]
after:
[{
    
    "repayTerm":"1","test":"2018-12-19"},
{
    
    "repayTerm":"2","test":"2018-12-13"},
{
    
    "repayTerm":"3","test":"2018-12-09"}]

第二版

        jta.sort(((a,b)->((JSONObject)a).getIntValue("repayTerm")-((JSONObject)b).getIntValue("repayTerm")));

befor:
[{
    
    "repayTerm":"3","test":"2018-12-09"},
{
    
    "repayTerm":"1","test":"2018-12-19"},
{
    
    "repayTerm":"2","test":"2018-12-13"}]
after:
[{
    
    "repayTerm":"1","test":"2018-12-19"},
{
    
    "repayTerm":"2","test":"2018-12-13"},
{
    
    "repayTerm":"3","test":"2018-12-09"}]

第三版

        jta.sort(Comparator.comparing(ele -> ((JSONObject) ele).getIntValue("repayTerm")));

befor:
[{
    
    "repayTerm":"3","test":"2018-12-09"},
{
    
    "repayTerm":"1","test":"2018-12-19"},
{
    
    "repayTerm":"2","test":"2018-12-13"}]
after:
[{
    
    "repayTerm":"1","test":"2018-12-19"},
{
    
    "repayTerm":"2","test":"2018-12-13"},
{
    
    "repayTerm":"3","test":"2018-12-09"}]

三版实际上都是做的同样的事情，寻求简洁优雅用第三版，求真务实追本溯源用第一版，第二版属于中庸，简洁且直观。