给定具有未知数量的不同字符串值的值列表，如何获取每个字符串的最新值？
我获得了一个具有以下三个属性的对象列表：Balance，BalanceType和CreatedDate。进入，我知道可以设置BalanceType的许多不同值，但我不能确定有多少不同的值。例如，这是一个有效的输入：
[{
"BalanceType":"Cash",
"Balance":350.03,
"CreatedDate":10-20-16
},
{
"BalanceType":"Cash",
"Balance":250.01,
"CreatedDate":10-20-15
},
{
"BalanceType":"Cash",
"Balance":450.21,
"CreatedDate":10-20-14
},
{
"BalanceType":"Securiites",
"Balance":350.03,
"CreatedDate":10-20-16
}]
如下：
[{
"BalanceType":"Cash",
"Balance":350.03,
"CreatedDate":10-20-16
},
{
"BalanceType":"Credit",
"Balance":250.01,
"CreatedDate":10-20-15
},
{
"BalanceType":"LoanAmount",
"Balance":450.21,
"CreatedDate":10-20-14
},
{
"BalanceType":"Securiites",
"Balance":350.03,
"CreatedDate":10-20-16
}]
我已经尝试过使用Max函数执行此操作，但我发现它只给出了最大指示值，而不是对象。我错过了什么？
Spring boot docs是相关的，但是在mysql中，它对我来说无法使用。
答案 0 :(得分：2)
如果您以var src1 = new[] {
new {
BalanceType = "Cash",
Balance = 350.03,
CreatedDate = new DateTime(2016, 10, 20)
},
new {
BalanceType = "Cash",
Balance = 250.01,
CreatedDate = new DateTime(2015, 10, 20)
},
new {
BalanceType = "Cash",
Balance = 450.21,
CreatedDate = new DateTime(2014, 10, 20)
},
new {
BalanceType = "Securiites",
Balance = 350.03,
CreatedDate = new DateTime(2016, 10, 20)
}
};
var src2 = new[] {
new {
BalanceType = "Cash",
Balance = 350.03,
CreatedDate = new DateTime(2016, 10, 20)
},
new {
BalanceType = "Credit",
Balance = 250.01,
CreatedDate = new DateTime(2015, 10, 20)
},
new {
BalanceType = "LoanAmount",
Balance = 450.21,
CreatedDate = new DateTime(2014, 10, 20)
},
new {
BalanceType = "Securiites",
Balance = 350.03,
CreatedDate = new DateTime(2016, 10, 20)
}
};
var ans1 = from r in src1
group r by r.BalanceType into rg
let latest = rg.Max(r => r.CreatedDate)
select new { BalanceType = rg.Key, Balance = rg.Where(r => r.CreatedDate == latest).FirstOrDefault().Balance, CreatedDate = latest };
var ans2 = from r in src2
group r by r.BalanceType into rg
let latest = rg.Max(r => r.CreatedDate)
select new { BalanceType = rg.Key, Balance = rg.Where(r => r.CreatedDate == latest).FirstOrDefault().Balance, CreatedDate = latest };
格式发布数据，那么它可以直接使用。翻译您的数据，这里是您的答案的查询：
BalanceType我认为如果有多个最新日期为DateTime，那么选择哪个并不重要，所以我使用了第一个。如果您的FirstOrDefault()实际上有时间，则可以将Single()替换为onCreateViewHolder，如果您的假设错误，则会导致您的计划崩溃。