我在library(data.table)
dt <- fread("listing_id date city type host_id availability
703451 25/03/2013 amsterdam Entire_home/apt 3542621 245
703451 20/04/2013 amsterdam Entire_home/apt 3542621 245
703451 28/05/2013 amsterdam Entire_home/apt 3542621 245
703451 15/07/2013 amsterdam Entire_home/apt 3542621 245
703451 30/07/2013 amsterdam Entire_home/apt 3542621 245
703451 19/08/2013 amsterdam Entire_home/apt 3542621 245")
dt$date <- as.Date(dt$date, "%d/%m/%Y")
dt[, .N, by=year(date)]
# year N
# 1: 2013 6
dt[, .N, by=.(year(date), month(date))]
# year month N
# 1: 2013 3 1
# 2: 2013 4 1
# 3: 2013 5 1
# 4: 2013 7 2
# 5: 2013 8 1
dt[, .N, by=date] # or: dt[, .N, by=.(year(date), month(date), day(date)]
# date N
# 1: 2013-03-25 1
# 2: 2013-04-20 1
# 3: 2013-05-28 1
# 4: 2013-07-15 1
# 5: 2013-07-30 1
# 6: 2013-08-19 1
Entities
VS 2015 Model
(家长)有Halaqat_Test
(Primary Key
)
MsqID
(儿童)有Halaqat_Test2
(FOREIGN Key
)
这两个实体之间没有链接或任何关联。 昨天我将每个实体链接到另一个实体。使用此代码
MsqID
但问题是,子 var result = db.Halaqati_Test.GroupJoin
(db.Halaqati_Test2,
c => c.msq_id,
s => s.MsqID,
(c, s) => new { Mosque = c, Rings = s });
return result;
列Table
(未包含在Halaqati_Test2
中的父资产中
看起来像:
Json
但是应该和预期的json是:
[
{
"Mosque": {
"msq_id": 3,
"MsqName": "Center 1",
"RowNum": 1
},
"Rings": [
{
"RingID": 613,
"RingName": "aaaa",
"MsqID": 3,
"RowNum": 231
},
{
"RingID": 614,
"RingName": "bbb",
"MsqID": 3,
"RowNum": 232
},
{
"RingID": 615,
"RingName": "ccc",
"MsqID": 3,
"RowNum": 233
}
]
}
]
更新
我想将Halaqati_Test2中的派生实体视为(父)到view_stdwithrings(Child)
[
{
"Mosque": {
"msq_id": 3,
"MsqName": "Center 1",
"RowNum": 1,
"Rings": [
{
"RingID": 613,
"RingName": "aaaa",
"MsqID": 3,
"RowNum": 231
},
{
"RingID": 614,
"RingName": "bbb",
"MsqID": 3,
"RowNum": 232
},
{
"RingID": 615,
"RingName": "ccc",
"MsqID": 3,
"RowNum": 233
}
]
}
}
]
我怎么能在戒指下制作这个级别,所以戒指会有更多的学生 我该怎么做才能确保孩子进入父节点。
答案 0 :(得分:1)
选择此行 - 从而创建生成的json结构。因为你在同一级别选择“清真寺”和“戒指”,你会得到一个反映它的json结构:
(c, s) => new { Mosque = c, Rings = s });
尝试将其更改为:
(c, s) => new { Mosque = new { msq_id = c.msq_id, MsqName = c.MsqName, RowNum = c.RowNum, Rings = s }});
(免责声明:我这里没有编码环境,所以我无法测试此代码)