如何解决Oracle SQL数据行迁移到按月列
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let PdfPrinter = require('pdfmake');
let printer = new PdfPrinter();
let docDefinition = {
content: .....
};
pdf = printer.createPdfKitDocument(docDefinition);
pdf.pipe(fs.createWriteStream('YOUR-PDFs/YOUR-PDF.pdf'));
pdf.end();
我正在将数据从旧系统迁移到新系统。 作为旧系统数据的一部分,每月维护一次,如果数据已更新并且表包含一个月的行,则该数据将更新同一行 我正在迁移到新闻系统,它包含开始日期和结束日期以进行活动记录。因此,在更新新数据时,需要插入和更新旧行的结束日期
我的预期数据
CODE1 CODE2 CODE3 RATE VALUE MONTH
A B C 1 1 202001
A B C 1 1 202002
A B C 1 1 202003
A B C 2 1 202004
A B C 2 1 202005
A B C 1 1 202006
A B C 1 1 202007
A B C 1 1 202008
A B C 1 1 202009
如果数据处于活动状态,我们会将日期更新为无穷大,那么999912
但是我只有两条记录,而我的查询在下面
CODE1 CODE2 CODE3 RATE VALUE START_DT END_DT
A B C 1 1 20200101 20200331
A B C 2 1 20200401 20200531
A B C 1 1 20200601 99991230
因为我要根据CODE1,CODE2,CODE3,RATE,VALUE进行分组,并根据分组获取最新数据,所以我无法获取旧数据
请帮助我获得预期的表结构。 预先感谢
如果需要更多详细信息,请发表评论
解决方法
这是一个空白问题,您希望将具有相同比率和值的“相邻”行分组在一起。
一种方法是使用行号之间的差异来构建组。假设这三个代码定义了基本组,并且您希望在速率或值发生变化时都进入新行:
import 'ol/ol.css';
import {Map,View} from 'ol';
import TileLayer from 'ol/layer/Tile';
import OSM from 'ol/source/OSM';
import {ScaleLine,defaults as defaultControls} from 'ol/control';
import TileSource from 'ol/source/Tile';
import TileWMS from 'ol/source/TileWMS';
var capa1 = new TileLayer({
source: new TileWMS({
url: 'http://localhost:8080/geoserver/wms',params: { 'LAYERS': 'earth:d39B19R','TILED': true },serverType: 'geoserver',})
});
var capa2 = new TileLayer({
source: new TileWMS({
url: 'http://localhost:8080/geoserver/wms',params: { 'LAYERS': 'earth:wrf49BR',})
});
var map = new Map({
target: 'map',controls: defaultControls().extend( [
new ScaleLine()
]),layers: [
new TileLayer({
source: new OSM()
}),capa1,capa2
],view: new View({
center: [-413106.304,4923069.399],zoom: 5
})
});
function funcionCapa(){
capa1.setVisible(false);
};
外部查询中的条件表达式将“最后”时间段的结束日期设置为select code1,code2,code3,rate,value,min(month) start_dt,case when row_number() over(partition by code1,code3 order by max(month) desc) = 1 then 999912 else max(month) end end_dt
from (
select t.*,row_number() over(partition by code1,code3 order by month) rn1,value order by month) rn2
from mytable t
) t
group by code1,rn1 - rn2
order by start_dt
。
here :
CODE1 | CODE2 | CODE3 | RATE | VALUE | START_DT | END_DT :---- | :---- | :---- | ---: | ----: | -------: | -----: A | B | C | 1 | 1 | 202001 | 202003 A | B | C | 2 | 1 | 202004 | 202005 A | B | C | 1 | 1 | 202006 | 999912,
您可以使用MATCH_RECOGNIZE
对数据进行逐行比较:
SELECT code1,start_dt,CASE end_dt
WHEN TO_NUMBER( TO_CHAR( SYSDATE,'YYYYMM' ) )
THEN 999912
ELSE end_dt
END AS end_dt
FROM table_name
MATCH_RECOGNIZE (
PARTITION BY code1,code3
ORDER BY month
MEASURES FIRST( rate ) AS rate,FIRST( value ) AS value,FIRST( month ) AS start_dt,LAST( month ) AS end_dt
ONE ROW PER MATCH
PATTERN (FIRST_ROW EQUAL_ROWS*)
DEFINE EQUAL_ROWS AS (
EQUAL_ROWS.rate = PREV(EQUAL_ROWS.rate)
AND EQUAL_ROWS.value = PREV(EQUAL_ROWS.value)
AND TO_DATE( EQUAL_ROWS.month,'YYYYMM' )
= ADD_MONTHS( TO_DATE( PREV(EQUAL_ROWS.month),'YYYYMM' ),1 )
)
)
因此,对于您的示例数据:
CREATE TABLE table_name ( CODE1,CODE2,CODE3,RATE,VALUE,MONTH ) AS
SELECT 'A','B','C',1,201912 FROM DUAL UNION ALL
SELECT 'A',202001 FROM DUAL UNION ALL
SELECT 'A',202002 FROM DUAL UNION ALL
SELECT 'A',202003 FROM DUAL UNION ALL
SELECT 'A',2,202004 FROM DUAL UNION ALL
SELECT 'A',202005 FROM DUAL UNION ALL
SELECT 'A',202006 FROM DUAL UNION ALL
SELECT 'A',202007 FROM DUAL UNION ALL
SELECT 'A',202008 FROM DUAL UNION ALL
SELECT 'A',202009 FROM DUAL;
这将输出:
CODE1 | CODE2 | CODE3 | RATE | VALUE | START_DT | END_DT :---- | :---- | :---- | ---: | ----: | -------: | -----: A | B | C | 1 | 1 | 201912 | 202003 A | B | C | 2 | 1 | 202004 | 202005 A | B | C | 1 | 1 | 202006 | 999912
db 提琴here
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