Question

这听起来很奇怪，但是我想用一些数据更新date列。请在下表中找到。而且我正在处理JSON格式的数据。

CityName | data1 | data2 | date

Mumbai   | 1.234 | 2.3456| Sat Jan 20 2018 12:00:00 GMT+0000 (UTC)
Mumbai   | 6.234 | 2.3456| Sat Jan 20 2018 18:00:00 GMT+0000 (UTC) 
Mumbai   | 1.234 | 2.3456| Sun Jan 21 2018 12:00:00 GMT+0000 (UTC) 
Mumbai   | 2.234 | 2.3456| Sun Jan 21 2018 18:00:00 GMT+0000 (UTC)
Mumbai   | 1.234 | 2.3456| Sun Jan 21 2018 24:00:00 GMT+0000 (UTC)
Mumbai   | 1.234 | 2.3456| Sun Jan 21 2018 00:00:00 GMT+0000 (UTC)
Mumbai   | 8.234 | 2.3456| Fri Jan 19 2018 01:00:00 GMT+0000 (UTC)
Mumbai   | 3.334 | 2.3456| Fri Jan 19 2018 12:00:00 GMT+0000 (UTC)
Mumbai   | 8.214 | 2.3456| Fri Jan 19 2018 14:00:00 GMT+0000 (UTC)
Mumbai   | 19.234| 2.3456| Thu Jan 18 2018 12:00:00 GMT+0000 (UTC)
Pune     | 1.234 | 2.3456| Wed Jan 17 2018 12:00:00 GMT+0000 (UTC)
Pune     | 1.234 | 2.3456| Wed Jan 17 2018 12:00:00 GMT+0000 (UTC)
Pune     | 1.234 | 2.3456| Sat Jan 20 2018 12:00:00 GMT+0000 (UTC)
Pune     | 1.234 | 2.3456| Sat Jan 20 2018 18:00:00 GMT+0000 (UTC) 
Pune     | 1.234 | 2.3456| Sun Jan 21 2018 12:00:00 GMT+0000 (UTC) 
Pune     | 1.234 | 2.3456| Sun Jan 21 2018 18:00:00 GMT+0000 (UTC)
Pune     | 1.234 | 2.3456| Sun Jan 21 2018 24:00:00 GMT+0000 (UTC)
Pune     | 1.234 | 2.3456| Sun Jan 21 2018 00:00:00 GMT+0000 (UTC)
Pune     | 1.234 | 2.3456| Fri Jan 19 2018 01:00:00 GMT+0000 (UTC)
Pune     | 1.234 | 2.3456| Fri Jan 19 2018 12:00:00 GMT+0000 (UTC)
Pune     | 1.234 | 2.3456| Fri Jan 19 2018 14:00:00 GMT+0000 (UTC)
Ahmadabad| 1.234 | 2.3456| Thu Jan 18 2018 12:00:00 GMT+0000 (UTC)
Ahmadabad| 1.234 | 2.3456| Wed Jan 17 2018 12:00:00 GMT+0000 (UTC)
Ahmadabad| 1.234 | 2.3456| Wed Jan 17 2018 12:00:00 GMT+0000 (UTC)

现在，根据城市名称，我想更新所有日期，例如，今天是2018年7月23日，我选择了Mumbai，现在我想先获取与所选城市有关的所有数据（例如“孟买”）按日期降序排列，并以昨天的日期（即，每4条记录的日期从2018年7月22日开始）以降序排列，从而开始更新记录，因此时间戳会相差6个小时。但是剩余的列数据不应更新。

然后针对孟买市，如果我按降序对数据进行排序，那么我会得到

Mumbai   | 1.234 | 2.3456| Sun Jan 21 2018 12:00:00 GMT+0000 (UTC) 
Mumbai   | 2.234 | 2.3456| Sun Jan 21 2018 18:00:00 GMT+0000 (UTC)
Mumbai   | 1.234 | 2.3456| Sun Jan 21 2018 24:00:00 GMT+0000 (UTC)
Mumbai   | 1.234 | 2.3456| Sun Jan 21 2018 00:00:00 GMT+0000 (UTC) 
Mumbai   | 1.234 | 2.3456| Sat Jan 20 2018 12:00:00 GMT+0000 (UTC)
Mumbai   | 6.234 | 2.3456| Sat Jan 20 2018 18:00:00 GMT+0000 (UTC) 
Mumbai   | 8.234 | 2.3456| Fri Jan 19 2018 01:00:00 GMT+0000 (UTC) 
Mumbai   | 3.334 | 2.3456| Fri Jan 19 2018 12:00:00 GMT+0000 (UTC)
Mumbai   | 8.214 | 2.3456| Fri Jan 19 2018 14:00:00 GMT+0000 (UTC)
Mumbai   | 19.234| 2.3456| Thu Jan 18 2018 12:00:00 GMT+0000 (UTC)

现在预期的o / p应该是

Mumbai   | 1.234 | 2.3456| Sun Jul 22 2018 00:00:00 GMT+0000 (UTC) 
Mumbai   | 2.234 | 2.3456| Sun Jul 22 2018 06:00:00 GMT+0000 (UTC)  
Mumbai   | 1.234 | 2.3456| Sun Jul 22 2018 12:00:00 GMT+0000 (UTC)
Mumbai   | 1.234 | 2.3456| Sun Jul 22 2018 18:00:00 GMT+0000 (UTC) 
Mumbai   | 1.234 | 2.3456| Sat Jul 21 2018 00:00:00 GMT+0000 (UTC)
Mumbai   | 6.234 | 2.3456| Sat Jul 21 2018 06:00:00 GMT+0000 (UTC)  
Mumbai   | 8.234 | 2.3456| Sat Jul 21 2018 12:00:00 GMT+0000 (UTC)
Mumbai   | 3.334 | 2.3456| Sat Jul 21 2018 18:00:00 GMT+0000 (UTC)
Mumbai   | 8.214 | 2.3456| Fri Jul 20 2018 00:00:00 GMT+0000 (UTC)
Mumbai   | 19.234| 2.3456| Fri Jul 20 2018 06:00:00 GMT+0000 (UTC)

我不应该更新其他列数据，而只更新我要更新的日期，方法是对同一日期但在不同时间段进行4条记录。

或者也可以接受任何中间件逻辑（最好使用javascript），它不会真正更新数据库中的数据，但可以在两者之间进行操作。

我们将不胜感激....！

Answer 1

这非常简单-只需对行编号，然后根据行号（SQLfiddle）计算时间

SELECT 
  IF(@city = cityname, @ctr := @ctr + 1, @ctr := 0) AS rownum, 
  @city := cityname AS cityName,
  DATE_SUB(CURRENT_DATE, INTERVAL CEIL((@ctr + 1) / 4) DAY) AS datum,
  SEC_TO_TIME((@ctr % 4) * 21600) AS vreme
FROM tblCity 
JOIN (SELECT @ctr := 0) AS tmp 
ORDER BY cityName,date DESC

编辑-说明

(SELECT @ctr := 0) AS tmp初始化一个自定义会话变量，其值为0。它将用于计算每个城市的行数。

然后您将获得城市和时间戳的列表-按升序对城市进行排序，但按降序对时间戳进行排序。

IF(@city = cityname, @ctr := @ctr + 1, @ctr := 0)确保每次您为下一个城市启动新的一组行时，将行号重置为0-我们使用会话变量@city来捕获切换。

DATE_SUB(CURRENT_DATE, INTERVAL CEIL((@ctr + 1) / 4) DAY)只需获取当前日期，并从中减去多少天，就如同当前城市有4行的分组一样。它将对0/1/2/3行减去1，然后对4/5/6/7行减去2，然后对8/9/10/11行减去3，依此类推。

SEC_TO_TIME((@ctr % 4) * 21600)只需将HH：MM：SS格式的秒数转换为时间，并且每行以6小时递增。

Answer 2

您可以使用窗口函数ROW_NUMBER。并使用它计算添加到当前日期的小时或秒。

但是在不同的DBMS中，日期时间函数有所不同。
因此，很难编写适用于任何数据库的SQL。

这是在MS Sql Server上运行的示例

select CityName, data1, data2 
, dateadd(hour, 
         (((row_number() over (partition by CityName order by "date")-1)%4)*6)
       -((floor((row_number() over (partition by CityName order by "date")-1)/4)+1)*24),
         cast(cast(current_timestamp as Date) as datetime))  as "date"
from grad as t
where CityName = 'Mumbai'
order by t."date";

测试here

我还没有找到类似的SAP HANA在线测试仪。

但是根据某些文档，它支持窗口功能。而且它具有ADD_SECONDS函数。

因此，此SQL也许可以在SAP HANA中使用：

select CityName, data1, data2 
, ADD_SECONDS(
        cast(cast(current_timestamp as Date) as datetime),
        ((mod((row_number() over (partition by CityName order by "date")-1),4)*6)
       -((floor((row_number() over (partition by CityName order by "date")-1)/4)+1)*24))*3600
         )  as "date"
from grad as t
where CityName = 'Mumbai'
order by t."date"

Answer 3

/* let the data is stored in the values key of json object i.e */
const jsonData = { values: [
  {
    "cityName": "Mumbai",
    "data1": 1.234,
    "data2": 2.3456,
    "date": "Sat Jan 20 2018 12:00:00 GMT+0000 (UTC)"
  },
  {
    "cityName": "Mumbai",
    "data1": 6.234,
    "data2": 2.3456,
    "date": "Sat Jan 20 2018 18:00:00 GMT+0000 (UTC)"
  },
  ...
]};

/* clone jsonData object into dataObj*/
 const dataObj = Object.assign({}, jsonData);

/* get the array of values from the cloned object i.e. dataObj */
const values = dataObj.values;

/* get name of selected city */
const selectedCity = 'mumbai';

/* find the records of the selected city from the array (convert to lowercase for matching to abort case sensitivity) */
let selectedCityRecords = values.filter(d => d.cityName.toLowerCase() === selectedCity.toLowerCase());

/* sort the records with the descending order of date */
selectedCityRecords.sort((a, b) => {
  var dateA = new Date(a.date);
  var dateB = new Date(b.date)
  if (dateA < dateB) {
    return 1;
  }
  if (dateA > dateB) {
    return -1;
  }

  return 0;
});

let i = 0;
var startDate = new Date();

/* Function to calculate the process date according to slot number */
function getdateTimeSlot(processDate, slotNumber) {
  processDate.setHours(slotNumber * 6, 0, 0);
  return processDate;
}

/* map the selected city data records */
selectedCityRecords.map(data => {
  updatedData = data;
  /* with 6 hours there are 4 slot for a day, after every slot date get decrease by one */
  if (i % 4 === 0) {
    startDate.setDate(startDate.getDate() - 1);
  }
  slotNumber = i % 4;
  newDate = getdateTimeSlot(startDate, slotNumber);
  updatedData.date = newDate.toString();
  i += 1;
  return updatedData;
});

console.log(selectedCityRecords);

想要更新日期列中的日期，以及过去日期从昨天的日期开始

3 个答案:

编辑-说明