SQL statements can be used to filter for duplicate billing items with the same name and date

select top 1 product name, time, from product table.

where Product Name ='Enter the name of the product'

order by **Time desc

Didn't you specify an ID, why do you need group by....

select

requestid,clientip,operatetime

from table where

requestid=78921

order by

operatetime desc

limit 1

For example, there is a table in your database with a duplicate name named A (name, logtime) and a duplicate name

select top 1 * from tab where name='a' order by logtime desc

If there is a time column, the largest time column will be used.

If there is an auto-incrementing ID, go to the maximum ID

If you want to return a recordset, and you have a time column, you can do this by selecting the repeat column, the time column, and the rest of the columns.

inner join (select repeating columns, max(time columns) as time columns from table group by repeating columns) as t2

on t1.Repeat column = t2Repeat columns and t1Time column = t2Time column.

from Table as t1

Since the data are all duplicated, the time is different, so just take max (time).

Select Column name 1 for duplicate data and column name 2 for duplicate data ,..max(time column) from table group by Column name 1 of duplicate data, column name of duplicate data 2,..

Group queries are fine.

select by stat date, province name, sum (field name).

group by stat date,province name:

1. Copy columns A and B from Table 1 to columns A and B from Table 3.

2. Enter the formula in C2 in Table 3 and copy it to C248 with the filling handle in the lower right corner

3. Select C2:C248 to clearly mark the negative value in red to warn of insufficient quota.

FYI. SQL language is an abbreviation for Structured Query Language. SQL is a database query and programming language for accessing data and querying, updating, and managing relational database systems; It is also an extension for database script files.

Does an employee only have one card? The following statement assumes that an employee has only one card:

select ,,time of first swipe, time of second swipe from table a1 join table a2 onwhere min(logtime) from table where logcard= and logtime>

and datediff(minute,,>40and between '2014-12-28 11:00:00' and '2014-12-28 13:00:00' and

between '2014-12-28 11:00:00' and '2014-12-28 13:00:00'

A query statement can be found out, which is a bit troublesome to write.

Let's do a function uf datecompare(@date, @strdates) return int

Then it's easy to deal with.

Split the fields first and compare them.

Let's say your date field is date, and the table name is huiyiselect * from huiyi where to char(date,'yyyy-mm-dd') >= '2016-11-29'

The query is greater than this 20161129 and the data after select * from huiyi where date >='2016-11-01' and <='2016-11-29'

The query shows all the meeting data from 2016-11-01 to 2016-11-29.

Your question is also vague, I hope it can help you.

create table test_price (

item varchar(10),itemprice int,price_date date

insert into test_price values('Apples', 5, to_date('2013-01-01', 'yyyy-mm-dd'));

insert into test_price values('Apples', 6, to_date('2013-03-01', 'yyyy-mm-dd'));

insert into test_price values('Apples', 7, to_date('2013-05-01', 'yyyy-mm-dd'));

insert into test_price values('Oranges', 3, to_date('2013-02-01', 'yyyy-mm-dd'));

insert into test_price values('Oranges', 4, to_date('2013-04-01', 'yyyy-mm-dd'));

insert into test_price values('Oranges', 5, to_date('2013-05-01', 'yyyy-mm-dd'));

select

item,itemprice,price_date

fromselect

row_number() over(partition by item order by price_date desc) as no,test_price.*

fromtest_price

whereno = 2;

Query result: item itemprice price date

Orange 4 01-Apr -13

Apple 6 01-Mar-13

If you have any questions, please feel free to ask.

select * from (select row number() over(partition by substance** order by date desc) as bz,* from table name ) aa where 2

Take out the second-to-last date first. Then take out ** as a condition. Just write it in one.

Sort by item**, date, and then take the second piece of data.