How to go about selecting all data from the most recent three records in table. This is specifically for a blog that on the home page will show the three most recently dated articles.
ID |title |date |...
---------------------------
1 |exampTitle1|2018-04-25|
2 |exampTitle2|2019-02-10|
3 |exampTitle3|2007-12-21|
4 |exampTitle4|2019-08-10|
The example table shows a subsection of the table, I need to select all data from each record but only from the most recent THREE records, I am aware of MAX() but as far as I am aware that can only get the most recent, not
In SQL, you can order the rows with ORDER BY. Then, you can LIMIT what you are selecting with the LIMIT 3 spesific to SQL of some DBMS products such as MySQL :
SELECT * FROM table ORDER BY date DESC LIMIT 3;
This should work, by sorting by ID in descending order and using LIMIT:
SELECT * FROM <TABLE_NAME> ORDER BY id DESC LIMIT 3
I have a MySQL Database with over 100k rows, so I need to make a search to fetch only the last 1000 rows , so if it is not found in the last 1000 rows the fetch ends (even if it is not found)
Example: if my table is like that
id name
1 AL
2 BL
...
1000 P12
1001 P15
And I do a fetch like this: SELECT * FROM myTable WHERE name = 'AL' ONLY LAST 1000 ROWS ORDER BY id DESC (Since I don't know what to use I invented the ONLY LAST 1000 ROWS)
This should return empty because I wanted my query to get the information only if it was on the last 1000 rows, not on the 1001th as specified.
Using LIMIT field doesn't work as it would LIMIT the FOUND ROWS not when they are not found.
Is there a way to implement this in MySQL ?
Thank you!
As touched on in the comments, you can use OFFSET to get the id of the 1000th last record, then SELECT records with an id larger than that record's id.
Something like this:
SELECT name
FROM myTable
WHERE id > (SELECT id FROM myTable ORDER BY id DESC LIMIT 1 OFFSET 1000)
AND name = 'AL'
I'm looking for a mysql select that will allow me to select (LIMIT 8) records after some changing number of first few matches;
select id
from customers
where name LIKE "John%"
Limit 8
So if i have a table with 1000 of johns with various last names
I want to be able to select records 500-508
You can send the offset to the limit statement, like this:
SELECT id
FROM customers
WHERE name LIKE "John%"
LIMIT 8 OFFSET 500
Notice the OFFSET 500 on the limit. That sets the 'start point' past the first 500 entries (at entry #501).
Therefor, entries #501, #502, #503, #504, #505, #506, #507 and #508 will be selected.
This can also be written:
LIMIT 500, 8
Personally, I don't like that as much and don't understand the order.
Pedantic point: 500-508 is 9 entries, so I had to adjust.
As a solution please try executing the following sql query
select id from customers where name LIKE "John%" Limit 500,8
I wonder if anyone could help with a MySQL query I am trying to write to return relevant results.
I have a big table of change log data, and I want to retrieve a number of record 'groups'. For example, in this case a group would be where two or more records are entered with the same timestamp.
Here is a sample table.
==============================================
ID DATA TIMESTAMP
==============================================
1 Some text 1379000000
2 Something 1379011111
3 More data 1379011111
3 Interesting data 1379022222
3 Fascinating text 1379033333
If I wanted the first two grouped sets, I could use LIMIT 0,2 but this would miss the third record. The ideal query would return three rows (as two rows have the same timestamp).
==============================================
ID DATA TIMESTAMP
==============================================
1 Some text 1379000000
2 Something 1379011111
3 More data 1379011111
Currently I've been using PHP to process the entire table, which mostly works, but for a table of 1000+ records, this is not very efficient on memory usage!
Many thanks in advance for any help you can give...
Get the timestamps for the filtering using a join. For instance, the following would make sure that the second timestamp is in a completed group:
select t.*
from t join
(select timestamp
from t
order by timestamp
limit 2
) tt
on t.timestamp = tt.timestamp;
The following would get the first three groups, no matter what their size:
select t.*
from t join
(select distinct timestamp
from t
order by timestamp
limit 3
) tt
on t.timestamp = tt.timestamp;
Is it possible to sort in MySQL by "order by" using a predefined set of column values (ID) like order by (ID=1,5,4,3) so I would get records 1, 5, 4, 3 in that order out?
UPDATE: Why I need this...
I want my records to change sort randomly every 5 minutes. I have a cron task to update the table to put different, random sort order in it.
There is just one problem! PAGINATION.
I will have visitors who come to my page, and I will give them the first 20 results. They will wait 6 minutes, go to page 2 and have the wrong results as the sort order has already changed.
So I thought that if I put all the IDs into a session on page 2, we get the correct records even if the sorting had already changed.
Is there any other better way to do this?
You can use ORDER BY and FIELD function.
See http://lists.mysql.com/mysql/209784
SELECT * FROM table ORDER BY FIELD(ID,1,5,4,3)
It uses Field() function, Which "Returns the index (position) of str in the str1, str2, str3, ... list. Returns 0 if str is not found" according to the documentation. So actually you sort the result set by the return value of this function which is the index of the field value in the given set.
You should be able to use CASE for this:
ORDER BY CASE id
WHEN 1 THEN 1
WHEN 5 THEN 2
WHEN 4 THEN 3
WHEN 3 THEN 4
ELSE 5
END
On the official documentation for mysql about ORDER BY, someone has posted that you can use FIELD for this matter, like this:
SELECT * FROM table ORDER BY FIELD(id,1,5,4,3)
This is untested code that in theory should work.
SELECT * FROM table ORDER BY id='8' DESC, id='5' DESC, id='4' DESC, id='3' DESC
If I had 10 registries for example, this way the ID 1, 5, 4 and 3 will appears first, the others registries will appears next.
Normal exibition
1
2
3
4
5
6
7
8
9
10
With this way
8
5
4
3
1
2
6
7
9
10
There's another way to solve this. Add a separate table, something like this:
CREATE TABLE `new_order` (
`my_order` BIGINT(20) UNSIGNED NOT NULL,
`my_number` BIGINT(20) NOT NULL,
PRIMARY KEY (`my_order`),
UNIQUE KEY `my_number` (`my_number`)
) ENGINE=INNODB;
This table will now be used to define your own order mechanism.
Add your values in there:
my_order | my_number
---------+----------
1 | 1
2 | 5
3 | 4
4 | 3
...and then modify your SQL statement while joining this new table.
SELECT *
FROM your_table AS T1
INNER JOIN new_order AS T2 on T1.id = T2.my_number
WHERE ....whatever...
ORDER BY T2.my_order;
This solution is slightly more complex than other solutions, but using this you don't have to change your SELECT-statement whenever your order criteriums change - just change the data in the order table.
If you need to order a single id first in the result, use the id.
select id,name
from products
order by case when id=5 then -1 else id end
If you need to start with a sequence of multiple ids, specify a collection, similar to what you would use with an IN statement.
select id,name
from products
order by case when id in (30,20,10) then -1 else id end,id
If you want to order a single id last in the result, use the order by the case. (Eg: you want "other" option in last and all city list show in alphabetical order.)
select id,city
from city
order by case
when id = 2 then city else -1
end, city ASC
If i had 5 city for example, i want to show the city in alphabetical order with "other" option display last in the dropdown then we can use this query.
see example other are showing in my table at second id(id:2) so i am using "when id = 2" in above query.
record in DB table:
Bangalore - id:1
Other - id:2
Mumbai - id:3
Pune - id:4
Ambala - id:5
my output:
Ambala
Bangalore
Mumbai
Pune
Other
SELECT * FROM TABLE ORDER BY (columnname,1,2) ASC OR DESC