SQL - ORDER table and SELECT all after row 25

Question

I am trying to create a SQL call that first order all the rows by date and returns all the rows starting from the number 25.

So far I have

"SELECT * FROM 'users' ORDER BY 'table'.'regdate' OFFSET 24"

But this is not working .. How can I do this ?

please check this : `SELECT * FROM 'users' ORDER BY 'table'.'regdate' limit 25,1000` — , Feb 16 '16 at 10:43
Yes this working.. However I limit the search to 1000 rows.. The table is containing more and always changing — SNos, Feb 16 '16 at 10:45
Possible duplicate of [MySQL skip first 10 results](http://stackoverflow.com/questions/2827029/mysql-skip-first-10-results) — CFreitas, Feb 16 '16 at 10:46
http://stackoverflow.com/questions/2827029/mysql-skip-first-10-results — Karthik N, Feb 16 '16 at 10:46

score 0 · Accepted Answer · answered Feb 16 '16 at 10:46

0

try like this:

first get count all data in table like this:

$count=select count(*) from users

then

SELECT * FROM 'users' ORDER BY 'table'.'regdate' limit 25,$count

answered Feb 16 '16 at 10:46

score 0 · Answer 2 · answered Feb 16 '16 at 11:25

There is nothing wrong with your query, the problem must be elsewhere.

Your query includes an offset which omits the first 24 results. This imposes no limits and should therefore return evrey row pat the 24 oldest dates. To omit the new dates instead you have to set your order as DESC

SELECT * FROM 'users' ORDER BY 'table'.'regdate' OFFSET 24

To do that and limit results to 100 records simply do this:

SELECT * FROM my_table ORDER BY date_field DESC OFFSET 24 LIMIT 100

SQL - ORDER table and SELECT all after row 25

2 Answers2