my mysql query dont humanize the ORDER BY.
SELECT COUNT(level) as count, level
from logtest
GROUP BY level
ORDER BY level;
Sample:
6
5
5
5
13
0
The correct would be
13
6
5
5
5
0
any help?
SELECT COUNT(level) as count, level
FROM logtest
GROUP BY level
ORDER BY cast(level as unsigned);
Near Dup of: Cast from VARCHAR to INT - MySQL
Not quite a dup because the issues on sort is that it's sorting by a text field (my guess is level is of character type) when you want it to sort by numeric type... so just cast which the above link describes how.
Related
I have the following table as an example:
numbers type
--------------
1 1
5 2
6 1
8 2
9 3
14 2
3 1
From this table I would like to select the closest number that is less or equal to 5 AND of type 1 and if there is no such row matching, then (and only then) I would like to return the first closest number larger than 5 of type 2
I can solve this by running two queries:
SELECT number FROM numbers WHERE number <= 5 AND type = 1 ORDER BY number LIMIT 1
and if above query returns 0 results, I simply run the second query:
SELECT number FROM numbers WHERE number > 5 AND type = 2 ORDER BY number LIMIT 1
But is it possible, to achieve the same result by only using one query?
I was thinking something like
SELECT number FROM numbers WHERE (number <= 5 AND type = 1) OR (number > 5 AND type = 2) ORDER BY number LIMIT 1
But that would only work, if mysql first checks the first conditional in the parentheses against all rows and if it finds a match, it returns it, and if not, then it checks all rows against the second parenthesed conditional. It will not work, if it checks each row against both parentheses and only then moves to the next row, which is how I suspect it works.
This query will do what you want. It selects all numbers that match your two query constraints, and orders the results first by type (so that if there is a result for type 1 it will appear first) and then by either -number or number dependent on type (so that numbers <= 5 sort in descending order but numbers > 5 sort in ascending order):
SELECT number
FROM numbers
WHERE ( number <= 5 AND type = 1 )
OR ( number > 5 AND type = 2 )
ORDER BY type, CASE WHEN type = 1 THEN -number ELSE number END
LIMIT 1
Output:
3
Demo on dbfiddle
Combine the two, and you always prefer type 1 over type 2, hence the ORDER BY and LIMIT. The ABS means whichever is first by type, is the closes to the number 5.
SELECT number, type
FROM numbers
WHERE (number <=5 AND type=1) OR
(number > 5 AND type=2)
ORDER BY type ASC, ABS(number-5) ASC
LIMIT 1
I have a database that lists a few area codes, area code + office codes and some whole numbers and a action. I want it to return a result by the digits given but I am not sure how to accomplish it. I have some MySQL knowledge but its not very deep.
Here is a example:
match | action
_____________________
234 | goto 1
333743 | goto 2
8005551212| goto 3
234843 | goto 4
I need to query the database with a full 10 digit number -
query 8005551212 gives "goto 3"
query 2345551212 gives "goto 1"
query 3337431212 gives "goto 2"
query 2348431212 gives "goto 4"
This would be similar to the LIKE selection, but I need to match against the database value instead of the query value. Matching the full number is easy,
SELECT * FROM database WHERE `match` = 8005551212;
First the number to query will always be 10 digits, so I am not sure how to format the SELECT statement to differentiate the match of 234XXXXXXX and 234843XXXX, as I can only have one match return. Basically if it does not match the 10 digits, then it checks 6 digits, then it will check the 3 digits.
I hope this makes sense, I do not have any other way to format the number and it has to be accomplished with just a single SQL query and return over a ODCB connection in Asterisk.
Try this
SELECT match, action FROM mytable WHERE '8005551212' like concat(match,'%')
The issue is that you will get two rows in one case .. given your data..
SELECT action
FROM mytable
WHERE '8005551212' like concat(match,'%')
order by length(match) desc limit 1
That should get the row that had the most digits matched..
try this:
SELECT * FROM (
SELECT 3 AS score,r.* FROM mytable r WHERE match LIKE CONCAT(SUBSTRING('1234567890',1,3),'%')
UNION ALL
SELECT 6 AS score,r.* FROM mytable r WHERE match LIKE CONCAT(SUBSTRING('1234567890',1,6),'%')
UNION ALL
SELECT 10 AS score,r.* FROM mytable r WHERE match LIKE CONCAT(SUBSTRING('1234567890',1,10),'%')
) AS tmp
ORDER BY score DESC
LIMIT 1;
What ended up working -
SELECT `function`,`destination`
FROM reroute
WHERE `group` = '${ARG2}'
AND `name` = 0
AND '${ARG1}' LIKE concat(`match`,'%')
ORDER BY length(`match`) DESC LIMIT 1
I have a database with 2 columns (ex: no_1, no_2) and then 5000 rows of data, numbers are between 1 - 20 , I need to find a pairs of numbers which where repeating the most of the times?
Any help please ?
Something like this maybe:-
SELECT no_1, no_2, COUNT(*) AS numbercount
FROM SomeTable
GROUP BY no_1, no_2
ORDER BY numbercount DESC
Hi I want to sort a table .The field contains numbers,alphabets and numbers with alphabets ie,
1
2
1a
11a
a
6a
b
I want to sort this to,
1
1a
2
6a
11a
a
b
My code is, SELECT * FROM t ORDER BY CAST(st AS SIGNED), st
But the result is,
a
b
1
1a
2
6a
11a
I found this code in this url "http://www.mpopp.net/2006/06/sorting-of-numeric-values-mixed-with-alphanumeric-values/"
Anyone please help me
This would do your required sort order, even in the presence of 0 in the table;
SELECT * FROM t
ORDER BY
st REGEXP '^[[:alpha:]].*',
st+0,
st
An SQLfiddle to test with.
As a first sort criteria, it sorts anything that starts with a letter after anything that doesn't. That's what the regexp does.
As a second sort criteria it sorts by the numerical value the string starts with (st+0 adds 0 to the numerical part the string starts with and returns an int)
As a last resort, it sorts by the string itself to get the alphabetical ones in order.
You can use this:
SELECT *
FROM t
ORDER BY
st+0=0, st+0, st
Using st+0 the varchar column will be casted to int. Ordering by st+0=0 will put alphanumeric rows at the bottom (st+0=0 will be 1 if the string starts with an alphanumeric character, oterwise it will be 0)
Please see fiddle here.
The reason that you are getting this output is that all the character like 'a', 'b' etc are converted to '0' and if you use order by ASC it will appear at the top.
SELECT CAST(number AS SIGNED) from tbl
is returning
1
2
1
11
0
6
0
Look at this fiddle:- SQL FIDDLE
I did small change in your query -
SELECT *, CAST(st AS SIGNED) as casted_column
FROM t
ORDER BY casted_column ASC, st ASC
this should work.
in theory your syntax should work but not sure why mysql doesn't accept these methods after from tag.
so created temp field and then sorted that one .
This should work as per my experience, and you can check it.
SQL FIDDLE
I want to count from the row with the least value to the row with a specific value.
For example,
Name / Point
--------------------
Pikachu / 7
Voltorb / 1
Abra / 4
Sunflora / 3
Squirtle / 8
Snorlax / 12
I want to count to the 7, so I get the returned result of '4' (counting the rows with values 1, 3, 4, 7)
I know I should use count() or mysql_num_rows() but I can't think of the specifics.
Thanks.
I think you want this :
select count(*) from mytable where Point<=7;
Count(*) counts all rows in a set.
If you're working with MySQL, then you could ORDER BY Point:
SELECT count(*) FROM table WHERE Point < 7 ORDER BY Point ASC
If you want to know all about ORDER BY, check out the w3schools page: http://www.w3schools.com/sql/sql_orderby.asp
Just in case you want to only count the rows based on the Point values:
SELECT count(*) FROM table WHERE Point < 7 GROUP BY Point
This may help you to get rows falling between range of values :
select count(*) from table where Point >= least_value and Point<= max_value