Can anyone lend me a hand as to what I should append to my ORDER BY statement to sort these values naturally:
1
10
2
22
20405-109
20405-101
20404-100
X
Z
D
Ideally I'd like something along the lines of:
1
2
10
22
20404-100
20405-101
20405-109
D
X
Z
I'm currently using:
ORDER BY t.property, l.unit_number
where the values are l.unit_number
I've tried doing l.unit_number * 1 and l.unit_number + 0 but they haven't worked.
Should I be doing sort of ORDER conditional, such as Case When IsNumeric(l.unit_number)?
Thank you.
This will do it:
SELECT value
FROM Table1
ORDER BY value REGEXP '^[A-Za-z]+$'
,CAST(value as SIGNED INTEGER)
,CAST(REPLACE(value,'-','')AS SIGNED INTEGER)
,value
The 4 levels of the ORDER BY:
REGEXP assigns any alpha line a 1 and non-alphas a 0
SIGNED INT Sorts all of the numbers by the portion preceding the dash.
SIGNED INT after removing the dash sorts any of the items with the same value before the dash by the portion after the dash. Potentially could replace number 2, but wouldn't want to treat 90-1 the same as 9-01 should the case arise.
Sorts the letters alphabetically.
Demo: SQL Fiddle
Related
I have a mysql database that contains a range of postcode district ranges.
I am trying to find the row that matches the given district, in this example 'SY18'.
I have mysql query below, but it returns both the values in the table because the fields are alphanumeric.
SELECT rate FROM table WHERE district_from <= 'SY18' AND district_to >= 'SY18'
Table example
id district_from district_to rate
1 SY1 SY9 10
2 SY16 SY22 20
3 AL1 AL99 37
4 B1 B99 37
5 BB1 BB99 37
6 CB1 CB99 40
How do I return only correct results?
You can use SUBSTR to skip first characters of the string and then +0 to consider the value as a number. Example:
SELECT v
FROM
(SELECT 'SY1' v UNION SELECT 'SY16') t
WHERE SUBSTR(t.v, 3)+0 > 10
;
CREATE FUNCTION normalize_district (district VARCHAR(4))
RETURNS VARCHAR(4) DETERMINISTIC
RETURN CONCAT( TRIM(TRAILING CASE WHEN 0 + RIGHT(district, 2) > 0
THEN RIGHT(district, 2)
ELSE RIGHT(district, 1) END FROM district),
LPAD(CASE WHEN 0 + RIGHT(district, 2) > 0
THEN 0 + RIGHT(district, 2)
ELSE 0 + RIGHT(district, 1) END, 2, '0') );
and then
SELECT *
FROM district_to_test
LEFT JOIN rate ON normalize_district(district_to_test.district)
BETWEEN normalize_district(district_from)
AND normalize_district(district_to);
fiddle
0 + RIGHT(district, 2) > 0 in the function thecks does the last 2 symbols in district are digits.
If true then 2-digit number wil be extracted by RIGHT(), and the whole expression value will be above zero (you claim that there is no values like 'AA0' or 'AA00').
If false, and only one digit is present, then RIGHT() will give a substring which is started from a letter, and the value will be zero.
Based on this I divide the whole value to a prefix and numeric postfix, and add a zero to the numeric part making it 2-digit unconditionally. Such value can be used for direct string comparing by BETWEEN operator.
The goal of a function is to convert the value to 'AA00' format. For this purposes we must divide the whole value to textual and numeric parts and normalize (zero-pad) the latter one.
You may use any other dividing method. For example, you may check does the value contains two digits using LIKE '%\\d\\d'. Or backward, you may determine does the value contains two letters using LIKE '\\D\\D%'... these methods have relatively equal difficulty.
I have a SQL Table in that i use BETWEEN Operater.
The BETWEEN Operater selects values within range. The values can be numbers, text , dates.
stu_id name city pin
1 Raj Ranchi 123456
2 sonu Delhi 652345
3 ANU KOLKATA 879845
4 K.K's Company Delhi 345546
5 J.K's Company Delhi 123456
I have a query like this:-
SELECT * FROM student WHERE stu_id BETWEEN 2 AND 4 //including 2 & 4
SELECT * FROM `student` WHERE name between 'A' and 'K' //including A & not K
Here My Question is why not including K.
but I want K also in searches.
Don't use between -- until you really understand it. That is just general advice. BETWEEN is inclusive, so your second query is equivalent to:
WHERE name >= 'A' AND
name <= 'K'
Because of the equality, 'K' is included in the result set. However, names longer than one character and starting with 'K' are not -- "Ka" for instance.
Instead, be explicit:
WHERE name >= 'A' AND
name < 'L'
Of course, BETWEEN can be useful. However, it is useful for discrete values, such as integers. It is a bit dangerous with numbers with decimals, strings, and date/time values. That is why I encourage you to express the logic as inequalities.
In supplement to gordon's answer, one way to get what you're expecting is to turn your name into a discrete set of values:
SELECT * FROM `student` WHERE LEFT(name, 1) between 'A' and 'K'
You need to appreciate that K.K's Company is alphabetically AFTER the letter K on its own so it is not BETWEEN, in the same way that 4.1 is not BETWEEN 2 and 4
By stripping it down to just a single character from the start of the string it will work like you expect, but take cautionary note, you should always avoid running functions on values in tables, because if you had a million names, thats a million strings that mysql has to strip out to just the first letter and it might no longer be able to use an index on name, battering the performance.
Instead, you could :
SELECT * FROM `student` WHERE name >= 'A' and name < 'L'
which is more likely to permit the use of an index as you aren't manipulating the stored values before comparing them
This works because it asks for everything up to but not including L.. Which includes all of your names starting with K, even kzzzzzzzz. Numerically it is equivalent to saying number >= 2 and number < 5 which gives you all the numbers starting with 2, 3 or 4 (like the 4.1 from before) but not the 5
Remember that BETWEEN is inclusive at both ends. Always revert to a pattern of a >= b and a < c, a >= c and a < d when you want to specify ranges that capture all possible values
Compare in lexicographical order, 'K.K's Company' > 'K'
We should convert the string to integer. You can try that mysql script with CAST and SUBSTRING. I've updated your script here. It will include the last record as well.
SELECT * FROM student WHERE name CAST(SUBSTRING(username FROM 1) AS UNSIGNED)
BETWEEN 'A' AND 'K';
The script will work. Hope it will helps to you.
Here I've attached my test sample.
In a MySQL database, prices are stored in a way like this:
98.06K
97.44K
929.14K
91.87K
2.66M
146.64K
14.29K
when i try to sort price ASC or Price DESC, it returns unexpected result.
Kindly suggest me how can i sort price when price is in
10K, 20M, 1.6B
I want result
14.29K
91.87K
97.44K
98.06K
146.64K
929.14K
2.66M
MySQL ignores trailing non-digits when casting string to numeric. This will return the correct price:
price *
case right(price,1)
when 'K' then 1000
when 'M' then 1000000
else 1
end
Of course, you can order by this, but you better apply it during load and store the price in a numeric column.
The problem lies in your data model. I understand that 2.66M is not necessarily exactly 2,660,000, which is why you don't want to store the whole number, but store '2.66M' instead to indicate the precision. This, however, is two pieces of information: the value and the precision, so use two columns:
mytable
value | unit
-------+-----
98.06 | K
97.44 | K
929.14 | K
91.87 | K
2.66 | M
146.64 | K
14.29 | K
Along with a lookup table:
units
unit | factor
-----+--------
K | 1000
M | 1000000
A possible query would be:
select *
from mytable
join units using (unit)
order by mytable.value * units.factor;
where you may want to extend the ORDER BY clause to something like
order by mytable.value * units.factor, units.factor;
or apply some rounding or whatever to consider precision of two seemingly equal values.
It is possible, though not advisable:
https://dbfiddle.uk/?rdbms=mariadb_10.3&fiddle=0a837287c7646823fa6657706f9ae634
SELECT *
, CAST(LEFT(price, LENGTH(price) - 1) AS DECIMAL(10,2)) AS value
, RIGHT(price, 1) AS unit
, CASE RIGHT(price,1)
WHEN 'K' THEN 1000
WHEN 'M' THEN 1000000
ELSE 1
END AS amount
FROM test1
ORDER BY amount, value;
Why not advisable? As the Explain in the dbfiddle shows, this query uses filesort for sorting, which is not very fast. If you do not have too many rows in your data, this should be no problem though.
I have two tables "A" and "B". Table "A" has two columns "Body" and "Number." The column "Number" is empty, the purpose is to populate it.
Table A: Body / Number
ABABCDEF /
IJKLMNOP /
QRSTUVWKYZ /
Table "B" only has one column:
Table B: Values
AB
CD
QR
Here is what I am looking for as a result:
ABABCDEF / 3
IJKLMNOP / 0
QRSTUVWKYZ / 1
In other words, I want to create a query that looks up, for each string in the "Body" column, how many times the substrings in the "Values" column appear.
How would you advise me to do that?
Here's the finished query; explanation will follow:
SELECT
Body,
SUM(
CASE WHEN Value IS NULL THEN 0
ELSE (LENGTH(Body) - LENGTH(REPLACE(Body, Value, ''))) / LENGTH(Value)
END
) AS Val
FROM (
SELECT TableA.Body, TableB.Value
FROM TableA
LEFT JOIN TableB ON INSTR(TableA.Body, TableB.Value) > 0
) CharMatch
GROUP BY Body
There's a SQL Fiddle here.
Now for the explanation...
The inner query matches TableA strings with TableB substrings:
SELECT TableA.Body, TableB.Value
FROM TableA
LEFT JOIN TableB ON INSTR(TableA.Body, TableB.Value) > 0
Its results are:
BODY VALUE
-------------------- -----
ABABCDEF AB
ABABCDEF CD
IJKLMNOP
QRSTUVWKYZ QR
If you just count these you'll only get a value of 2 for the ABABCDEF string because it just looks for the existence of the substrings and doesn't take into consideration that AB occurs twice.
MySQL doesn't appear to have an OCCURS type function, so to count the occurrences I used the workaround of comparing the length of the string to its length with the target string removed, divided by the length of the target string. Here's an explanation:
REPLACE('ABABCDEF', 'AB', '') ==> 'CDEF'
LENGTH('ABABCDEF') ==> 8
LENGTH('CDEF') ==> 4
So the length of the string with all AB occurrences removed is 8 - 4, or 4. Divide the 4 by 2 (LENGTH('AB')) to get the number of AB occurrences: 2
String IJKLMNOP will mess this up. It doesn't have any of the target values so there's a divide by zero risk. The CASE inside the SUM protects against this.
You want an update query:
update A
set cnt = (select sum((length(a.body) - length(replace(a.body, b.value, '')) / length(b.value))
from b
)
This uses a little trick for counting the number of occurrence of b.value in a given string. It replaces each occurrence with an empty string and counts the difference in length of the strings. This is divided by the length of the string being replaced.
If you just wanted the number of matches (so the first value would be "2" instead of "3"):
update A
set cnt = (select count(*)
from b
where a.body like concat('%', b.value, '%')
)
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