I have a table with a locale field like this:
Id, locale
1, "en-US"
2, "en-BR"
3, "en-SK"
4, "fr-FR"
5, "fr-FS"
I want to do a select on this table, and group based on "en" or "fr" (part of the locale field string), what should i write to accomplish this?
Alternatively, you may also use the LEFT() function
SELECT left(locale, 2), count(id) FROM table group by left(locale, 2)
At the end of the day, the important part is to make use of a group by statement which contains a function.
Note that this works in MySQL very well. Other DBMS might not like this.
You can use a combination of substring and instr.
select substring(locale,1,instr(locale,'-')-1), count(*)
from tablename
group by substring(locale,1,instr(locale,'-')-1)
You can group by substring function result
SELECT SUBSTR(locale,1,2) FROM table GROUP BY SUBSTR(locale,1,2)
I would use substring_index():
Select substring_index(locale, '-', 1) as grp, count(*)
From t
Group by grp
This is simple and general, in the sense of taking all characters before the first separator.
Related
I need to get maximum number from a part of the value that generally start with year followed by slash(/). So I need a maximum number after the slash(/) but year should be 2016
2016/422
2016/423
2016/469
2016/0470
2014/777
2015/123
2015/989
I tried this query
SELECT columname FROM tablename WHERE columname LIKE '2016/%' ORDER BY id DESC
the above query always giving '2016/469' as first record, how to get '2016/0470' as the maximum number?
any help will be much appreciated.
Thank you.
If columname follows that pattern YEAR/0000, you can use SUBSTRING function from MySQL to remove the part of the string you don't want.
SELECT value FROM (
SELECT CAST(SUBSTRING(columname, 0, 4) AS UNSIGNED) as year, CAST(SUBSTRING(columname FROM 6) AS UNSIGNED) as value FROM tablename
) total
ORDER BY year DESC, value DESC
LIMIT 1;
You need to split the string into 2 parts and evaluate them as numbers, instead of strings. The following formula will return the number after the / in the fieldname. All functions used below are described in the string functions section of the MySQL documentation. This way you can get the number after the / character, even if it is not year before the /, but sg else. The + 0 converts the string to a number, eliminating any leading 0.
select right(columnname, char_length(columnname)-locate('/',columnname)) + 0
from tablename
Just take the max() of the above expression to get the expected results.
UPDATE:
If you need the original number and the result has to be restricted to a specific year, then you need to join back the results to the original table:
select columnname
from tablename t1
inner join (select max(right(t.columnname, char_length(t.columnname)-locate('/',t.columnname)) + 0) as max_num
from tablename t
where left(t.columnname,4)='2016'
) t2
on right(t1.columnname, char_length(1t.columnname)-locate('/',t1.columnname)) + 0 = t2.max_num
where left(t1.columnname,4)='2016'
There are lots of suggestions given as answers already. But some of those seem overkill to me.
Seems like the only change needed to the OP query is the expression in the ORDER BY clause.
Instead of:
ORDER BY id
We just need to order by the numeric value following the slash. And there are several approaches, several expressions, that will get that from the example data.
Since the query already includes a condition columname LIKE '2016/%'
We can get the characters after the first five characters, and then convert that string to a numeric value by adding zero.
ORDER BY SUBSTRING(columname,6) + 0 DESC
If we only want to return one row, add
LIMIT 1
http://dev.mysql.com/doc/refman/5.7/en/string-functions.html#function_substring
If we only want to return the numeric value, we could use the same expression in the SELECT list, in addition columnname.
This isn't the only approach. There are lots of other approaches that will work, and don't use SUBSTRING.
Try like this:
SELECT
MAX(CAST(SUBSTRING(t.name,
LOCATE('/', t.name) + 1)
AS UNSIGNED)) AS max_value
FROM
tablename AS t;
You can try with this little uggly approach:
SELECT t.id, t2.secondNumber FROM table AS t
JOIN (SELECT id,
CONCAT(SUBSTRING(field,1,5),
if(SUBSTRING(SUBSTRING(field, 6),1,1)='0',
SUBSTRING(field, 6),
SUBSTRING(field,7)
)
) as secondNumber FROM table ) AS t2 ON t2.id=t.id
ORDER BY t2.secondNumber DESC
Would be valid only if the 0 (zeroes) before the second number (after the slash) are no more than 1.
Or if the year doesn`t matter you can try to order them only by the second number if it is ok:
SELECT t.id, t2.secondNumber FROM table AS t
JOIN (SELECT id,
if(SUBSTRING(SUBSTRING(field, 6),1,1)='0',
SUBSTRING(field, 6),
SUBSTRING(field,7)
) as secondNumber FROM table ) AS t2 ON t2.id=t.id
ORDER BY t2.secondNumber DESC
I want to calculate percentage for test groups.
I have group A,B and C. And I want to know how much success percentage each group have.
My first query is counting total test ran in each group by doing the following:
SELECT type, count(type) as total_runs
From mytable
Where ran_at > '2015-09-11'
Group by type
Second query is counting success for each group:
SELECT type, count(type) as success
FROM mytable
where run_status like '%success%' and ran_at> '2015-09-11'
Group by type
Now I need to divide one in the other and multiply in 100.
how do I do this in one query in an efficient way, I guess nested query is not so efficient- but anyway I can't see how I can uses nested query to solve it.
I would appreciate answer which include simple way, maybe not so efficient, and an efficient way with explanations
You can just use conditional aggregation:
SELECT type, sum(run_status like '%success%') as success,
100 * avg(run_status like '%success%') as p_success
FROM mytable
where ran_at> '2015-09-11'
Group by type;
In a numeric context, MySQL treats boolean expressions as integers with 1 for true and 0 for false. The above works assuming that run_status is not NULL. If it can be NULL, then you need an explicit case statement for the avg().
I had this one, but Gordon have a better solution if run_status is not NULL.
Select type, sum(if(run_status like '%success%',1,0)) / count(1) * 100) as p_success
From mytable
Where ran_at > '2015-09-11'
Group by type
I need to calculate the sum of one column(col2) , but the column has both numbers and text. How do I exclude the text alone before I use sum()?
The table has around 1 million rows, so is there any way other than replacing the text first?
My query will be :
Select col1,sum(col2) from t1 group by col1,col2
Thanks in advance
You can use regexp to filter the column:
Select col1,sum(col2) from t1 WHERE col2 REGEXP '^[0-9]+$' group by col1,col2
You could use MySQL built in REGEXP function.
to learn more visit : https://dev.mysql.com/doc/refman/5.1/en/regexp.html
Or another way is using CAST or CONVERT function
to learn in detail : https://dev.mysql.com/doc/refman/5.0/en/cast-functions.html
Hope this is helpful
Assuming you mean the number is at the beginning of the tex, the easiest way is simply to use implicit conversion:
Select col1, sum(col2 + 0)
from t1
group by col1, col2;
If col2 starts with a non-numeric character, then MySQL will return 0. Otherwise, it will convert the leading numeric characters to a number.
Note that your query doesn't really make sense, because you are aggregating by col2 as well as including it in the group by. I suspect you really want:
Select col1, sum(col2 + 0)
from t1
group by col1;
Is it a way in mysql to alphabetically ordering a string ?
I am looking for a function who does that :
select alphabeticallyorder('cba')
will return me
'abc'
A query like this should return the values that you need. I know it's not a nice query, and you also need a numbers table, filled with numbers:
SELECT col, GROUP_CONCAT(SUBSTRING(col, n, 1)
ORDER BY SUBSTRING(col, n, 1)
SEPARATOR '') AS ordered_col
FROM
tablename INNER JOIN numbers
ON LENGTH(tablename.col)>=numbers.n
GROUP BY
id, col
Also it will work only if LENGTH(col)=CHAR_LENGTH(col). Please see fiddle here.
Please see the REVERSE function.
http://dev.mysql.com/doc/refman/5.5/en/string-functions.html#function_reverse
mysql> SELECT REVERSE('abc');
-> 'cba'
I've got a database table mytable with a column name in Varchar format, and column date with Datetime values. I'd like to count names with certain parameters grouped by date. Here is what I do:
SELECT
CAST(t.date AS DATE) AS 'date',
COUNT(*) AS total,
SUM(LENGTH(LTRIM(RTRIM(t.name))) > 4
AND (LOWER(t.name) LIKE '%[a-z]%')) AS 'n'
FROM
mytable t
GROUP BY
CAST(t.date AS DATE)
It seems that there's something wrong with range syntax here, if I just do LIKE 'a%' it does count properly all the fields starting with 'a'. However, the query above returns 0 for n, although should count all the fields containing at least one letter.
You write:
It seems that there's something wrong with range syntax here
Indeed so. MySQL's LIKE operator (and SQL generally) does not support range notation, merely simple wildcards.
Try MySQL's nonstandard RLIKE (a.k.a. REGEXP), for fuller-featured pattern matching.
I believe LIKE is just for searching for parts of a string, but it sounds like you want to implement a regular expression to search for a range.
In that case, use REGEXP instead. For example (simplified):
SELECT * FROM mytable WHERE name REGEXP "[a-z]"
Your current query is looking for a string of literally "[a-z]".
Updated:
SELECT
CAST(t.date AS DATE) AS 'date',
COUNT(*) AS total,
SUM(LENGTH(LTRIM(RTRIM(t.name))) > 4
AND (LOWER(t.name) REGEXP '%[a-z]%')) AS 'n'
FROM
mytable t
GROUP BY
CAST(t.date AS DATE)
I believe you want to use WHERE REGEXP '^[a-z]$' instead of LIKE.
You have regex in your LIKE statement, which doesn't work. You need to use RLIKE or REGEXP.
SELECT CAST(t.date AS DATE) AS date,
COUNT(*) AS total
FROM mytable AS t
WHERE t.name REGEXP '%[a-zA-Z]%'
GROUP BY CAST(t.date AS DATE)
HAVING SUM(LENGTH(LTRIM(RTRIM(t.name))) > 4
Also just FYI, MySQL is terrible with strings, so you really should trim before you insert into the database. That way you don't get all that crazy overhead everytime you want to select.