How to count the number of lines in a table and the number of lines where a certain condition is true without resorting to subselects like this:
create table t (a integer);
insert into t (a) values (1), (2), (null);
select
(select count(*) from t) as total_lines,
(select count(*) from t where a = 1) as condition_true
;
total_lines | condition_true
-------------+----------------
3 | 1
select count(*) as total_lines, count(a = 1 or null) as condition_true
from t
;
total_lines | condition_true
-------------+----------------
3 | 1
It works because:
First while count(*) counts all lines regardless of anything, count(my_column) will count only those lines where my_column is not null:
select count(a) as total
from t
;
total
-------
2
Second (false or null) returns null so whenever my condition is not met it will return null and will not be counted by count(condition or null) which only counts not nulls.
Use SUM(condition)!
select
count(*) as total_lines,
sum(a = 1) as condition_true
from t
See it working here.
This works because in mysql, true is 1 and false is 0, so the sum() of a condition will add 1 when it's true and 0 when it's false - which effectively counts the number of times the condition is true.
Many people falsely believe you need a case statement, but you don't with mysql (you do with some other databases)
this can be easily done using a condition inside count. I don't know if its the optimized method of doing it but it gets the work done
you can do it as follows
select count(*) as total_lines, COUNT(CASE WHEN a = 1 THEN 1 END) as condition_true from t
you can check it here
sqlFiddle
Related
I have a table like so:
id_type id_option
"1" "1"
"1" "5"
"2" "1"
"2" "5"
"2" "8"
I am trying to write a query that given a list of option IDs finds the "type" that matches the list, but only those ID's
For example, if given 1 and 5 as options, it should return the type 1 but only the type 1 as the 8 required to match type 2 is not present.
I have tried the following:
SELECT *
FROM my_table
WHERE id_option IN (1, 5)
GROUP BY id_type
HAVING COUNT(DISTINCT id_option) = 2
This returns both "types" - I had hoped that the COUNT restriction of 2 would have helped but I now understand why it doesn't, but I can't think of a clever way to limit this.
I could just pull the first record as typically the types with less options are saved first but I don't think I can rely on this 100%
Thank you for your time
Here's a solution:
SELECT *
FROM my_table
GROUP BY id_type
HAVING SUM(id_option IN (1,5)) = COUNT(*)
It relies on a trick specific to MySQL: boolean true is literally the integer 1. So you can use SUM() to count the rows where a condition is true, but putting a boolean expression inside SUM().
For folks reading this who use other databases besides MySQL, you'd have to use an expression to convert the boolean condition to the integer 1:
HAVING SUM(CASE WHEN id_option IN (1,5) THEN 1 ELSE 0 END) = COUNT(*)
In this case, let all rows become part of the groups. That is, do not use a WHERE clause to restrict the query to rows where the id_option is 1 or 5. Then count the total rows in the group, and "count" (i.e. use the SUM() trick) the rows where the id_options is 1 or 5. Comparing these counts will be equal if there are no id_options values besides 1 or 5.
If you also want to make sure that both 1 and 5 are found, you need another condition:
SELECT *
FROM my_table
GROUP BY id_type
HAVING SUM(id_option IN (1,5)) = COUNT(*)
AND COUNT(DISTINCT CASE WHEN id_option IN (1,5) THEN id_option END) = 2
The CASE expression will return 1 or 5, or if there are any other values, those are converted to NULL. The COUNT() function ignores NULLs.
If you can pass the options as a sorted comma separated list string, then use GROUP_CONCAT():
SELECT id_type
FROM my_table
GROUP BY id_type
HAVING GROUP_CONCAT(id_option ORDER BY id_option) = '1,5'
If there are duplicate options for each type, use DISTINCT:
HAVING GROUP_CONCAT(DISTINCT id_option ORDER BY id_option) = '1,5'
While I can't comment yet, here's a tiny adjustment to Bill Karwin's last example (in the accepted solution):
SELECT *
FROM my_table
GROUP BY id_type
HAVING SUM(id_option IN (1,5)) = COUNT(*)
AND COUNT(DISTINCT id_option) = 2
Suppose I have a table :
start_range end_range
1 4
4 8
I want the result to be true if it is greater than any of the value of start_range and less than any of the corresponding end_range.
Eg.
value 2 should return true , as 2>1 and 2<4
but value 4 should return false in this case as 4>1 but 4<4 becomes false, as well as 4>4 becomes false for the second case.
I cannot use the query
SELECT Sumthing
FROM XYZ
WHERE value> SOME(start_range) AND value < SOME(end_range)
The problem with the above query is let say value = 4.
Now 4> SOME(start_range) will become true as 4>1. AND
4< SOME(end_range) will also become true as 4<8.
But in actual the comparison should be like (((4>1)AND(4<4)) OR ((4>4)AND(4<8))) . It should return false.
One more thing , the above table is not persistent , I have been creating it in a subquery.Thats why i have been using SOME.
if still my question isn't clear, mention in comments.
Assuming that xyz is your table:
select (count(*) > 0) as HasMatch
from xyz
where value > start_range and value < end_range;
I'm not sure why you are using some.
EDIT:
It occurs to me that you want to use subqueries, and xyz is not the table in question. Perhaps this is what you want:
select xyz.*
from xyz
where exists (select 1
from (<your query here>) t
where xyz.value > t.start_range and xyz.value < t.end_range
);
you can do something like this
SELECT CASE WHEN start_range<value and end_range>value
THEN 'true'
ELSE 'false'
END here_name_to_this_column(optional)
FROM table_name
tutorial link
select (count(*) > 0) as HasMatch
from (select IF(start_range<value and end_range>value, true, false ) as value
from XYZ having value =1) as MatchTable
DEMO
I have a set of records which return flags in individual records. Based on that flags I want to perform and operation within those flags (e.g.if any single flag is "true" final result should return me "true")
Following are the set of records which is returned by normal SQL query ;
select ID,flag from tablename;
--------------------------
ID(int) | flag (varchar(5))
--------------------------
1 | true
2 | false
3 | true
4 | false
5 | false
I would like to know the SQL query which returns result as shown below (such a way that final result should be and operation within each of those 5 records)
-------
flag
-------
true
I'm not sure I completely understand what you want to do, but if you want to define your AND operation as "at least one of the records must have as its flag value 'true'", you can use the following SQL statement which will return 1 if there is at least one 'true' value and otherwise will return 0:
SELECT IF(COUNT(*) > 0, 'true', 'false')
FROM Table1
WHERE flag = 'true';
See the SQL Fiddle: http://sqlfiddle.com/#!2/1e6cd7/11
i came up with this solution but maybe you will get a short and better one :) .
select if(count(*) = 1 , 'mixed' , flag) flag from(
select flag from table1
where flag = 'true'
having count(*) = 5
union
select flag from table1
where flag = 'false'
having count(*) = 5
union
select 'mixed' from table1
)t
DEMO TO PLAY WITH
I have a table with 5 fields. Each field can store a number from 1 - 59.
Similar to countif in Excel, how do I count the number of times a number from 1 - 59 shows up in all 5 fields?
Here's an example for the count of occurances for the number 1 in all five fields:
SELECT SUM(pick_1 = 1 OR pick_2 = 1 OR pick_3 = 1 OR pick_4 = 1 OR pick_5 = 1) AS total_count_1
FROM tbldraw
Hopefully I made sense.
There was an answer here that had a solution. I think this is just a variation.
Step1: Create a numbers table (1 field, called id, 59 records (values 1 -59))
Step2:
SELECT numbers_table.number as number
, COUNT(tbldraw.pk_record)
FROM numbers_table
LEFT JOIN tbldraw
ON numbers_table.number = tbldraw.pick_1
OR numbers_table.number = tbldraw.pick_2
OR numbers_table.number = tbldraw.pick_3
OR numbers_table.number = tbldraw.pick_4
OR numbers_table.number = tbldraw.pick_5
GROUP BY number
ORDER BY number
How about a two step process? Assuming a table called summary_table ( int id, int ttl), for each number you care about...
insert into summary_table values (1,
(select count(*)
from table
where field1 = 1 or field2 = 1 or field3 = 1 or field4 = 1 or field5 = 1))
do that 59 times, once for each value. You can use a loop in most cases. Then you can select from the summary_table
select *
from summary_table
order by id
That will do it. I leave the coversion of this SQL into a stored procedure for those that know what database is in use.
The ALL() function, which returns true if the preceding operator is true for all parameters, makes the query particularly elegant and succinct.
To find the count a particular number (eg 3):
select count(*)
from tbldraw
where 3 = all (pick_1, pick_2, pick_3, pick_4, pick_5)
To find the count of all such numbers:
select pick_1, count(*)
from tbldraw
where pick_1 = all (pick_2, pick_3, pick_4, pick_5)
group by pick_1
I need to compare 2 columns in a table and give 3 things:
Count of rows checked (Total Rows that were checked)
Count of rows matching (Rows in which the 2 columns matched)
Count of rows different (Rows in which the 2 columns differed)
I've been able to get just rows matching using a join on itself, but I'm unsure how to get the others all at once. The importance of getting all of the information at the same time is because this is a very active table and the data changes with great frequency.
I cannot post the table schema as there is a lot of data in it that is irrelevant to this issue. The columns in question are both int(11) unsigned NOT NULL DEFAULT '0'. For purposes of this, I'll call them mask and mask_alt.
select
count(*) as rows_checked,
sum(col = col2) as rows_matching,
sum(col != col2) as rows_different
from table
Note the elegant use of sum(condition).
This works because in mysql true is 1 and false is 0. Summing these counts the number of times the condition is true. It's much more elegant than case when condition then 1 else 0 end, which is the SQL equivalent of coding if (condition) return true else return false; instead of simply return condition;.
Assuming you mean you want to count the rows where col1 is or is not equal to col2, you can use an aggregate SUM() coupled with CASE:
SELECT
COUNT(*) AS total,
SUM(CASE WHEN col = col2 THEN 1 ELSE 0 END )AS matching,
SUM(CASE WHEN col <> col2 THEN 1 ELSE 0 END) AS non_matching
FROM table
It may be more efficient to get the total COUNT(*) in a subquery though, and use that value to subtract the matching to get the non-matching, if the above is not performant enough.
SELECT
total,
matching,
total - matching AS non_matching
FROM
(
SELECT
COUNT(*) AS total,
SUM(CASE WHEN col = col2 THEN 1 ELSE 0 END )AS matching
FROM table
) sumtbl