Suppose I have a table as follows:
id name length
1 A 21.5
2 B 12.4
3 C 0
4 D 17
5 E 1
I wish to get:
id name length
1 A 21.5
5 E 1
Meaning all rows that hase length that ends up with 1.
length is a numeric column.
It's very simple thing to do with programing languages but it seems quite not natural for SQL. How can I do that efficiently and simply?
My only thought is to convert the field to Text and then lose eveything after the . then convert it to array and choose the letter in the position of array length. This will probebly work but it seems like a very bad solution.
You can use FLOOR and modulo division:
SELECT *
FROM tab
WHERE FLOOR(length) % 10 = 1;
SqlFiddleDemo
Related
The following SQL query is supposed to return the max consecutive numbers in a set.
WITH RECURSIVE Mystery(X,Y) AS (SELECT A AS X, A AS Y FROM R)
UNION (SELECT m1.X, m2.Y
FROM Mystery m1, Mystery m2
WHERE m2.X = m1.Y + 1)
SELECT MAX(Y-X) + 1 FROM Mystery;
This query on the set {7, 9, 10, 14, 15, 16, 18} returns 3, because {14 15 16} is the longest chain of consecutive numbers and there are three numbers in that chain. But when I try to work through this manually I don't see how it arrives at that result.
For example, given the number set above I could create two columns:
m1.x
m2.y
7
7
9
9
10
10
14
14
15
15
16
16
18
18
If we are working on rows and columns, not the actual data, as I understand it WHERE m2.X = m1.Y + 1 takes the value from the next row in Y and puts it in the current row of X, like so
m1.X
m2.Y
9
7
10
9
14
10
15
14
16
15
18
16
18
Null?
The main part on which I am uncertain is where in the SQL recursion actually happens. According to Denis Lukichev recursion is the R part - or in this case the RECURSIVE Mystery(X,Y) - and stops when the table is empty. But if the above is true, how would the table ever empty?
Since I don't know how to proceed with the above, let me try a different direction. If WHERE m2.X = m1.Y + 1 is actually a comparison, the result should be:
m1.X
m2.Y
14
14
15
15
16
16
But at this point, it seems that it should continue recursively on this until only two rows are left (nothing else to compare). If it stops here to get the correct count of 3 rows (2 + 1), what is actually stopping the recursion?
I understand that for the above example the MAX(Y-X) + 1 effectively returns the actual number of recursion steps and adds 1.
But if I have 7 consecutive numbers and the recursion flows down to 2 rows, should this not end up with an incorrect 3 as the result? I understand recursion in C++ and other languages, but this is confusing to me.
Full disclosure, yes it appears this is a common university question, but I am retired, discovered this while researching recursion for my use, and need to understand how it works to use similar recursion in my projects.
Based on this db<>fiddle shared previously, you may find it instructive to alter the CTE to include an iteration number as follows, and then to show the content of the CTE rather than the output of final SELECT. Here's an amended CTE and its content after the recursion is complete:
Amended CTE
WITH RECURSIVE Mystery(X,Y) AS ((SELECT A AS X, A AS Y, 1 as Z FROM R)
UNION (SELECT m1.X, m2.A, Z+1
FROM Mystery m1
JOIN R m2 ON m2.A = m1.Y + 1))
CTE Content
x
y
z
7
7
1
9
9
1
10
10
1
14
14
1
15
15
1
16
16
1
18
18
1
9
10
2
14
15
2
15
16
2
14
16
3
The Z field holds the iteration count. Where Z = 1 we've simply got the rows from the table R. The, values X and Y are both from the field A. In terms of what we are attempting to achieve these represent sequences consecutive numbers, which start at X and continue to (at least) Y.
Where Z = 2, the second iteration, we find all the rows first iteration where there is a value in R which is one higher than our Y value, or one higher than the last member of our sequence of consecutive numbers. That becomes the new highest number, and we add one to the number of iterations. As only three numbers in our original data set have successors within the set, there are only three rows output in the second iteration.
Where Z = 3, the third iteration, we find all the rows of the second iteration (note we are not considering all the rows of the first iteration again), where there is, again, a value in R which is one higher than our Y value, or one higher than the last member of our sequence of consecutive numbers. That, again, becomes the new highest number, and we add one to the number of iterations.
The process will attempt a fourth iteration, but as there are no rows in R where the value is one more than the Y values from our third iteration, no extra data gets added to the CTE and recursion ends.
Going back to the original db<>fiddle, the process then searches our CTE content to output MAX(Y-X) + 1, which is the maximum difference between the first and last values in any consecutive sequence, plus one. This finds it's value from the record produced in the third iteration, using ((16-14) + 1) which has a value of 3.
For this specific piece of code, the output is always equivalent to the value in the Z field as every addition of a row through the recursion adds one to Z and adds one to Y.
I have a column using bits to record status of every mission. The index of bits represents the number of mission while 1/0 indicates if this mission is successful and all bits are logically isolated although they are put together.
For instance: 1010 is stored in decimal means a user finished the 2nd and 4th mission successfully and the table looks like:
uid status
a 1100
b 1111
c 1001
d 0100
e 0011
Now I need to calculate: for every mission, how many users passed this mission. E.g.: for mission1: it's 0+1+1+0+1 = 5 while for mission2, it's 0+1+0+0+1 = 2.
I can use a formula FLOOR(status%POWER(10,n)/POWER(10,n-1)) to get the bit of every mission of every user, but actually this means I need to run my query by n times and now the status is 64-bit long...
Is there any elegant way to do this in one query? Any help is appreciated....
The obvious approach is to normalise your data:
uid mission status
a 1 0
a 2 0
a 3 1
a 4 1
b 1 1
b 2 1
b 3 1
b 4 1
c 1 1
c 2 0
c 3 0
c 4 1
d 1 0
d 2 0
d 3 1
d 4 0
e 1 1
e 2 1
e 3 0
e 4 0
Alternatively, you can store a bitwise integer (or just do what you're currently doing) and process the data in your application code (e.g. a bit of PHP)...
uid status
a 12
b 15
c 9
d 4
e 3
<?php
$input = 15; // value comes from a query
$missions = array(1,2,3,4); // not really necessary in this particular instance
for( $i=0; $i<4; $i++ ) {
$intbit = pow(2,$i);
if( $input & $intbit ) {
echo $missions[$i] . ' ';
}
}
?>
Outputs '1 2 3 4'
Just convert the value to a string, remove the '0's, and calculate the length. Assuming that the value really is a decimal:
select length(replace(cast(status as char), '0', '')) as num_missions as num_missions
from t;
Here is a db<>fiddle using MySQL. Note that the conversion to a string might look a little different in Hive, but the idea is the same.
If it is stored as an integer, you can use the the bin() function to convert an integer to a string. This is supported in both Hive and MySQL (the original tags on the question).
Bit fiddling in databases is usually a bad idea and suggests a poor data model. Your data should have one row per user and mission. Attempts at optimizing by stuffing things into bits may work sometimes in some programming languages, but rarely in SQL.
I was thinking about simple reordering rows in relational database's table.
I would like to avoid method described here:
How can I reorder rows in sql database
My simple idea was to use as ListOrder column of type double-precision 64-bit IEEE 754 floating point.
At inserting a row between two existing rows we calculate listOrder value as average of these sibling elements.
Example:
1. Starting state:
value, listOrder
a 1
b 2
c 3
d 4
e 5
f 6
2. Moving "e" two rows up
One simple sql update on e-row: update mytable set listorder=2.5 where value='e'
value, listOrder
a 1
b 2
e 2.5
c 3
d 4
f 6
3. Moving "a" one position down
value, listOrder
b 2
a 2.25
e 2.5
c 3
d 4
f 6
I have a question. How many insertions can I perform (in the edge situation) to have properly ordered list.
For the 64 bit integer there is less than 64 insertions in the same place.
Is floating point types allows to more insertions?
There are other problems with described approach?
Do you see any patches/adjustments to make this idea safe and usable in applications?
This is similar to a lexical order, which can also be done with varchar columns:
A
B
C
D
E
F
becomes
A
B
BM
C
D
F
becomes
B
BF
BM
C
D
F
I prefer the two step process, where you update every row in the table after the one you move to be one larger. Sql is efficient about this, where updating the rows following a change is not as bad as it seems. You preserve something that's more human readable, the storage size for your ordinal value scales in a linear rather with your data size, and you don't risk coming to a point where you don't have enough precision to put an item in between two values
Every letter has a value
a b c d e f g h i j k l m n o p q r s t u v w x y z
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
TableA
String Length Value Subwords
exampledomain 13 132 #example-domain#example-do-main#
creditcard 10 85 #credit-card#credit-car-d#
TableB
Words Length Value
example 7 76
do 2 19
main 4 37
domain 6 56
credit 6 59
card 4 26
car 3 22
d 1 4
Explanation
TableA has string based over milion rows, and it will be new added 100k rows/daily to tableA.
And also "string" column has no whitespaces
TableB has words based over milion rows,there is every letter and words in 1-2 languages
What i want to do
i want to split strings in TableA to its subwords, as you see in example; "creditcard" i search in TableB all words and try to find which words when comes together matches the string
What i did,and couldnt solve my question
i took the string and JOIN the TableB with INNER JOINS i made 2-3 times INNER JOINS because there can be 3word 4word strings too, and that WORKED!! but it takes too much time even doing it for 100-200 strings. Guess i want to do it for 100k/everyday???
Now what i try to do
i gave values to everyletter as you see above,
Took the strings one by one and from their including letters i count the value of strings..
And the same for the words too in TableB..
Now i have every string in TableA and everyword in TableB with their VALUES..
_
1- i will take the string,length and value of it (Exmple; creditcard - 10 - 85)
2- and make a search in TableB to find the possible words when they come together, with their SUM(length), and SUM(value) matches the strings length and value, and write theese possibilities to a new column.
At last even their sum of length and sum of values matches each other there can be some posibilities that doesnt match the whole string i will elliminate theese ones (Example; "doma-in" can be "moda-in" too and their lengths and values are same but not same words)
I dont know but,i guess with that value method i can solve the time proplem??? , or if there is another ways to do that, i will be gratefull taking your advices.
Thanks
You could try to find the solutions recursively by looking always at the next letter. For example for the word DOMAIN
D - no
DO - is a word!
M - no
MA - no
MAI - no
MAIN - is a word!
No more letters --> DO + MAIN
DOM - is a word!
A - no
AI - no
AIN - no
Finished without result
DOMA - no
DOMAI - no
DOMAIN - is a word!
No more letters --> DOMAIN
Total newbie here, regarding sqlite, so don't flame too hard :)
I have a table:
index name length L breadth B height H
1 M-1234 10 5 2
2 M-2345 20 10 3
3 ....
How do I put some tabular data (let' say ten x,y values) corresponding to index 1, then another table to index 2, and then another, etc. In short, so that I have a table of x and y values that is "connected" to first row, then another that is connected to second row.
I'm reading some tutorials on sqlite3 (which I'm using), but am having trouble finding this. If anyone knows a good newbie tutorial or a book dealing with sqlite3 (CLI) I'm all ears for that too :)
You are just looking for information on joins and the concept of foreign key, that although SQLite3 doesn't enforce, is what you need. You can go without it, anyway.
In your situation you can either add two "columns" to your table, being one x and another y, or create a new table with 3 "columns": foreign_index, x and y. Which one to use depends on what you are trying to accomplish, performance and maintainability.
If you go the linked table route, you'd end up with two tables, like this:
MyTable
index name length L breadth B height H
1 M-1234 10 5 2
2 M-2345 20 10 3
3 ....
XandY
foreign_index x y
1 12 9
2 8 7
3 ...
When you want the x and y values of your element, you just use something like SELECT x, y FROM XandY WHERE foreign_index = $idx;
To get all the related attributes, you just do a JOIN:
SELECT index, name, length, breadth, height, x, y FROM MyTable INNER JOIN XandY ON MyTable.index = XandY.foreign_index;