I'm trying to sum up all the values in this script below but not sure how to do it, I searched the net and found array_sum() but not sure where to use it...
while($row = mysql_fetch_array($result))
{
$a = $row['aa'];
$b = $row['bb'];
$c = $row['cc'];
}
in the script above, all the variables has a value of either 1 or 0, I can manually add them by using $a + $b + $c but if the list gets longer it will take some time. Is there a faster way so that i can add up everything?
illustration of table
ID NAME AA BB CC
1 YOU 1 0 1
2 ME 1 1 1
So what i want is that "YOU" will have the value of 2 while "ME" will have the value of 3
SELECT (aa+bb+cc) AS yoursum FROM yourtable
According to your comment, I guess you want something like this:
SELECT SUM(aa), SUM(bb), SUM(cc) AS yoursum FROM yourtable
Or this, if there name column can have duplicates.
SELECT name, SUM(aa+bb+cc) AS yoursum
FROM yourtable
GROUP BY name
EDIT:
A slightly modified version of #cularis solution (in case you want only one value in the end):
SELECT (SUM(aa) + SUM(bb) + SUM(cc)) AS yoursum FROM yourtable
I myself would probably use one of the aggregate functions build into MySQL (most likely the SUM()). You can read more of these functions here: http://dev.mysql.com/doc/refman/5.0/en/group-by-functions.html
Related
I am trying to do a search that would be sorted by relevance.
Let's say the search term contains 3 words: A, B and C. What I am trying to do is to check if the search term is present in the SELECT result and if yes that would increase its rank.
ORDER BY CASE
(
WHEN search_word_A_is_present THEN +1
WHEN search_word_B_is_present THEN +1
WHEN search_word_C_is_present THEN +1
ELSE 0
END
)
DESC
While there is no syntax error and the search runs and sorts by something (that seems different from what I want) but I am not sure what is being added up if anything. How would I go about seeing what the final rank (sum) is at the end for each result? Is this the correct way to do it?
Since in MySQL boolean conditions result in 1 and 0, you can simply add those up
ORDER BY search_word_A_is_present + search_word_B_is_present + search_word_C_is_present
DESC
A more practical example:
ORDER BY col1 = 1 + col2 = 'A' + col3 = 44 DESC
I am trying to find a reliable query which returns the first instance of an acceptable insert range.
Research:
some of the below links adress similar questions, but I could get none of them to work for me.
Find first available date, given a date range in SQL
Find closest date in SQL Server
MySQL difference between two rows of a SELECT Statement
How to find a gap in range in SQL
and more...
Objective Query Function:
InsertRange(1) = (StartRange(i) - EndRange(i-1)) > NewValue
Where InsertRange(1) is the value the query should return. In other words, this would be the first instance where the above condition is satisfied.
Table Structure:
Primary Key: StartRange
StartRange(i-1) < StartRange(i)
StartRange(i-1) + EndRange(i-1) < StartRange(i)
Example Dataset
Below is an example User table (3 columns), with a set range distribution. StartRanges are always ordered in a strictly ascending way, UserID are arbitrary strings, only the sequences of StartRange and EndRange matters:
StartRange EndRange UserID
312 6896 user0
7134 16268 user1
16877 22451 user2
23137 25142 user3
25955 28272 user4
28313 35172 user5
35593 38007 user6
38319 38495 user7
38565 45200 user8
46136 48007 user9
My current Query
I am trying to use this query at the moment:
SELECT t2.StartRange, t2.EndRange
FROM user AS t1, user AS t2
WHERE (t1.StartRange - t2.StartRange+1) > NewValue
ORDER BY t1.EndRange
LIMIT 1
Example Case
Given the table, if NewValue = 800, then the returned answer should be 23137. This means, the first available slot would be between user3 and user4 (with an actual slot size = 813):
InsertRange(1) = (StartRange(i) - EndRange(i-1)) > NewValue
InsertRange = (StartRange(6) - EndRange(5)) > NewValue
23137 = 25955 - 25142 > 800
More Comments
My query above seemed to be working for the special case where StartRanges where tightly packed (i.e. StartRange(i) = StartRange(i-1) + EndRange(i-1) + 1). This no longer works with a less tightly packed set of StartRanges
Keep in mind that SQL tables have no implicit row order. It seems fair to order your table by StartRange value, though.
We can start to solve this by writing a query to obtain each row paired with the row preceding it. In MySQL, it's hard to do this beautifully because it lacks the row numbering function.
This works (http://sqlfiddle.com/#!9/4437c0/7/0). It may have nasty performance because it generates O(n^2) intermediate rows. There's no row for user0; it can't be paired with any preceding row because there is none.
select MAX(a.StartRange) SA, MAX(a.EndRange) EA,
b.StartRange SB, b.EndRange EB , b.UserID
from user a
join user b ON a.EndRange <= b.StartRange
group by b.StartRange, b.EndRange, b.UserID
Then, you can use that as a subquery, and apply your conditions, which are
gap >= 800
first matching row (lowest StartRange value) ORDER BY SB
just one LIMIT 1
Here's the query (http://sqlfiddle.com/#!9/4437c0/11/0)
SELECT SB-EA Gap,
EA+1 Beginning_of_gap, SB-1 Ending_of_gap,
UserId UserID_after_gap
FROM (
select MAX(a.StartRange) SA, MAX(a.EndRange) EA,
b.StartRange SB, b.EndRange EB , b.UserID
from user a
join user b ON a.EndRange <= b.StartRange
group by b.StartRange, b.EndRange, b.UserID
) pairs
WHERE SB-EA >= 800
ORDER BY SB
LIMIT 1
Notice that you may actually want the smallest matching gap instead of the first matching gap. That's called best fit, rather than first fit. To get that you use ORDER BY SB-EA instead.
Edit: There is another way to use MySQL to join adjacent rows, that doesn't have the O(n^2) performance issue. It involves employing user variables to simulate a row_number() function. The query involved is a hairball (that's a technical term). It's described in the third alternative of the answer to this question. How do I pair rows together in MYSQL?
I am currently using a database with different entries like dates, names, but also one column with time "ranges". This basically means that there can be a definite number like "10" in this cell, but also a value like "10-15" or "5-10".
So what I want to do here is to sort them by an "average" value ((Lowest+Highest)/2). So in case of the 3 mentioned values it should be
5-10
10
10-15
I am wondering if it is possible to embed this into the SQL statement in some way.
And if it is not possible, I'd like to know the easiest way to implement it otherwise.
Right now I am putting the $SQL_statement together via several conditions, then putting everything into $resultset which is then used with "while". Here are some snippets:
$resultset=mysql_query($SQL_statement);
while ($currententry=mysql_fetch_array($resultset))
{
echo $currententry['Platform'];
echo $currententry['PlaytimeInH']."h";
}
You can do this with substring_index() and arithmetic:
order by (substring_index(col, '-', 1) + 0 +
substring_index(col, '-', -1) + 0
) / 2
The division by 2 is unnecessary, but you do specify the average in your question.
Note that the above will work even if col has no hyphen in it.
You could use a select with a case when clause in order by
select col1, col2, cole
from your_table
order by case
when your_column = '5-10' then 1
when your_column = '10-15' then 2
when your_column = '15-20' then 3
else 4
end
I'm having trouble with this SQL:
$sql = mysql_query("SELECT $menucompare ,
(COUNT($menucompare ) * 100 / (SELECT COUNT( $menucompare )
FROM data WHERE $ww = $button )) AS percentday FROM data WHERE $ww >0 ");
$menucompare is table fields names what ever field is selected and contains data bellow
$button is the week number selected (lets say week '6')
$ww table field name with row who have the number of week '6'
For example, I have data in $menucompare like that:
123456bool
521478bool
122555heel
147788itoo
and I want to select those, who have same word in the last of the data and make percentage.
The output should be like that:
bool -- 50% (2 entries)
heel -- 25% (1 entry)
itoo -- 25% (1 entry)
Any clearness to my SQL will be very appreciated.
I didn't find anything like that around.
Well, keeping data in such format probably not the best way, if possible, split the field into 2 separate ones.
First, you need to extract the string part from the end of the field.
if the length of the string / numeric parts is fixed, then it's quite easy;
if not, you should use regular expressions which, unfortunately, are not there by default with MySQL. There's a solution, check this question: How to do a regular expression replace in MySQL?
I'll assume, that numeric part is fixed:
SELECT s.str, CAST(count(s.str) AS decimal) / t.cnt * 100 AS pct
FROM (SELECT substr(entry, 7) AS str FROM data) AS s
JOIN (SELECT count(*) AS cnt FROM data) AS t ON 1=1
GROUP BY s.str, t.cnt;
If you'll have regexp_replace function, then substr(entry, 7) should be replaced to regexp_replace(entry, '^[0-9]*', '') to achieve the required result.
Variant with substr can be tested here.
When sorting out problems like this, I would do it in two steps:
Sort out the SQL independently of the presentation language (PHP?).
Sort out the parameterization of the query and the presentation of the results after you know you've got the correct query.
Since this question is tagged 'SQL', I'm only going to address the first question.
The first step is to unclutter the query:
SELECT menucompare,
(COUNT(menucompare) * 100 / (SELECT COUNT(menucompare) FROM data WHERE ww = 6))
AS percentday
FROM data
WHERE ww > 0;
This removes the $ signs from most of the variable bits, and substitutes 6 for the button value. That makes it a bit easier to understand.
Your desired output seems to need the last four characters of the string held in menucompare for grouping and counting purposes.
The data to be aggregated would be selected by:
SELECT SUBSTR(MenuCompare, -4) AS Last4
FROM Data
WHERE ww = 6
The divisor in the percentage is the count of such rows, but the sub-stringing isn't necessary to count them, so we can write:
SELECT COUNT(*) FROM Data WHERE ww = 6
This is exactly what you have anyway.
The divdend in the percentage will be the group count of each substring.
SELECT Last4, COUNT(Last4) * 100.0 / (SELECT COUNT(*) FROM Data WHERE ww = 6)
FROM (SELECT SUBSTR(MenuCompare, -4) AS Last4
FROM Data
WHERE ww = 6
) AS Week6
GROUP BY Last4
ORDER BY Last4;
When you've demonstrated that this works, you can re-parameterize the query and deal with the presentation of the results.
Consider the situation i have a table name "test"
-------
content (varchar(30))
-------
1
abc
2
bcd
-------
if i use order by
Select * from test order by content asc
i could get result like
--------
content
--------
1
2
abc
bcd
---------
but is there any way i could get the following result using query
--------
content
--------
abc
bcd
1
2
---------
To get by the collation, you can do by testing the first character... it appears you want anything starting with a numeric to be after anything alhpa oriented... something like the ISNUMERIC() representation by Ted, but my quick check doesn't show such function in MySQL.. So an alternative... because numerics in ASCII list are less than "A" (char 65)
Select *
from test
order by
case when left( content, 1 ) < "A" then 2 else 1 end,
content
Although I've seen different CONVERT() calls, I don't have MySQL available to confirm. However, in addition to the above case/when, you can add a SECOND case/when and call some UDF() or other convert function on the "content" value. If the string starts as alpha, it should return a zero value so the first case/when will keep them to the top of the list, then since all are all non-convertible to numeric would have a value of zero... no impact on the sort, then finally the content itself which will keep in alpha order.
HOWEVER, if your second case/when / convert function call DOES return a numeric value, then it will be properly sorted within the numeric grouping segment... which will then supercede that of the content... However, if content was something like
100 smith rd and
100 main st
they will sort in the same "100" category numeric value, but then alphabetically by the content as
100 main st
100 smith rd
100
this will do it:
SELECT *
FROM test
ORDER BY CAST(field AS UNSIGNED), field ASC
select * from sometable order by content between '0' and '9', content
Not sure on MySql but on SQL Server you can do this...
SELECT * FROM test
ORDER BY IsNumeric(content), content
The order of results is defined by collation used, so if you can find the right collation then yes.
http://dev.mysql.com/doc/refman/5.0/en/charset-collate.html
//edit
This is tricky. I've done some research and it seems that no currently available collation can do that. However there's also possibility to add new collation to MySQL. Here's how.