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问题 Here is my query: SELECT COUNT(*) FROM Similarities WHERE T1Similarity = 0 OR T2Similarity = 0 Here is the result: Msg 8115, Level 16, State 2, Line 1 Arithmetic overflow error converting expression to data type int. The table has 4 billion rows. I don't except this query to be fast, but after about 5mins, it fails with an overflow error. Is there a COUNT function for bigger data than int? Thanks. 回答1: Use COUNT_BIG SELECT COUNT_BIG(*) FROM Similarities WHERE T1Similarity = 0 OR T2Similarity

问题 I have two tables one Employee and mailing Subscriptions Employee looks like this: Name (pk) | Surname | Age mailing Subsriptions MailId (pk)| EmployeeName (fk)|Description | Date I wanted to subscription number for each customer, therefore I tried the following query: Select COUNT(c.Name) From Employee INNER JOIN mailingSubscriptions as m ON c.Name = m.EmployeeName; It will give me all counts for each Employee that has an entry in the mailing subscription. My problem is that I want to see

问题 I need some help with counting both unique and duplicate values in MySQL. I want to know how many records there are total, and also how many is there two times and three times and so on... Do I need to use UNION or something? I think SUM would be the best solution for me because of I might use some joins with this in future. Sample data: | custId | name | |--------|--------| | 1001 | Alex | | 1001 | Alex | | 1002 | Daniel | | 1003 | Mark | | 1002 | Daniel | Sample results: | total | twoTimes

问题 Say I have one column called colors with 1000 cells populated with values. Some of the cells have the word blue in it. In another column I have unique identifiers that correspond with the colors column. For example Blue can have a value associated to it of 01, 02, 04, or 05. The word blue appears 20 times within my name column. What is one way I can find how many unique identifiers are associated with the word blue ? In the example listed above the answer should return 4. The current method I

问题 I have three tables: Member MemberBranch Branch The Member can have many Branch es through MemberBranch . And a Branch can have many Member s etc. What I want to be able to do is get a count of how many members a branch has. so $branch = Branch::find_by_title('London'); $branch->number_of_members; // Will equal how many members have the branch through MemberBranch How would I go about doing this? 回答1: Try this: SELECT B.*, (SELECT count(*) as total from MembersBranch where branch_id = B.id)

问题 I want to increment the count and display the value in a message box. I'm using a nested if statement. This is my code if sum( abs( f1(:) - f2(:))) == 0.0 i = i + 1; elseif sum(abs(f2(:) - f3(:))) == 0.0 i = i+ 1; elseif sum(abs(f3(:) - f4(:))) == 0.0 i = i + 1; else i = 1; end h = msgbox('Perfect = %d',i); Here f1 , f2 , f3 , and f4 contains the difference between two images in float. I have declared i = 0; before if statement. Still I'm not able to see the message box in the output. I tried

问题 i have an issue with query 1st table (Master) Name :MainCategory with fields (Category_id,Category_name) 2nd Table (Transation) Name : Incident with fields (Incident_id,Category_id,subject,description) i want to query to count the appearance of category_id in the table Transation for eg result can be Category_name Number_of_Faults Hardware 10 Software 22 Network 17 thanks Kumar 回答1: Try this: SELECT a.Category_Name, COUNT(b.Incident_Id) Number_of_Faults FROM MainCategory a JOIN Incident b ON

问题 I have a SPARQL query which returns result with a LIMIT of 20.. In this query I also want to know the total number of results without running the query two times (one with LIMIT and one without LIMIT ).. For Example-- On running a query total possible results are 500 with LIMIT it displays only 20 at a time, but in my response I want a field which displays total result count, i.e., 500 ... Updated question Suppose Now if i do a query where sequence = abc_11 with LIMIT=2 I will get something

问题 If I have 3 models: Model Section Model User has_many :votes Model Vote belongs_to :user and inside Model Section has_many :users has_many :votes, :through => :users How to get the Sections list ordered by votes quantity using the AR associations? 回答1: The most reasonable way to do this is to use a subquery written as raw SQL for ordering the result as follows... Section.order( '(select count(1) from votes inner join users on votes.user_id=users.id where users.section_id=sections.id)' ) 回答2:

问题 Ok so my requirement is fairly simple. I have a table like so. images | videos | reviews -------------------------------------------------- 0 | 2 | 1 4 | 5 | 0 11 | 1 | 4 0 | 1 | 7 I need to write a query that counts the rows in each column that is not equal to 0 , and also that gives the actual sum of the numbers in the columns. So the result will look something like this. Columns | Counts | Sum ---------------------------------------------------- images | 2 | 15 videos | 4 | 9 reviews | 3 |