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  • show array size with will_paginate

    - by chief
    @users = User.paginate( :page => params[:page], :per_page => 2, I would like to return the total number of results. @users.size simply gives me the number of results on a page. If page 1 has 2 users @users.size = 2, going to page 2 with say 1 user @users.size = 1. How can I show the size of all the users?

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  • Window resolution handling without javascript

    - by Sai Sasdhar
    Is there any CSS property to deal window resolution. I created a HTML page taking total resolution width as 1424px.When I open it on different resolution I see a scroll bar & resolution of HTML does not adjust automatically. I don't want the scrollbar to be seen in maximized size of the page. Is it possible without using javascript.At the same time how to increase the resolution when itself is created in lower res.

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  • How to get multiple counts with one SQL query?

    - by Crobzilla
    I am wondering how to write this query. I know this actual syntax is bogus, but it will help you understand what I am wanting. I need it in this format, because it is part of a much bigger query. SELECT distributor_id, COUNT(*) AS TOTAL, COUNT(*) WHERE level = 'exec', COUNT(*) WHERE level = 'personal' I need this all returned in one query. Also, it need to be in one row, so the following won't work: 'SELECT distributor_id, COUNT(*) GROUP BY distributor_id'

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  • How many posibilities on a binary ?

    - by Val
    in hexadecimal "10 10 10 10" system you have 0-255 posibilities right? in total 256 different posibilities as there are 8 1s and 0s. how many different posibilities would i get? if i had 10 digits. instead of 8? or how would i calculate that in php ?

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  • Recreating the Apple Store's summary that moves with the page scrolling

    - by Darryl Hein
    I'm looking to create something like what Apple has on their online store for displaying the summary of your computer and total as seen here. I'm guessing this is JavaScript, but it'd be even cooler if it could be done in CSS. Basically I'd like something scrolls with the page like position:fixed, but I don't want it to start moving until the user has scroll past a certain position. I'm using jQuery, so a jQuery plugin would be perfect as well.

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  • Tomcat memory issue

    - by user305210
    Hello, I have noticed that my application that is running on Tomcat 5 starts with 1gig of memory and as soon as it starts receiving requests from client, the memory starts dropping until it is down to 100MBs and troubles start from there. I am looking at /manager/status page of tomcat under JVM section where "Free Memory", "Total Memory", "Max Memory" is listed. Is this an indicator of memory leak? Memory does not seem to be freed-up automatically even if there are no requests coming from client machines.

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  • Cannot .Count() on IQueryable (NHibernate)

    - by Bruno Reis
    Hello, I'm with an irritating problem. It might be something stupid, but I couldn't find out. I'm using Linq to NHibernate, and I would like to count how many items are there in a repository. Here is a very simplified definition of my repository, with the code that matters: public class Repository { private ISession session; /* ... */ public virtual IQueryable<Product> GetAll() { return session.Linq<Product>(); } } All the relevant code in the end of the question. Then, to count the items on my repository, I do something like: var total = productRepository.GetAll().Count(); The problem is that total is 0. Always. However there are items in the repository. Furthermore, I can .Get(id) any of them. My NHibernate log shows that the following query was executed: SELECT count(*) as y0_ FROM [Product] this_ WHERE not (1=1) That must be that "WHERE not (1=1)" clause the cause of this problem. What can I do to be able .Count() the items in my repository? Thanks! EDIT: Actually the repository.GetAll() code is a little bit different... and that might change something! It is actually a generic repository for Entities. Some of the entities implement also the ILogicalDeletable interface (it contains a single bool property "IsDeleted"). Just before the "return" inside the GetAll() method I check if if the Entity I'm querying implements ILogicalDeletable. public interface IRepository<TEntity, TId> where TEntity : Entity<TEntity, TId> { IQueryable<TEntity> GetAll(); ... } public abstract class Repository<TEntity, TId> : IRepository<TEntity, TId> where TEntity : Entity<TEntity, TId> { public virtual IQueryable<TEntity> GetAll() { if (typeof (ILogicalDeletable).IsAssignableFrom(typeof (TEntity))) { return session.Linq<TEntity>() .Where(x => (x as ILogicalDeletable).IsDeleted == false); } else { return session.Linq<TEntity>(); } } } public interface ILogicalDeletable { bool IsDeleted {get; set;} } public Product : Entity<Product, int>, ILogicalDeletable { ... } public IProductRepository : IRepository<Product, int> {} public ProductRepository : Repository<Product, int>, IProductRepository {} Edit 2: actually the .GetAll() is always returning an empty result-set for entities that implement the ILogicalDeletable interface (ie, it ALWAYS add a WHERE NOT (1=1) clause. I think Linq to NHibernate does not like the typecast.

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  • Ant variable does not exists in Ubuntu 10.10

    - by Nishat Baig
    I am trying to set up ANT build. However when I invoke build command helloworld_15/${NAME} does not exist. BUILD FAILED (total time: 0 seconds) Also the configure variables does not seems to be assigned. However i have set them into /etc/envitonment I tried echo $<varaiable_name> and value get displayed. Tried to google but not solutions seems am the first one having this issue. PS: OS Ubuntu 10.10

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  • Zend_Paginator / Doctrine 2

    - by Kevin
    I'm using Doctrine 2 with my Zend Framework application and a typical query result could yield a million (or more) search results. I want to use Zend_Paginator in line with this result set. However, I don't want to return all the results as an array and use the Array adapter as this would be inefficient, instead I would like to supply the paginator the total amount of rows then and array of results based on limit/offset amounts. Is this doable using the Array adapter or would I need to create my own pagination adapter?

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  • I need to modify a program to use arrays and a method call. Should I modify the running file, the data collection file, or both?

    - by g3n3rallyl0st
    I have to have multiple classes for this program. The problem is, I don't fully understand arrays and how they work, so I'm a little lost. I will post my program I have written thus far so you can see what I'm working with, but I don't expect anyone to DO my assignment for me. I just need to know where to start and I'll try to go from there. I think I need to use a double array since I will be working with decimals since it deals with money, and my method call needs to calculate total price for all items entered by the user. Please help: RUNNING FILE package inventory2; import java.util.Scanner; public class RunApp { public static void main(String[] args) { Scanner input = new Scanner( System.in ); DataCollection theProduct = new DataCollection(); String Name = ""; double pNumber = 0.0; double Units = 0.0; double Price = 0.0; while(true) { System.out.print("Enter Product Name: "); Name = input.next(); theProduct.setName(Name); if (Name.equalsIgnoreCase("stop")) { return; } System.out.print("Enter Product Number: "); pNumber = input.nextDouble(); theProduct.setpNumber(pNumber); System.out.print("Enter How Many Units in Stock: "); Units = input.nextDouble(); theProduct.setUnits(Units); System.out.print("Enter Price Per Unit: "); Price = input.nextDouble(); theProduct.setPrice(Price); System.out.print("\n Product Name: " + theProduct.getName()); System.out.print("\n Product Number: " + theProduct.getpNumber()); System.out.print("\n Amount of Units in Stock: " + theProduct.getUnits()); System.out.print("\n Price per Unit: " + theProduct.getPrice() + "\n\n"); System.out.printf("\n Total cost for %s in stock: $%.2f\n\n\n", theProduct.getName(), theProduct.calculatePrice()); } } } DATA COLLECTION FILE package inventory2; public class DataCollection { String productName; double productNumber, unitsInStock, unitPrice, totalPrice; public DataCollection() { productName = ""; productNumber = 0.0; unitsInStock = 0.0; unitPrice = 0.0; } //setter methods public void setName(String name) { productName = name; } public void setpNumber(double pNumber) { productNumber = pNumber; } public void setUnits(double units) { unitsInStock = units; } public void setPrice(double price) { unitPrice = price; } //getter methods public String getName() { return productName; } public double getpNumber() { return productNumber; } public double getUnits() { return unitsInStock; } public double getPrice() { return unitPrice; } public double calculatePrice() { return (unitsInStock * unitPrice); } }

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  • Haskell Add Function Return to List Until Certain Length

    - by kienjakenobi
    I want to write a function which takes a list and constructs a subset of that list of a certain length based on the output of a function. If I were simply interested in the first 50 elements of the sorted list xs, then I would use fst (splitAt 50 (sort xs)). However, the problem is that elements in my list rely on other elements in the same list. If I choose element p, then I MUST also choose elements q and r, even if they are not in the first 50 elements of my list. I am using a function finderFunc which takes an element a from the list xs and returns a list with the element a and all of its required elements. finderFunc works fine. Now, the challenge is to write a function which builds a list whose total length is 50 based on multiple outputs of finderFunc. Here is my attempt at this: finish :: [a] -> [a] -> [a] --This is the base case, which adds nothing to the final list finish [] fs = [] --The function is recursive, so the fs variable is necessary so that finish -- can forward the incomplete list to itself. finish ps fs -- If the final list fs is too small, add elements to it | length fs < 50 && length (fs ++ newrs) <= 50 = fs ++ finish newps newrs -- If the length is met, then add nothing to the list and quit | length fs >= 50 = finish [] fs -- These guard statements are currently lacking, not the main problem | otherwise = finish [] fs where --Sort the candidate list sortedps = sort ps --(finderFunc a) returns a list of type [a] containing a and all the -- elements which are required to go with it. This is the interesting -- bit. rs is also a subset of the candidate list ps. rs = finderFunc (head sortedps) --Remove those elements which are already in the final list, because -- there can be overlap newrs = filter (`notElem` fs) rs --Remove the elements we will add to the list from the new list -- of candidates newps = filter (`notElem` rs) ps I realize that the above if statements will, in some cases, not give me a list of exactly 50 elements. This is not the main problem, right now. The problem is that my function finish does not work at all as I would expect it to. Not only does it produce duplicate elements in the output list, but it sometimes goes far above the total number of elements I want to have in the list. The way this is written, I usually call it with an empty list, such as: finish xs [], so that the list it builds on starts as an empty list.

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  • Sorting a list of numbers with modified cost

    - by David
    First, this was one of the four problems we had to solve in a project last year and I couldn’t find a suitable algorithm so we handle in a brute force solution. Problem: The numbers are in a list that is not sorted and supports only one type of operation. The operation is defined as follows: Given a position i and a position j the operation moves the number at position i to position j without altering the relative order of the other numbers. If i j, the positions of the numbers between positions j and i - 1 increment by 1, otherwise if i < j the positions of the numbers between positions i+1 and j decreases by 1. This operation requires i steps to find a number to move and j steps to locate the position to which you want to move it. Then the number of steps required to move a number of position i to position j is i+j. We need to design an algorithm that given a list of numbers, determine the optimal (in terms of cost) sequence of moves to rearrange the sequence. Attempts: Part of our investigation was around NP-Completeness, we make it a decision problem and try to find a suitable transformation to any of the problems listed in Garey and Johnson’s book: Computers and Intractability with no results. There is also no direct reference (from our point of view) to this kind of variation in Donald E. Knuth’s book: The art of Computer Programing Vol. 3 Sorting and Searching. We also analyzed algorithms to sort linked lists but none of them gives a good idea to find de optimal sequence of movements. Note that the idea is not to find an algorithm that orders the sequence, but one to tell me the optimal sequence of movements in terms of cost that organizes the sequence, you can make a copy and sort it to analyze the final position of the elements if you want, in fact we may assume that the list contains the numbers from 1 to n, so we know where we want to put each number, we are just concerned with minimizing the total cost of the steps. We tested several greedy approaches but all of them failed, divide and conquer sorting algorithms can’t be used because they swap with no cost portions of the list and our dynamic programing approaches had to consider many cases. The brute force recursive algorithm takes all the possible combinations of movements from i to j and then again all the possible moments of the rest of the element’s, at the end it returns the sequence with less total cost that sorted the list, as you can imagine the cost of this algorithm is brutal and makes it impracticable for more than 8 elements. Our observations: n movements is not necessarily cheaper than n+1 movements (unlike swaps in arrays that are O(1)). There are basically two ways of moving one element from position i to j: one is to move it directly and the other is to move other elements around i in a way that it reaches the position j. At most you make n-1 movements (the untouched element reaches its position alone). If it is the optimal sequence of movements then you didn’t move the same element twice.

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  • SQL: count days in date range?

    - by John Isaacks
    I have a query like this: SELECT COUNT(*) AS amount FROM daily_individual_tracking WHERE sales = 'YES' AND daily_individual_tracking_date BETWEEN '2010-01-01' AND '2010-03-31' I am selected from a date range. Is there a way to also get the total days in the date range?

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  • How to get a list of unrepeatable date from my db in PHP?

    - by SzamDev
    Hi in my db there are 5 fields (id, list_date, amount, total, m_from) and contain data, for example : 1 - 1/1/2010 - 10 - 50 - 'example111' 1 - 1/1/2010 - 10 - 50 - 'example111' 1 - 1/4/2010 - 10 - 50 - 'example154 1 - 1/1/2010 - 10 - 50 - 'example111' 1 - 1/5/2010 - 10 - 50 - 'test' I need to know how I can get dates from list_date but without repeatable dates like '1/1/2010, 1/5/2010, 1/4/2010' Thanks in Advance.

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  • mysql query if else statemnet?

    - by user253530
    I have this sql query: SELECT S.SEARCH, S.STATUS, C.TITLE AS CategoryName, E.SEARCH_ENGINES AS Engine, S.RESULTS, S.DATE, S.TOTAL_RESULTS AS Total, S.ID FROM PLD_SEARCHES AS S Join PLD_CATEGORY AS C ON C.ID = S.CATEGORY_ID Join PLD_SEARCH_ENGINES AS E ON S.SEARCH_ENGINES_ID = E.ID ORDER BY S.DATE ASC I want to identify if S.STATUS is either 1 or 0 and according to those values to return COMPLETE or PENDING in the query results

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  • Counting comma seperated values in php,how to?

    - by Jay
    Hey folks, I have a variable holding values separated by a comma (Implode), I'm trying to get the total count of the values in that variable however, count() is just returning 1. I've tried converting the comma separated values to a properly formatted array which still spits out1. So heres the quick snippet where the sarray session equals to value1,value2,value3: $schools = $_SESSION['sarray']; $result = count($schools); Any help would be appreciated.

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