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  • Good firefox extension to download .flv flash video?

    - by Tamás Szelei
    I'm looking for a decent flash video downloader for Firefox. There are tons of them. I'd like to use it for a custom site with some random flash video player, not Youtube. How can it be accomplished? Edit: None of these methods you guys suggested worked so far; I looked up various places where the temporary file might end up - it just doesn't. It is indeed a flash player, but it seems like it's not caching - is that possible?

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  • Tout savoir sur Office 365, la nouvelle offre Cloud professionnelle complète de Microsoft, avec la Directrice Marketing des Produits Office

    TechDays 2011 : Office 365, la nouvelle offre applicative professionnelle complète De Microsoft en mode Cloud, avec la Directrice Marketing des Produits Office Non, Office 365 n'est pas la version Cloud de Office 2010. C'est un peu plus que cela. Office 365 est le nouveau fer de lance de Microsoft sur le marché des applications professionnelles hébergées. Il s'agit d'une suite complète qui se compose de : SharePoint Online (et d'un espace en ligne), pour la création de portails, de sites web ou intranet, etc. Exchange Online, pour la gestion des courriels, agendas et contacts partagés. ...

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  • Feed aggregator with E-mail/RSS channels

    - by Toc
    Which feed aggregators, besides FriendFeed, allow RSS and e-mail as input and output channels? That is, allow to suscribe external RSS feeds and to write a post by e-mail, and allow to be notified both by RSS feed and by e-mail?

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  • Recommend a Linux video editor. [closed]

    - by joeforker
    Possible Duplicate: Looking For Video Editing Software for Ubuntu I'd like a working video editor for 720p MPEG4 (.mov) with 16Khz ulaw audio in Linux. I've already tried pitivi (audio encoding issues) and kdenlive (crashes almost immediately after a clip is imported). What should I try?

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  • Fusion Human Capital Management - Enterprise Grade Software As a Service

    Tune into this conversation with Anand Subbaraman, Senior Director of Product Strategy for Fusion HCM and Technology, to learn how Oracle is delivering offer a complete HCM SaaS application with single-vendor accountability. Unlike other vendors, which rely on other partners to complete their solutions, Oracle Fusion HCM includes integrated modules for HR, Payroll, Benefits, Compensation, Performance, along with industry-firsts such as Workforce Predictions, Network at Work, and Talent Review - all available on the Cloud.

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  • From the Tips Box: Location-based To-Do Reminders, DIY Floppy Drive Music, and Easy Access to Product Manuals

    - by Jason Fitzpatrick
    Once a week we round up some great tips from the HTG tip box and share them with you; this week we’re looking at location based to-do reminders for Android phones, how to make your own floppy drive symphony, and an easy way to enjoy anywhere access to your manuals and product documentation. HTG Explains: What The Windows Event Viewer Is and How You Can Use It HTG Explains: How Windows Uses The Task Scheduler for System Tasks HTG Explains: Why Do Hard Drives Show the Wrong Capacity in Windows?

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  • Actual High Speed USB flash drive

    - by CSkau
    I'm looking to upgrade my EEE 1000H by possibly replacing the HDD with simple (internal) usb connected storage. The problem I'm having now is that I can't seem to find any actual high speed usb sticks. They all proclaim high speeds, but usually turn out to be ~30 mb/s - much lower than the 60 mb/s (480 mbit/s / 8 ) I understand USB 2.0 is at - no ? Can anyone enlighten me as to why no USB sticks seem to go past that low bar or alternatively point me in the direction of some actual high speed usb sticks ? Any help is greatly appreciated :) Cheers!

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  • Un grand quotidien américain bloque l'accès à son site pour vendre son application iPad, censure ou marketing ?

    Un grand quotidien américain bloque l'accès à son site pour favoriser son application iPad Censure ou marketing ? Alors que chaque jour ressorts des usages inédits des navigateurs modernes, rendus possibles par leurs capacités et performances sans précédent, le plus ancien journal des États-Unis vient d'interdire l'accès à sa version électronique au navigateur Safari de l'iPad. Le New York Post force les possesseurs de la tablette populaire d'Apple à télécharger son application payante s'ils veulent consulter son contenu pourtant disponible gratuitement sur le Web, en affichant une landing-page sans issue en remplacement de toutes ses pages. [IMG]http://idelways.dev...

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  • Oracle???????????47??????????

    - by user758881
    Oracle???2014?5?31???,??????,40?Oracle???????47????Oracle??? Oracle Accelerate ????? ?Oracle 2014?????????47???????????????????????Oracle????,??Oracle Financials Cloud, Oracle Sales Cloud ? Oracle Service Cloud –???? Oracle CX Cloud, ?? Oracle Human Capital Management (HCM) Cloud. ???Oracle Accelerate??????????????????? ???????????????????, ??, ???, ??, ??, ???????????????????,????????????????? ???????????????????????????????,Oracle??????????????????????Oracle???Oracle????????????? l   ??????????,???????????????——Oracle ???? eVerge Group, Certus Solutions, Presence of IT, CSolutor, Grant Thornton, ? KBACE Technologies ?????Oracle HCM Cloud ?Oracle Accelerate ????????????????????????,???????????????????,???????????????? l   ???????????????????????????——DAZ, Inc., Frontera Consulting?Inoapps ?????Oracle Financials Cloud????????????????????????? l   ?????????????????????——Capricorn Ventis, Enigen, Fellow Consulting, Solveso Interactive, CSolutor, Birchman Consulting,BPI On Demand, Business Technology Services (BizTech)? eVerge Group?????Oracle CX Cloud?????????????????????????? ??,Oracle???????????????????????????????????: l   ?????? BPI On Demand ??????????????????????Oracle Sales Cloud????? ?????????? ·          “??????????????????? ???Oracle Financials Cloud?Oracle Accelerate???? ?????????????????????????????????????????????????”–Phil Wilson, Business Development & Alliances,Inoapps ·          “KBACE?Oracle Accelerate???????KBACE ????????????????????????????????????????KBACE? Oracle Accelerate????,??Oracle HCM???,????????????????????”–Mike Peterson, President & COO, KBACE Technologies ·          “???????Oracle Financials Cloud,??????????????????????????????????????????????Oracle Accelerate????,????????????????????”—Deborah Arnold, President, DAZ Systems, Inc. ·          “????????????Oracle ERP Cloud????Oracle Accelerate?????????????????” - Sean Moore, Principal. C3Biz ·          “????,????Oracle HCM????????????????????????????eVerge Group??Oracle HCM????Oracle Accelerate???????????????????????” - John Peketz, Vice President, Marketing, eVerge Group

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  • How Important is Keyword Research to My Marketing Campaign?

    Engaging in a good deal of keyword research will help you ensure that your website obtains the type of attention that you desire. The way that the internet is set up these days, everything is done through keywords, if your website content does not contain relevant keywords then your website may not be able to get the respectable attention that you desire. Keywords are commonly defined as one word or a phrase of words that describes the type of product or service that you are opting to promote.

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  • Google lance Tag Manager, un outil gratuit qui facilite la gestion du suivi et des balises de marketing des sites Web

    Google lance Tag Manager un outil gratuit qui facilite la gestion du suivi et des balises de marketing des sites Web Google vient d'annoncer le lancement de Google Tag Manager, son nouvel outil pour la gestion des différentes balises dans un site Web. Pour mieux monétiser leur site Web et contrôler la manière dont le contenu est utilisé, les gestionnaires de sites ont recours à des outils de suivi de statistiques comme Google Analytics. Pour chaque service, un morceau de code doit être intégré dans chaque page du site. Bien que d'une utilisation relativement simple, la multiplication de ces bouts de code sur une page peut rendre leur gestion fastidieuse. De plus, les requêtes entr...

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  • Problem with duplicates in a SQL Join

    - by Chris Ballance
    I have the following result set from a join of three tables, an articles table, a products table, an articles to products mapping table. I would like to have the results with duplicates removed similar to a select distinct on content id. Current result set: [ContendId] [Title] [productId] 1 article one 2 1 article one 3 1 article one 9 4 article four 1 4 article four 10 4 article four 14 5 article five 1 6 article six 8 6 article six 10 6 article six 11 6 article six 13 7 article seven 14 Desired result set: [ContendId] [Title] [productId] 1 article one * 4 article four * 5 article five * 6 article six * 7 article seven * Here is condensed example of the relevant SQL: IF EXISTS (SELECT * FROM sys.objects WHERE object_id = OBJECT_ID(N'tempdb.dbo.products') AND type = (N'U')) drop table tempdb.dbo.products go CREATE TABLE tempdb.dbo.products ( productid int, productname varchar(255) ) go IF EXISTS (SELECT * FROM sys.objects WHERE object_id = OBJECT_ID(N'articles') AND type = (N'U')) drop table tempdb.dbo.articles go create table tempdb.dbo.articles ( contentid int, title varchar(255) ) IF EXISTS (SELECT * FROM sys.objects WHERE object_id = OBJECT_ID(N'articles') AND type = (N'U')) drop table tempdb.dbo.articles go create table tempdb.dbo.articles ( contentid int, title varchar(255) ) IF EXISTS (SELECT * FROM sys.objects WHERE object_id = OBJECT_ID(N'articleproducts') AND type = (N'U')) drop table tempdb.dbo.articleproducts go create table tempdb.dbo.articleproducts ( contentid int, productid int ) insert into tempdb.dbo.products values (1,'product one'), (2,'product two'), (3,'product three'), (4,'product four'), (5,'product five'), (6,'product six'), (7,'product seven'), (8,'product eigth'), (9,'product nine'), (10,'product ten'), (11,'product eleven'), (12,'product twelve'), (13,'product thirteen'), (14,'product fourteen') insert into tempdb.dbo.articles VALUES (1,'article one'), (2, 'article two'), (3, 'article three'), (4, 'article four'), (5, 'article five'), (6, 'article six'), (7, 'article seven'), (8, 'article eight'), (9, 'article nine'), (10, 'article ten') INSERT INTO tempdb.dbo.articleproducts VALUES (1,2), (1,3), (1,9), (4,1), (4,10), (4,14), (5,1), (6,8), (6,10), (6,11), (6,13), (7,14) GO select DISTINCT(a.contentid), a.title, p.productid from articles a JOIN articleproducts ap ON a.contentid = ap.contentid JOIN products p ON a.contentid = ap.contentid AND p.productid = ap.productid ORDER BY a.contentid

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  • FreeText COUNT query on multiple tables is super slow

    - by Eric P
    I have two tables: **Product** ID Name SKU **Brand** ID Name Product table has about 120K records Brand table has 30K records I need to find count of all the products with name and brand matching a specific keyword. I use freetext 'contains' like this: SELECT count(*) FROM Product inner join Brand on Product.BrandID = Brand.ID WHERE (contains(Product.Name, 'pants') or contains(Brand.Name, 'pants')) This query takes about 17 secs. I rebuilt the FreeText index before running this query. If I only check for Product.Name. They query is less then 1 sec. Same, if I only check the Brand.Name. The issue occurs if I use OR condition. If I switch query to use LIKE: SELECT count(*) FROM Product inner join Brand on Product.BrandID = Brand.ID WHERE Product.Name LIKE '%pants%' or Brand.Name LIKE '%pants%' It takes 1 secs. I read on MSDN that: http://msdn.microsoft.com/en-us/library/ms187787.aspx To search on multiple tables, use a joined table in your FROM clause to search on a result set that is the product of two or more tables. So I added an INNER JOINED table to FROM: SELECT count(*) FROM (select Product.Name ProductName, Product.SKU ProductSKU, Brand.Name as BrandName FROM Product inner join Brand on product.BrandID = Brand.ID) as TempTable WHERE contains(TempTable.ProductName, 'pants') or contains(TempTable.BrandName, 'pants') This results in error: Cannot use a CONTAINS or FREETEXT predicate on column 'ProductName' because it is not full-text indexed. So the question is - why OR condition could be causing such as slow query?

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  • Is it possible to save the product key of Windows 8?

    - by Dibya Ranjan
    I have Windows 8 activated in my system. I don't have the product key of windows right now. Now I want to format my system again. Is there any way so that I can reuse the key? Is there any way I can get the key from an activated windows machine? Edit: I am not able to find the product key because I have used a MAK as my product key. Now I want the same to use it after formatting my disk. I found a software Volume Activation Manager tool on the windows website. I am not sure how to use it. Please tell me how can I reuse my key?

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  • Problems with :uniq => true/Distinct option in a has_many_through association w/ named scope (Rails)

    - by MikeH
    I had to make some tweaks to my app to add new functionality, and my changes seem to have broken the :uniq option that was previously working perfectly. Here's the set up: #User.rb has_many :products, :through = :seasons, :uniq = true has_many :varieties, :through = :seasons, :uniq = true #product.rb has_many :seasons has_many :users, :through = :seasons, :uniq = true has_many :varieties #season.rb belongs_to :product belongs_to :variety belongs_to :user named_scope :by_product_name, :joins = :product, :order = 'products.name' #variety.rb belongs_to :product has_many :seasons has_many :users, :through = :seasons, :uniq = true First I want to show you the previous version of the view that is now breaking, so that we have a baseline to compare. The view below is pulling up products and varieties that belong to the user. In both versions below, I've assigned the same products/varieties to the user so the logs will looking at the exact same use case. #user/show <% @user.products.each do |product| %> <%= link_to product.name, product %> <% @user.varieties.find_all_by_product_id(product.id).each do |variety| %> <%=h variety.name.capitalize %></p> <% end %> <% end %> This works. It displays only one of each product, and then displays each product's varieties. In the log below, product ID 1 has 3 associated varieties. And product ID 43 has none. Here's the log output for the code above: Product Load (11.3ms) SELECT DISTINCT `products`.* FROM `products` INNER JOIN `seasons` ON `products`.id = `seasons`.product_id WHERE ((`seasons`.user_id = 1)) ORDER BY name, products.name Product Columns (1.8ms) SHOW FIELDS FROM `products` Variety Columns (1.9ms) SHOW FIELDS FROM `varieties` Variety Load (0.7ms) SELECT DISTINCT `varieties`.* FROM `varieties` INNER JOIN `seasons` ON `varieties`.id = `seasons`.variety_id WHERE (`varieties`.`product_id` = 1) AND ((`seasons`.user_id = 1)) ORDER BY name Variety Load (0.5ms) SELECT DISTINCT `varieties`.* FROM `varieties` INNER JOIN `seasons` ON `varieties`.id = `seasons`.variety_id WHERE (`varieties`.`product_id` = 43) AND ((`seasons`.user_id = 1)) ORDER BY name Ok, so everything above is the previous version which was working great. In the new version, I added some columns to the join table called seasons, and made a bunch of custom methods that query those columns. As a result, I made the following changes to the view code that you saw above so that I could access those methods on the seasons model: <% @user.seasons.by_product_name.each do |season| %> <%= link_to season.product.name, season.product %> #Note: I couldn't get this loop to work at all, so I settled for the following: #<% @user.varieties.find_all_by_product_id(product.id).each do |variety| %> <%=h season.variety.name.capitalize %> <%end%> <%end%> Here's the log output for that: SQL (0.9ms) SELECT count(DISTINCT "products".id) AS count_products_id FROM "products" INNER JOIN "seasons" ON "products".id = "seasons".product_id WHERE (("seasons".user_id = 1)) Season Load (1.8ms) SELECT "seasons".* FROM "seasons" INNER JOIN "products" ON "products".id = "seasons".product_id WHERE ("seasons".user_id = 1) AND ("seasons".user_id = 1) ORDER BY products.name Product Load (0.7ms) SELECT * FROM "products" WHERE ("products"."id" = 43) ORDER BY products.name CACHE (0.0ms) SELECT "seasons".* FROM "seasons" INNER JOIN "products" ON "products".id = "seasons".product_id WHERE ("seasons".user_id = 1) AND ("seasons".user_id = 1) ORDER BY products.name Product Load (0.4ms) SELECT * FROM "products" WHERE ("products"."id" = 1) ORDER BY products.name Variety Load (0.4ms) SELECT * FROM "varieties" WHERE ("varieties"."id" = 2) ORDER BY name CACHE (0.0ms) SELECT * FROM "products" WHERE ("products"."id" = 1) ORDER BY products.name Variety Load (0.4ms) SELECT * FROM "varieties" WHERE ("varieties"."id" = 8) ORDER BY name CACHE (0.0ms) SELECT * FROM "products" WHERE ("products"."id" = 1) ORDER BY products.name Variety Load (0.4ms) SELECT * FROM "varieties" WHERE ("varieties"."id" = 7) ORDER BY name CACHE (0.0ms) SELECT * FROM "products" WHERE ("products"."id" = 43) ORDER BY products.name CACHE (0.0ms) SELECT count(DISTINCT "products".id) AS count_products_id FROM "products" INNER JOIN "seasons" ON "products".id = "seasons".product_id WHERE (("seasons".user_id = 1)) CACHE (0.0ms) SELECT "seasons".* FROM "seasons" INNER JOIN "products" ON "products".id = "seasons".product_id WHERE ("seasons".user_id = 1) AND ("seasons".user_id = 1) ORDER BY products.name CACHE (0.0ms) SELECT * FROM "products" WHERE ("products"."id" = 1) ORDER BY products.name CACHE (0.0ms) SELECT * FROM "products" WHERE ("products"."id" = 1) ORDER BY products.name CACHE (0.0ms) SELECT * FROM "varieties" WHERE ("varieties"."id" = 8) ORDER BY name I'm having two problems: (1) The :uniq option is not working for products. Three distinct versions of the same product are displaying on the page. (2) The :uniq option is not working for varieties. I don't have validation set up on this yet, and if the user enters the same variety twice, it does appear on the page. In the previous working version, this was not the case. The result I need is that only one product for any given ID displays, and all varieties associated with that ID display along with such unique product. One thing that sticks out to me is the sql call in the most recent log output. It's adding 'count' to the distinct call. I'm not sure why it's doing that or whether it might be an indication of an issue. I found this unresolved lighthouse ticket that seems like it could potentially be related, but I'm not sure if it's the same issue: https://rails.lighthouseapp.com/projects/8994/tickets/2189-count-breaks-sqlite-has_many-through-association-collection-with-named-scope I've tried a million variations on this and can't get it working. Any help is much appreciated!

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  • Any good Open Source or Cheap ASP.NET Catalog Applications

    - by Zhaph - Ben Duguid
    We're looking for a cheap-to-free "off the shelf" ASP.NET catalogue application, that will meet the following requirements: Support two kinds of listings: Suppliers of Services Suppliers of Products, and their Products Suppliers can be categorised by: Area of specialisation - including sub-categories Location Other data, e.g. where listing came from Versioning of supplier/product details Easy to use management interface Use masterpages so we can drop it into our existing site layout Run on a Windows 2003 server, with .NET 3.5 installed In an ideal world, the following additional requirements might be met: Suppliers can manage their own listings Other products that are available to us (that will obviously need some additional development to meet these requirements) are: Content Management System (MS) Commerce Server - bear in mind we're not selling the products/suppliers, just listing them Simple DB application. I'm happy to knock something up in MCMS/simple DB, I'm just looking to see if anyone's had any experience with off the shelf apps that could save us some time. I'm also happy to receive "Don't use this because" type answers. Thanks.

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  • Sum of path products in DAG

    - by Jules
    Suppose we have a DAG with edges labeled with numbers. Define the value of a path as the product of the labels. For each (source,sink)-pair I want to find the sum of the values of all the paths from source to sink. You can do this in polynomial time with dynamic programming, but there are still some choices that can be made in how you decompose the problem. In my case I have one DAG that has to be evaluated repeatedly with different labelings. My question is: for a given DAG, how can we pre-compute a good strategy for computing these values for different labelings repeatedly. It would be nice if there was an algorithm that finds an optimal way, for example a way that minimizes the number of multiplications. But perhaps this is too much to ask, I would be very happy with an algorithm that just gives a good decomposition.

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  • Is this type of calculation to be put in Model or Controller?

    - by Hadi
    i have Product and SalesOrder model (to simplify, 1 sales_order only for 1 product) Product has_many SalesOrder SalesOrder belongs_to Product pa = Product A #2000 so1 = SalesOrder 1 order product A #1000, date:yesterday so2 = SalesOrder 2 order product A #999, date:yesterday so3 = SalesOrder 3 order product A #1000, date:now Based on the date, pa.find_sales_orders_that_can_be_delivered will give: SalesOrder 1 order product A #1000, date:yesterday SalesOrder 2 order product A #999, date:yesterday SalesOrder 3 order product A #1, date:now <-- the newest The question is: is find_sales_orders_that_can_be_delivered should be in the Model? i can do it in controller. and the general question is: what goes in Model and what goes in Controller. Thank you

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  • Best similarity metric for collaborative filtering?

    - by allclaws
    I'm trying to decide on the best similarity metric for a product recommendation system using item-based collaborative filtering. This is a shopping basket scenario where ratings are binary valued - the user has either purchased an item or not - there is no explicit rating system (eg, 5-stars). Step 1 is to compute item-to-item similarity, though I want to look at incorporating more features later on. Is the Tanimoto coefficient the best way to go for binary values? Or are there other metrics that are appropriate here? Thanks.

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  • Php - Looping For Different CSS

    - by bob
    MySQL $selectSize = "SELECT * FROM products"; $querySize = $db->select($selectSize); while ($product = $db->fetcharray($querySize)) { HTML <ul> <li>Product A</li> <li>Product B</li> <li class='right'>Product C</li> <li>Product D</li> <li>Product E</li> <li class='right'>Product F</li> </ul> Question While getting the product, I want the Product C and Product F or any product after 3 loops will have class='right' to the list style. Let me know Thanks

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