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  • The internal storage of a DATETIME2 value

    - by Peter Larsson
    Today I went for investigating the internal storage of DATETIME2 datatype. What I found out was that for a datetime2 value with precision 0 (seconds only), SQL Server need 6 bytes to represent the value, but stores 7 bytes. This is because SQL Server add one byte that holds the precision for the datetime2 value. Start with this very simple repro declare @now datetime2(7) = '2010-12-15 21:04:03.6934231'   select  cast(cast(@now as datetime2(0)) as binary(7)),         cast(cast(@now as datetime2(1)) as binary(7)),         cast(cast(@now as datetime2(2)) as binary(7)),         cast(cast(@now as datetime2(3)) as binary(8)),         cast(cast(@now as datetime2(4)) as binary(8)),         cast(cast(@now as datetime2(5)) as binary(9)),         cast(cast(@now as datetime2(6)) as binary(9)),         cast(cast(@now as datetime2(7)) as binary(9)) Now we are going to copy and paste these binary values and investigate which value is representing what time part. Prefix  Ticks       Ticks         Days    Days    Original value ------  ----------  ------------  ------  ------  -------------------- 0x  00  442801             75844  A8330B  734120  0x00442801A8330B 0x  01  A5920B            758437  A8330B  734120  0x01A5920BA8330B  0x  02  71BA73           7584369  A8330B  734120  0x0271BA73A8330B 0x  03  6D488504        75843693  A8330B  734120  0x036D488504A8330B 0x  04  46D4342D       758436934  A8330B  734120  0x0446D4342DA8330B 0x  05  BE4A10C401    7584369342  A8330B  734120  0x05BE4A10C401A8330B 0x  06  6FEBA2A811   75843693423  A8330B  734120  0x066FEBA2A811A8330B 0x  07  57325D96B0  758436934231  A8330B  734120  0x0757325D96B0A8330B Let us use the following color schema Red - Prefix Green - Time part Blue - Day part What you can see is that the date part is equal in all cases, which makes sense since the precision doesm't affect the datepart. What would have been fun, is datetime2(negative) just like round accepts a negative value. -1 would mean rounding to 10 second, -2 rounding to minute, -3 rounding to 10 minutes, -4 rounding to hour and finally -5 rounding to 10 hour. -5 is pretty useless, but if you extend this thinking to -6, -7 and so on, you could actually get a datetime2 value which is accurate to the month only. Well, enough ranting about this. Let's get back to the table above. If you add 75844 second to midnight, you get 21:04:04, which is exactly what you got in the select statement above. And if you look at it, it makes perfect sense that each following value is 10 times greater when the precision is increased one step too. //Peter

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  • SQLBeat Podcast – Episode 4 – Mark Rasmussen on Machine Guns,Jelly Fish and SQL Storage Engine

    - by SQLBeat
    In this this 4th SQLBeat Podcast I talk with fellow Dane Mark Rasmussen on SQL, machine guns and jelly fish fights; apparently they are common in our homeland. Who am I kidding, I am not Danish, but I try to be in this podcast. Also, we exchange knowledge on SQL Server storage engine particulars as well as some other “internals” like password hashes and contained databases. And then it just gets weird and awesome. There is lots of background noise from people who did not realize we were recording. And I call them out and make fun of them as they deserve; well just one person who is well known in these parts. I also learn the correct (almost) pronunciation of “fjord”. Seriously, a word with an “F” followed by a “J”. And there are always the hippies and hipsters to discuss. Should be fun.

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  • The internal storage of a DATETIMEOFFSET value

    - by Peter Larsson
    Today I went for investigating the internal storage of DATETIME2 datatype. What I found out was that for a datetime2 value with precision 0 (seconds only), SQL Server need 6 bytes to represent the value, but stores 7 bytes. This is because SQL Server add one byte that holds the precision for the datetime2 value. Start with this very simple repro declare    @now datetimeoffset(7) = '2010-12-15 21:04:03.6934231 +03:30'   select     cast(cast(@now as datetimeoffset(0)) as binary(9)),            cast(cast(@now as datetimeoffset(1)) as binary(9)),            cast(cast(@now as datetimeoffset(2)) as binary(9)),            cast(cast(@now as datetimeoffset(3)) as binary(10)),            cast(cast(@now as datetimeoffset(4)) as binary(10)),            cast(cast(@now as datetimeoffset(5)) as binary(11)),            cast(cast(@now as datetimeoffset(6)) as binary(11)),            cast(cast(@now as datetimeoffset(7)) as binary(11)) Now we are going to copy and paste these binary values and investigate which value is representing what time part. Prefix  Ticks       Ticks         Days    Days    Suffix  Suffix  Original value ------  ----------  ------------  ------  ------  ------  ------  ------------------------ 0x  00  0CF700             63244  A8330B  734120  D200       210  0x000CF700A8330BD200 0x  01  75A609            632437  A8330B  734120  D200       210 0x0175A609A8330BD200 0x  02  918060           6324369  A8330B  734120  D200       210  0x02918060A8330BD200 0x  03  AD05C503        63243693  A8330B  734120  D200       210  0x03AD05C503A8330BD200 0x  04  C638B225       632502470  A8330B  734120  D200       210  0x04C638B225A8330BD200 0x  05  BE37F67801    6324369342  A8330B  734120  D200       210  0x05BE37F67801A8330BD200 0x  06  6F2D9EB90E   63243693423  A8330B  734120  D200       210  0x066F2D9EB90EA8330BD200 0x  07  57C62D4093  632436934231  A8330B  734120  D200       210  0x0757C62D4093A8330BD200 Let us use the following color schema Red - Prefix Green - Time part Blue - Day part Purple - UTC offset What you can see is that the date part is equal in all cases, which makes sense since the precision doesn't affect the datepart. If you add 63244 seconds to midnight, you get 17:34:04, which is the correct UTC time. So what is stored is the UTC time and the local time can be found by adding "utc offset" minutes. And if you look at it, it makes perfect sense that each following value is 10 times greater when the precision is increased one step too. //Peter

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  • Strategy for backwards compatibility of persistent storage

    - by Baqueta
    In my experience, trying to ensure that new versions of an application retain compatibility with data storage from previous versions can often be a painful process. What I currently do is to save a version number for each 'unit' of data (be it a file, database row/table, or whatever) and ensure that the version number gets updated each time the data changes in some way. I also create methods to convert from v1 to v2, v2 to v3, and so on. That way, if I'm at v7 and I encounter a v3 file, I can do v3-v4-v5-v6-v7. So far this approach seems to be working out well, but I haven't had to make use of it extensively yet so there may be unforseen problems. I'm also concerned that if the objects I'm loading change significantly, I'll either have to keep around old versions of the classes or face updating all my conversion methods to handle the new class definition. Is my approach sound? Are there other/better approaches I could be using? Are there any design patterns applicable to this problem?

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  • Product Support News for Oracle Solaris, Systems, and Storage

    - by user12244613
    Hi System Support Customers, April Newsletter is now available The April, 2012 Newsletter for Oracle Solaris, Systems, and Storage is now available via document 1363390.1 *Requires a My Oracle Support account to access. Please take a few minutes to read the newsletter. The newsletter is the primary method of communication about what we in support would like you to be aware of. If you are not receiving the newsletter, it could be due to: (a) Your Oracle profile does not have the allow Oracle Communication selected (on oracle.com Sign In, or if logged in select "Account" and under your Job Role, check you have selected this box : [ ] Yes, send me e-mails in Oracle Products.... (b) you have not logged a service request during the last 12 months. Oracle is working to improve the distribution process and changes are coming and once they are ready I will write more about that. But today if you don't automatically receive the newsletter all you can do is save it as a favorite within My Oracle Support and come back on the 2nd of each month to check out the changes. This month I am really interested to find out from you is the Newsletter providing you the type of items that you are interested in. To gather some data on that, I have a small 2minute survey running on the newsletter or you can access it [ here ] Finally, if you think I am missing a topic in the Newsletter, let me know by taking the survey or suggesting a topic via this blog. Get Proactive Don't forget about being Proactive. The latest updates for Systems and Solaris pages in the Get Proactive area are now available. Check out document 432.1 and learn what proactive features are available for Systems and Solaris.

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  • Storage of leftover values in a situation of having to round down

    - by jt0dd
    I'm writing an app (client and server side) where the number of sales required by each employee must be kept track of in round-number form. Each month, the employees are required to sell a certain number, and this app needs to keep track of how many sales must be made for each 12 hour interval during the work week. Because I have to round the values down to a whole number, I must keep track of leftovers in the rounding process and ensure that they are always carried over. My method must ensure the storage of the leftover value even when client and server side crash, restart, close, etc. Right now, I'm working on doing this by storing the leftovers in a field in the user's account row in the database each time a value is rounded, reading the stored value, removing any portion that is used (when a whole number is reached, most of the leftover is used up), and storing the new value. This practice seems weird because while the leftovers are calculated on the client side, it's the same number for each employee, and every employee using the app is storing a copy of the same leftover data. Alternatively, I could have all clients store the data at once into the same data field on a general table, but this is just as weird. Is there a better way that this can be handled or is my method correct?

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  • Where to store things like user pictures using Azure? Blob Storage?

    - by n26
    I have just migrated a project of mine for test cases to Microsoft's azure. But for functionalities similar to an avatar upload I need write access to the files on the harddrive. But this is a cloud, so this is not possible. How can I build such functionalities instead? Should I use the Blob Storage or is there a better solution? Does it make sense to store all website images (f.e. layout images) in the Blob Storage? So I would have a Cookie-free Domain for my static content?

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  • Are there any USB flash drives or SD cards which use RAID or redundant storage for additional reliability?

    - by Luke Dennis
    I'm looking to get a fault-tolerant USB flash drive, which saves data to multiple independent locations, whether using RAID or some other means to back up data. Has a product like this ever been created, or are my only options to hack something together? (By the way: I'm aware that RAID doesn't prevent data corruption from software or the file system. I'm just looking for something that can handle one of the memory sticks going dead.)

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  • Using Google Docs as a Storage System like S3. Is there any limit?

    - by mickthomp
    Hi all, I'm considering to upgrade to a Google Docs Premium Account (gDrive)? I'm wondering if that can be used as I'm using Amazon S3 at the moment. I'd like to upload images. Do you know if there is there any limit on the number of images I can upload on my 200GB Google Docs account? I think it could be really useful to have something like that and we could save some money on our webapps. Thank you ;)

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  • Need a solution to store images (1 billion, 1000,000,000) which users will upload to a website via php or javascript upload [on hold]

    - by wish_you_all_peace
    I need a solution to store images (1 billion) which users will upload to a website via PHP or Javascript upload (website will have 1 billion page views a month using Linux Debian distros) assuming 20 photos per user maximum (10 thumbnails of size 90px by 90px and 10 large, script resized images of having maximum width 500px or maximum height 500px depending on shape of image, meaning square, rectangle, horizontal, vertical etc). Assume this to be a LEMP-stack (Linux Nginx MySQL PHP) social-media or social-matchmaking type application whose content will be text and images. Since everyone knows storing tons of images (website users uploaded images in this case) are bad inside a single directory or NFS etc, please explain all the details about the architecture and configuration of the entire setup of storage solution, to store 1 billion images on any method you recommend (no third-party cloud storage like S3 etc. It has to be within the private data center using our own hardware and resources.). The solution has to include both the storage solution and organizing the images uploaded by users. How will we organize the users images if a single user will not have more than 20 images (10 thumbs and 10 large of having either width or height 500px)? Please consider that this has to be organized in a structural way so we can fetch a single user's images via PHP/Javascript or API programmatically through some type of user's unique identifier(s).

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  • STOP PRESS: FY15 Q1 Oracle ZS3 Contest for Partners

    - by Cinzia Mascanzoni
    04 JUNE 2014 Oracle EMEA Partners Stop Press Stay Connected Oracle Media Network   OPN on PartnerCast   STOP PRESS: FY15 Q1 Oracle ZS3 Contest for PartnersShare an unforgettable experience at the Teatro Alla Scala in Milan Dear valued Partner, We are pleased to launch a partner contest exclusive to our partners dedicated to promoting and selling Oracle Systems! You are essential to the success of Oracle and we want to recognize your contribution and effort in driving Oracle Storage to the market. To show our appreciation we are delighted to announce a contest, giving the winners the opportunity to attend a roundtable chaired by Senior Oracle Executives and spend an unforgettable evening at the magnificent Teatro Alla Scala in Milan, followed by a stay at the Grand Hotel et de Milan, courtesy of Oracle. Recognition will be given to 12 partner companies (10 VARs & 2 VADs) who will be recognized for their ZFS storage booking achievement in the broad market between June 1st and July 18th 2014. Criteria of Eligibility A minimum deal value of $30k is required for qualification Partners who are wholly or partially owned by a public sector organization are not eligible for participation Winners The winning VARs will be: The highest ZS3 or ZBA bookings achievers by COB on July 18th, 2014 in each Oracle EMEA region (1) The highest Oracle on Oracle (2) ZS3 or ZBA bookings achievers by COB on July 18th, 2014 in each Oracle EMEA region The winning VADs (3) will be: The highest ZS3 or ZBA bookings achiever by COB on July 18th 2014 in EMEA The highest Oracle on Oracle (2) ZS3 or ZBA bookings achiever by COB on July 18th 2014 in EMEA (1) Two VAR winners for each EMEA region – Eastern Europe & CIS, Middle East & Africa, South Europe, North Europe, UK/Ireland & Israel - as per the criteria outlined above(2) Oracle on Oracle, in this instance, means ZS3 or ZBA storage attached to DB or DB options, Engineered Systems or Sparc servers sold to the same customer by the same partner within the contest timelines.(3) Two VAD winners, one for each of the criteria outlined above, will be selected from across EMEA. Oracle shall be the final arbiter in selecting the winners. All winners will be notified via their Oracle account manager. Full details about the contest, expenses covered by Oracle and timetable of events can be found on the Oracle EMEA Hardware (Servers & Storage) Partner Community workspace (FY15 Q1 ZFS Partner Contest). Access to the community workspace requires membership. If you are not a member please register here. The Prize Winners will be invited to participate to a roundtable chaired by Oracle on Monday September 8th 2014 in Milan and to be guests of Oracle in the evening of September 8th, 2014 at the Teatro Alla Scala. The evening will comprise of a private tour of the Scala museum, cocktail reception at the elegant museum rooms and attending the performance by the renowned Soprano, Maria Agresta. Our guests will then retire for the evening to the Grand Hotel et de Milan, courtesy of Oracle. Good Luck!! For more information, please contact Sasan Moaveni. Regards, Olivier TordoSenior Director - Systems Business DevelopmentOracle EMEA Alliances & Channels Resources EMEA Hardware Partner Community EMEA Oracle Partner Days Find Partner Events EMEA Partner News Blog EMEA Partner Enablement Blog Oracle PartnerNetwork Copyright © 2014, Oracle and/or its affiliates.All rights reserved. Contact Us | Legal Notices and Terms of Use | Privacy Statement

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  • Hyper-V 2012 and P2000 SAS SAN

    - by user155950
    Hi I am having major problems setting up a Hyper-V 2012 cluster on a P2000 SAS SAN. Running System Center VMM 2012 SP1 I am unable to see any storage to create my cluster. Has anyone had experienced anything similar? Under fabric and storage I can't add the P2000, all I can do is use storage spaces in server manager to create a storage pool and virtual disk. This allows me to create a file share which I can add to VMM but I still can't see any disk to create a cluster. I am just about at the point where I want to tear my hair out wipe the servers and stick VMware on them because I know it works as I have set several systems up like this in the past. The Hyper-V servers can see the storage and in server manager on my management machine it seems to know both servers can see the same disk. VMM is running on the same machine and it can't see any disk. Help..... Thanks Mike

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  • Oracle ZS3 Contest for Partners: Share an unforgettable experience at the Teatro Alla Scala in Milan

    - by Claudia Caramelli-Oracle
    12.00 Dear valued Partner, We are pleased to launch a partner contest exclusive to our partners dedicated to promoting and selling Oracle Systems! You are essential to the success of Oracle and we want to recognize your contribution and effort in driving Oracle Storage to the market. To show our appreciation we are delighted to announce a contest, giving the winners the opportunity to attend a roundtable chaired by Senior Oracle Executives and spend an unforgettable evening at the magnificent Teatro Alla Scala in Milan, followed by a stay at the Grand Hotel et de Milan, courtesy of Oracle. Recognition will be given to 12 partner companies (10 VARs & 2 VADs) who will be recognized for their ZFS storage booking achievement in the broad market between June 1st and July 18th 2014. Criteria of Eligibility A minimum deal value of $30k is required for qualification Partners who are wholly or partially owned by a public sector organization are not eligible for participation  Winners The winning VARs will be: The highest ZS3 or ZBA bookings achievers by COB on July 18th, 2014 in each Oracle EMEA region (1) The highest Oracle on Oracle (2) ZS3 or ZBA bookings achievers by COB on July 18th, 2014 in each Oracle EMEA region The winning VADs (3) will be: The highest ZS3 or ZBA bookings achiever by COB on July 18th 2014 in EMEA The highest Oracle on Oracle (2) ZS3 or ZBA bookings achiever by COB on July 18th 2014 in EMEA  The Prize Winners will be invited to participate to a roundtable chaired by Oracle on Monday September 8th 2014 in Milan and to be guests of Oracle in the evening of September 8th, 2014 at the Teatro Alla Scala. The evening will comprise of a private tour of the Scala museum, cocktail reception at the elegant museum rooms and attending the performance by the renowned Soprano, Maria Agresta. Our guests will then retire for the evening to the Grand Hotel et de Milan, courtesy of Oracle. Oracle shall be the final arbiter in selecting the winners and all winners will be notified via their Oracle account manager.Full details about the contest, expenses covered by Oracle and timetable of events can be found on the Oracle EMEA Hardware (Servers & Storage) Partner Community workspace (FY15 Q1 ZFS Partner Contest). Remember: access to the community workspace requires membership. If you are not a member please register here. Good Luck!! For more information, please contact Sasan Moaveni. (1) Two VAR winners for each EMEA region – Eastern Europe & CIS, Middle East & Africa, South Europe, North Europe, UK/Ireland & Israel - as per the criteria outlined above (2) Oracle on Oracle, in this instance, means ZS3 or ZBA storage attached to DB or DB options, Engineered Systems or Sparc servers sold to the same customer by the same partner within the contest timelines.(3) Two VAD winners, one for each of the criteria outlined above, will be selected from across EMEA. Normal 0 14 false false false IT X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0cm 5.4pt 0cm 5.4pt; mso-para-margin-top:0cm; mso-para-margin-right:0cm; mso-para-margin-bottom:10.0pt; mso-para-margin-left:0cm; line-height:115%; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi; mso-fareast-language:EN-US;}

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  • Define a positive number to be isolated if none of the digits in its square are in its cube [closed]

    - by proglaxmi
    Define a positive number to be isolated if none of the digits in its square are in its cube. For example 163 is n isolated number because 69*69 = 26569 and 69*69*69 = 4330747 and the square does not contain any of the digits 0, 3, 4 and 7 which are the digits used in the cube. On the other hand 162 is not an isolated number because 162*162=26244 and 162*162*162 = 4251528 and the digits 2 and 4 which appear in the square are also in the cube. Write a function named isIsolated that returns 1 if its argument is an isolated number, it returns 0 if its not an isolated number and it returns -1 if it cannot determine whether it is isolated or not (see the note below). The function signature is: int isIsolated(long n) Note that the type of the input parameter is long. The maximum positive number that can be represented as a long is 63 bits long. This allows us to test numbers up to 2,097,151 because the cube of 2,097,151 can be represented as a long. However, the cube of 2,097,152 requires more than 63 bits to represent it and hence cannot be computed without extra effort. Therefore, your function should test if n is larger than 2,097,151 and return -1 if it is. If n is less than 1 your function should also return -1. Hint: n % 10 is the rightmost digit of n, n = n/10 shifts the digits of n one place to the right. The first 10 isolated numbers are N n*n n*n*n 2 4 8 3 9 27 8 64 512 9 81 729 14 196 2744 24 576 13824 28 784 21952 34 1156 39304 58 3364 195112 63 3969 250047

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  • Serve up PC hard drive as USB mass storage

    - by sheepsimulator
    Is there a software package available that can serve up a hard-drive internal to a PC and make it available over USB to other USB Master nodes as mass storage? Ex: take your C: or /dev/hda drive on a PC (let's call the computer PC-A), and run a driver program which makes your C: or /dev/hda drive available to external devices as USB mass storage. When you'd hook up another PC (PC-B) to PC-A via USB, it would detect a USB mass storage device, which is C: or /dev/hda on PC-A. Is this even possible? EDIT: I know that there are other ways of making data on a drive available between two different computers (eg. putting PC-A's hdd in a USB-drive-enclosure, or having PC-A make the hdd available via a network share). But I'd like to know if the method that I describe above is even technically possible.

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  • Amazon S3 - Storage Class and Server Side Encryption

    - by Steven
    Ahhh! I am using Amazon S3 for some low price storage to clear down out SAN. I created the bucket and created a root folder. I set the storage class to standard and server side encryption AES. I started a copy job to move the files, some files copied over and i checked the files: Reduced Redundancy Encryption set to none WTF? So i deleted all files and folders. I manuallyed created the folder structure and again set the storage class and encryption level. I coped some files and bamm, still showing (at a file level as Reduced and no encryption). So my question is this, is it really raid'd and encrypted just not showing it properly (as the root folder is, how can the file not be??) or (b) am i being a huge tool and missing something?

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  • Dropbox’s Great Space Race Delivers Additional Space to Students

    - by Jason Fitzpatrick
    If you’re a student or faculty member (or still have an active .edu email account) now’s the time to cash in on some free cloud storage courtesy of Dropbox’s Great Space Race. Just by linking your .edu address with your Dropbox account you’ll get an extra 3GB of storage for the next two years. The more people from your school that sign up, the higher the total climbs–up to an extra 25GB for two years. The Space Race lasts for the next eight weeks, you can read more about the details here or just jump to the signup page at the link below. The Great Space Race [Dropbox] Why Enabling “Do Not Track” Doesn’t Stop You From Being Tracked HTG Explains: What is the Windows Page File and Should You Disable It? How To Get a Better Wireless Signal and Reduce Wireless Network Interference

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  • Recommendations for distributed processing/distributed storage systems

    - by Eddie
    At my organization we have a processing and storage system spread across two dozen linux machines that handles over a petabyte of data. The system right now is very ad-hoc; processing automation and data management is handled by a collection of large perl programs on independent machines. I am looking at distributed processing and storage systems to make it easier to maintain, evenly distribute load and data with replication, and grow in disk space and compute power. The system needs to be able to handle millions of files, varying in size between 50 megabytes to 50 gigabytes. Once created, the files will not be appended to, only replaced completely if need be. The files need to be accessible via HTTP for customer download. Right now, processing is automated by perl scripts (that I have complete control over) which call a series of other programs (that I don't have control over because they are closed source) that essentially transforms one data set into another. No data mining happening here. Here is a quick list of things I am looking for: Reliability: These data must be accessible over HTTP about 99% of the time so I need something that does data replication across the cluster. Scalability: I want to be able to add more processing power and storage easily and rebalance the data on across the cluster. Distributed processing: Easy and automatic job scheduling and load balancing that fits with processing workflow I briefly described above. Data location awareness: Not strictly required but desirable. Since data and processing will be on the same set of nodes I would like the job scheduler to schedule jobs on or close to the node that the data is actually on to cut down on network traffic. Here is what I've looked at so far: Storage Management: GlusterFS: Looks really nice and easy to use but doesn't seem to have a way to figure out what node(s) a file actually resides on to supply as a hint to the job scheduler. GPFS: Seems like the gold standard of clustered filesystems. Meets most of my requirements except, like glusterfs, data location awareness. Ceph: Seems way to immature right now. Distributed processing: Sun Grid Engine: I have a lot of experience with this and it's relatively easy to use (once it is configured properly that is). But Oracle got its icy grip around it and it no longer seems very desirable. Both: Hadoop/HDFS: At first glance it looked like hadoop was perfect for my situation. Distributed storage and job scheduling and it was the only thing I found that would give me the data location awareness that I wanted. But I don't like the namename being a single point of failure. Also, I'm not really sure if the MapReduce paradigm fits the type of processing workflow that I have. It seems like you need to write all your software specifically for MapReduce instead of just using Hadoop as a generic job scheduler. OpenStack: I've done some reading on this but I'm having trouble deciding if it fits well with my problem or not. Does anyone have opinions or recommendations for technologies that would fit my problem well? Any suggestions or advise would be greatly appreciated. Thanks!

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  • Silverlight 4.0: How to increase quota in Isolated File Storage

    - by xscape
    Got this line of code here but its not working. using (IsolatedStorageFile isf = IsolatedStorageFile.GetUserStoreForApplication()) { long newSpace = isf.Quota + spaceRequest; try { if(true == isf.IncreaseQuotaTo(newSpace)) { Debug.WriteLine("success"); } else { Debug.WriteLine("unsuccessful"); } } catch (Exception e) { throw; } }

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  • How to get .NET HttpContext for isolated code block

    - by Matt Thrower
    Hi, In .NET is it possible to get the HttpContext of the current page from within an external class? So, for example in my page test1.aspx codebehind I've got: Dim blah As New FeedWriter() blah.Run() But inside FeedWriter.vb, can I get the HttpContext of test1.aspx? Or would I have to pass it in to Run()? (I'm unwilling to do the latter because FeedWriter implements an interface which will need to be re-written if it's to take arguments) Cheers, Matt

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  • Confused with the Isolated Storage with Multiple Assemblies Access

    - by Peter Lee
    I googled and searched a lot, but I got no luck. I have a WindowsFormsApplication.exe and ConsoleApplication.exe. I want both of them to access to the same IsolatedStorage, is it possible? I tried using this in ConsoleApplication.exe: IsolatedStorageFile isoStore = IsolatedStorageFile.GetMachineStoreForApplication(); but I got: IsolatedStorageException: Unable to determine application identity of the caller. How can I fix this? Or can I use this way? P.S.: This is NOT a ClickOnce app.

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  • Regular Expression to Match Specific "Values" in Isolated Group

    - by Gandarez
    If have this regular expression to test (\&TRUNC)[\(]{1,}(.+)[\)]{1,} And I have this "tester" ((((&TRUNC((1800,000 / 510)) * 510) * 920) + (2 * (510 * 700)) + ((&TRUNC((1800,000 / 510)) - 1) * 2 * 510 * 80)) / 1000000) * 85,715 My expected value is (inside the personal command "&TRUNC(command)") (1800,000 / 510) I got this value 1800,000 / 510)) * 510) * 920) + (2 * (510 * 700)) + ((&TRUNC((1800,000 / 510)) - 1) * 2 * 510 * 80)) / 1000000 How can I get only expected value in a separated group? PS:. The expressions inside the command called for me as "&TRUNC(command)" is variable.

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  • Unit-testing a directive with isolated scope and bidirectional value

    - by unludo
    I want to unit test a directive which looks like this: angular.module('myApp', []) .directive('myTest', function () { return { restrict: 'E', scope: { message: '='}, replace: true, template: '<div ng-if="message"><p>{{message}}</p></div>', link: function (scope, element, attrs) { } }; }); Here is my failing test: describe('myTest directive:', function () { var scope, compile, validHTML; validHTML = '<my-test message="message"></my-test>'; beforeEach(module('myApp')); beforeEach(inject(function($compile, $rootScope){ scope = $rootScope.$new(); compile = $compile; })); function create() { var elem, compiledElem; elem = angular.element(validHTML); compiledElem = compile(elem)(scope); scope.$digest(); return compiledElem; } it('should have a scope on root element', function () { scope.message = 'not empty'; var el = create(); console.log(el.text()); expect(el.text()).toBeDefined(); expect(el.text()).not.toBe(''); }); }); Can you spot why it's failing? The corresponding jsFiddle Thanks :)

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  • Establish SSH Connection Between Two Isolated Machines Using a 3rd System

    - by FurryHead
    I'd like to do the following with Python: Computer 1 starts SSH server (probably using twisted or paramiko) Computer 1 connects to Server 1 (idle connection) Computer 2 connects to Server 1 Server 1 forwards Computer 2's connection to Computer 1 (connection no longer idle) Computer 1 forwards Server 1's connection to listening SSH port (on computer 1) Result being Computer 2 now has a SSH session with Computer 1, almost as if Computer 2 had started a normal SSH session (but with Server 1's IP instead of Computer 1's) I need this because I can't port forward on Computer 1's network (the router doesn't support it).

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