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  • SQL SERVER – Find Max Worker Count using DMV – 32 Bit and 64 Bit

    - by pinaldave
    During several recent training courses, I found it very interesting that Worker Thread is not quite known to everyone despite the fact that it is a very important feature. At some point in the discussion, one of the attendees mentioned that we can double the Worker Thread if we double the CPU (add the same number of CPU that we have on current system). The same discussion has triggered this quick article. Here is the DMV which can be used to find out Max Worker Count SELECT max_workers_count FROM sys.dm_os_sys_info Let us run the above query on my system and find the results. As my system is 32 bit and I have two CPU, the Max Worker Count is displayed as 512. To address the previous discussion, adding more CPU does not necessarily double the Worker Count. In fact, the logic behind this simple principle is as follows: For x86 (32-bit) upto 4 logical processors  max worker threads = 256 For x86 (32-bit) more than 4 logical processors  max worker threads = 256 + ((# Procs – 4) * 8) For x64 (64-bit) upto 4 logical processors  max worker threads = 512 For x64 (64-bit) more than 4 logical processors  max worker threads = 512+ ((# Procs – 4) * 8) In addition to this, you can configure the Max Worker Thread by using SSMS. Go to Server Node >> Right Click and Select Property >> Select Process and modify setting under Worker Threads. According to Book On Line, the default Worker Thread settings are appropriate for most of the systems. Reference: Pinal Dave (http://blog.SQLAuthority.com) Filed under: Pinal Dave, SQL, SQL Authority, SQL Query, SQL Scripts, SQL Server, SQL System Table, SQL Tips and Tricks, T SQL, Technology Tagged: SQL DMV

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  • Is there a Windows 7 equivalent to the *NIX ability to create a hard link to /dev/null?

    - by minameismud
    I saw another question here that the Windows equivalent to /dev/null is simply NUL. I also know that you can use the mklink command to make sym links (shortcuts) from the command line: MKLINK [[/D] | [/H] | [/J]] Link Target /D Creates a directory symbolic link. Default is a file symbolic link. /H Creates a hard link instead of a symbolic link. /J Creates a Directory Junction. Link specifies the new symbolic link name. Target specifies the path (relative or absolute) that the new link refers to. When I try to use the /j switch to make a hard link ("junction") instead of a simple shortcut to NUL, I get: C:\>mklink /j "C:\Program Files\MyNewHardlinkFolder" NUL Local volumes are required to complete the operation. I can create shortcuts to NUL all day long using the /d switch, but I would much prefer the hard link. Any ideas?

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  • SQL - Count grouped entries and then get the max values grouped by date

    - by Marcus
    hello, I am out of any logic how to write the right sql statment. I've got a sqlite table holding every played track in a row with played date/time Now I will count the plays of all artists, grouped by day and then find the artist with the max playcount per day. I used this Query SELECT COUNT(ARTISTID) AS artistcount, ARTIST AS artistname,strftime('%Y-%m-%d', playtime) AS day_played FROM playcount GROUP BY artistname to get this result "93"|"The Skygreen Leopards"|"2010-06-16" "2" |"Arcade Fire" |"2010-06-15" "2" |"Dead Kennedys" |"2010-06-15" "2" |"Wolf People" |"2010-06-15" "3" |"16 Horsepower" |"2010-06-15" "3" |"Alela Diane" |"2010-06-15" "46"|"Motorama" |"2010-06-15" "1" |"Ariel Pink's Haunted" |"2010-06-14" I tried then to query this virtual table but I always get false results in artistname. SELECT MAX(artistcount), artistname , day_played FROM ( SELECT COUNT(ARTISTID) AS artistcount, ARTIST AS artistname,strftime('%Y-%m-%d', playtime) AS day_played FROM playcount GROUP BY artistname ) GROUP BY strftime('%Y-%m-%d',day_played) result in this "93"|"lilium" |"2010-06-16" "46"|"Wolf People"|"2010-06-15" "30"|"of Montreal"|"2010-06-14" but the artist name is false. I think through the grouping by day, it just use the last artist, or so. I tested stuff like INNER JOIN or GROUP BY ... HAVING in trial and error, I read examples of similar issues but always get lost in columnnames and stuff (I am a bit burned out) I hope someone can give me a hint. thanks m

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  • If I create a link to a folder, how can I get from that linked folder to the "real" folder from within Nautilus?

    - by snowguy
    Say I have a folder several layers down in my documents folder. And I want easy access to it from my desktop. To do that I: Go to the parent folder in Nautilus. Right click on the folder's Icon and choose Make Link Cut / Paste the new "Link to ..." folder onto my desktop. Great. And mostly this works fine for me. But suppose I want to get to that folder's parent. I can of course get there using the original path--what Nautilus calls the "link path" which I can see in the properties of the folder. But that seems harder than it ought to be. How can I click on the folder and go to the link path directly?

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  • translate stored procedure - to Linq2SQL (count, max, group, orderby)

    - by Walter
    I've two tables (1:N) CREATE TABLE master (idMaster int identity (1,1) not null, TheName varchar( 100) null, constraint pk_master primary key(idMaster) clustered) and - CREATE TABLE lnk (idSlave int not null, idMaster int not null, constraint pk_lnk_master_slave(idSlave) primary key clustered) link between Master.idMaster and lnk.idMaster I've a SQL query: select max (master.idMaster) as idMaster, master.theName, count (lnk.idSlave) as freq from lnk inner join master ON lnk.idMaster = master.idMaster Group by master.theName order by freq desc, master.theName I need to translate this T-SQL query to a Linq-to-SQL statement, preferably in C#

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  • how to create a linq query using join and max

    - by geoff swartz
    I have 2 tables in my linq dbml. One is people with a uniqueid called peopleid and the other is a vertical with a foreign key for peopleid and a uniqueid called id. I need to create a type of linq query that does a left outer join on people and gets the latest record in the vertical table based off the max(id) column. Can anyone suggest what this should look like? Thanks.

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  • Scenario - NTFS Symbolic Link or Junction?

    - by Unsigned
    Differences Absolute Relative File Directory UNC Symbolic link ? ? ? ? ? Junction ? x x ? x Scenario Let's assume we're creating a reparse point to create the redirect C:\SomeDir => D:\SomeDir Since this scenario only requires local, absolute paths, either a junction or symlink would work. In this situation, is there any advantage to using one or the other? Assume Windows 7 for the OS, disregarding backward-compatibility (prior to Vista, symlinks are not supported). Update I have found another difference. Symbolic Link - Link's permissions only affect delete/rename operations on the link itself, read/write access (to the target) is governed by the target's permissions Junction - Junction's permissions affect enumeration, revoking permissions on the junction will deny file listing through that junction, even if the target folder has more permissive ACLs The permissions make it interesting, as symlinks can allow legacy applications to access configuration files in UAC-restricted areas (such as %ProgramFiles%) without changing existing access permissions, by storing the files in a non-restricted location and creating symlinks in the restricted directory.

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  • Using max-width = 100% and max-height = 100% on an image, calculate the display width/height

    - by NatalieMac
    I am creating a slideshow for images of various sizes to display centered vertically and horizontally within a canvas area. In my CSS, I set the width and height of the image to 100% so that each image would proportionally fill the canvas. I want the canvas to auto-size itself to fit within the viewer's screensize as the original size of the images is quite large (up to 800 pixels tall). I am using jQuery 1.4, and using the height of the image to calculate the top value for absolute positioning it to the middle of the canvas. I have tried using jQuery to get the .height(), innerHeight(), and outerHeight(), but it always gets the full size of the image. I extracted the DOM element from the jQuery object and tried using .width, .offsetWidth, and .clientWidth, but that too always seems to return the full size of the image. Firebug displays the correct dimensions, so I know there's some way of calculating the actual display height of the image, I just can't figure out what it is. How do you get the actual display height of an image if you've set max-height = 100%? I didn't want to have to calculate and set the height of each image in the js, but if I have to, I will. It just seems like I should be able to set the canvas size and have the images auto-adjust.

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  • MySQL SELECT MAX multiple tables : foreach parent return eldest son's picture

    - by Guillermo
    **Table parent** parentId | name **Table children** childId | parentId | pictureId | age **Table childrenPictures** pictureId | imgUrl no i would like to return all parent names with their eldest son's picture (only return parents that have children, and only consider children that have pictures) so i thought of something like : SELECT c.childId AS childId, p.name AS parentName, cp.imgUrl AS imgUrl, MAX(c.age) AS age FROM parent AS p RIGHT JOIN children AS c ON (p.parentId = c.parentId) RIGHT JOIN childrenPictures AS cp ON (c.pictureId = cp.pictureId)) GROUP BY p.name This query will return each parent's eldest son's age, but the childId will not correspond to the eldest sons id, so the output does not show the right sons picture. Well if anyone has a hint i'd appreciate very much Thank you very much, G

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  • max count with joins

    - by trixet
    I have 3 tables: users: Id Login 1 John 2 Bill 3 Jim computers: Id Name 1 Computer1 2 Computer2 3 Computer3 4 Computer4 5 Computer5 sessions: UserId ComputerId Minutes 1 2 47 2 1 32 1 4 15 2 5 5 1 2 7 1 1 40 2 5 31 I would like to display this resulting table: Login Total_sess Total_min Most_freq_computer Sess_on_most_freq Min_on_most_freq John 4 109 Computer2 2 54 Bill 3 68 Computer5 2 36 Jim - - - - - Myself I can only cover first 3 columns with: SELECT Login, COUNT(sessions.UserId), SUM(Minutes) FROM users LEFT JOIN sessions ON users.Id = sessions.UserId GROUP BY users.Id And some kind of other columns with: SELECT main.* FROM (SELECT UserId, ComputerId, COUNT(*) AS cnt ,SUM(Minutes) FROM sessions GROUP BY UserId, ComputerId) AS main INNER JOIN ( SELECT ComputerId, MAX(cnt) AS maxCnt FROM ( SELECT ComputerId, UserId, COUNT(*) AS cnt FROM sessions GROUP BY ComputerId, UserId ) AS Counts GROUP BY ComputerId) AS maxes ON main.ComputerId = maxes.ComputerId AND main.cnt = maxes.maxCnt But I need to get whole resulting table in one query. I feel I'm doing something completely wrong. Need help.

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  • Select *, max(date) works in phpMyAdmin but not in my code

    - by kdobrev
    OK, my statement executes well in phpMyAdmin, but not how I expect it in my php page. This is my statement: SELECT egid , group_name , limit , MAX( date ) FROM employee_groups GROUP BY egid ORDER BY egid DESC ; This is may table: CREATE TABLE employee_groups ( egid int(10) unsigned NOT NULL, date date NOT NULL, group_name varchar(50) NOT NULL, limit smallint(5) unsigned NOT NULL, PRIMARY KEY (egid,date) ) ENGINE=MyISAM DEFAULT CHARSET=cp1251; I want to extract the most recent list of groups, e.g. if a group has been changed I want to have only the last change. And I need it as a list (all groups).

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  • brew link jpeg issues

    - by y2p
    I am trying to install opencv on Mac OSX Lion. brew install opencv I get the following error (and a few other similar ones) Error: The linking step did not complete successfully The formula built, but is not symlinked into /usr/local You can try again using `brew link jpeg' When I do brew link jpeg Linking /usr/local/Cellar/jpeg/8d... ln: wrjpgcom: File exists I do not understand what this means? What should I be doing? Thanks

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  • Toshiba laptop won't connect to D-link router

    - by user3314725
    This is Team 4786 Nicolet Fear FIRST FRC robotics, our problem is we cant get our D-Link (DAP-1522) to connect to our Toshiba (TECRA R950) laptop wireless. It has connected before in the past and we don't know why it is not working anymore. The D-links still function correctly and we think our problem lies within the Toshiba. EDIT:The Toshiba connects to the school Wifi, and other things (small FRC drive station) connect to the D-link, but they won't connect to each other.

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  • Dereference symbolic link in OS X?

    - by ithinkihaveacat
    In OS X, how can I dereference a symbolic link to a canonical file name? (i.e. one starting with /.) That is, I'm after the equivalent of GNU readlink's -f option: kapow:~$ greadlink -f .bash_profile /Users/mjs/.config/home/.bash_profile OS X's readlink instead returns a relative link: kapow:~$ readlink .bash_profile .config/home/.bash_profile stat(1) does have an amazing number of options, but I couldn't figure out the right combination to do what I want.

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  • How to get max row from table

    - by Odette
    HI I have the following code and a massive problem: WITH CALC1 AS ( SELECT OTQUOT, OTIT01 AS ITEMS, ROUND(OQCQ01 * OVRC01,2) AS COST FROM @[email protected] WHERE OTIT01 <> '' UNION ALL SELECT OTQUOT, OTIT02 AS ITEMS, ROUND(OQCQ02 * OVRC02,2) AS COST FROM @[email protected] WHERE OTIT02 <> '' UNION ALL SELECT OTQUOT, OTIT03 AS ITEMS, ROUND(OQCQ03 * OVRC03,2) AS COST FROM @[email protected] WHERE OTIT03 <> '' UNION ALL SELECT OTQUOT, OTIT04 AS ITEMS, ROUND(OQCQ04 * OVRC04,2) AS COST FROM @[email protected] WHERE OTIT04 <> '' UNION ALL SELECT OTQUOT, OTIT05 AS ITEMS, ROUND(OQCQ05 * OVRC05,2) AS COST FROM @[email protected] WHERE OTIT05 <> '' ORDER BY OTQUOT ASC ) SELECT OTQUOT, ITEMS, MAX(COST) FROM CALC1 WHERE OTQUOT = '04886471' GROUP BY OTQUOT, ITEMS result: 04886471 FEPO5050WCGA24 13.21 04886471 GFRK1650SGL 36.21 04886471 FRA7500GA 12.6 04886471 CGIFESHAZ 11.02 04886471 CGIFESHPDPR 11.79 04886471 GFRK1350DBL 68.23 04886471 RET1.63825GP 32.55 04886471 FRSA 0.12 04886471 GFRK1350SGL 55.94 04886471 GFRK1650DBL 71.89 04886471 FEPO6565WCGA24 16.6 04886471 PCAP5050GA 0.28 04886471 FEPO6565NCPAG24 0.000000 How can I get the result of the row with the Itemcode that has the highest value? In this case I need the result: 04886471 GFRK1650DBL 71.89 but i dont know how to change my code to get that - can anybody please help me?

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  • Iam unable to convert magnet link into a torrent link.on a company owned laptop

    - by najam
    Firefox magnet link issue or if right click on the magnet link and copy this is the result of paste (http://proxychina.net/browse.php?u=Oi8vdGhlcGlyYXRlYmF5LnN4L3RvcnJlbnQvODY0MjU1MC9tYWduZXQ6P3h0PXVybjpidGloOmQzNzMyMmQxMGFmNDY4Yzg4NTk3ZWZiNWU4NWQ5ZmRhMzU1MzllYTQmZG49RmFzdCtBbmQrRnVyaW91cys2JTVCMjAxMyU1RFdFQlJpcCtYdmlELUVUUkcmdHI9dWRwJTNBJTJGJTJGdHJhY2tlci5vcGVuYml0dG9ycmVudC5jb20lM0E4MCZ0cj11ZHAlM0ElMkYlMkZ0cmFja2VyLnB1YmxpY2J0LmNvbSUzQTgwJnRyPXVkcCUzQSUyRiUyRnRyYWNrZXIuaXN0b2xlLml0JTNBNjk2OSZ0cj11ZHAlM0ElMkYlMkZ0cmFja2VyLmNjYy5kZSUzQTgwJnRyPXVkcCUzQSUyRiUyRm9wZW4uZGVtb25paS5jb20lM0ExMzM3&b=31) Strange but two different links Please a prompt reply is appreciated..many thanks in advance

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  • Wireless Bridge with NetGear and TP-Link

    - by Tiago Cruz
    I have a wireless NetGear WGR614 v7 (little old) router connected to the internet, but I can't get a good signal in the other end of my house. I have another new one, model TP-Link TL-WR941ND wireless router. I was able to do the stuff works using a wired cable, but now, I would like to do the same using wireless connections (bridge mode, some like WDS?) Now, the computer connected to TP LINK was able to ping my computer connected to NETGEAR, but we cannot go IP ADDRESS outside my network, only internals ones. What can I do to configure this? Is needed that BOTH wireless routers support BRIDGE mode or only one its good enough? Thanks a lot!!

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  • Ubuntu make symbolic link between new folder in Home to existing folder

    - by Fath
    Hello, To the point. I have Ubuntu Maverick running on my Lenovo G450. Before, it was Windows 7. All my data are inside another partition, its NTFS. FSTAB line to mount that partition : /dev/sda5 /data ntfs auto,users,uid=1000,gid=1000,utf8,dmask=027,fmask=137 0 0 Inside /data there are folder Musics, Graphics, Tools, Cores, etc. If I'm about to create new folder, let see, GFX on /home/apronouva/GFX and make it link or pointing to /data/Graphics, how do I do that ? So when I open /home/apronouva/GFX the content will be the same as inside /data/Graphics .. and whatever changes I made inside GFX, it will also affect /data/Graphics I tried : $ ln -s /data/Graphics /home/apronouva/GFX it resulted : error, cannot make symbolic link between folder Thanks in advance, Fath

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  • FTP Error 550 when trying to access a folder via symbolic link

    - by OrangeTux
    I'm configuring svftp on a linux machine. At the moment local users can login via ftp and they will see listened their home dir. They have write acces to it. No I want the users to write in de /var/www/ dir. Therefore I created an new group apache. Added users to the group and gave the group write access to /var/www. Via the terminal all users can write .var/www/. I created a link in the home directory to /var/www via ln -s /var/www/ /home/user/www ls gives: drwxr-xr-x 2 orangetux orangetux 4096 Jun 23 15:06 ftp lrwxrwxrwx 1 orangetux orangetux 21 Jun 23 15:00 www -> /var/www/ But when I use FTP I see the link but I cannot follow it. Error 550 which means file not found or bad access. How can I solve this, so that the users have access to /var/www via their home dir?

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  • Reasons for sticking with TEXT, NTEXT and IMAGE instead of (N)VARCHAR(max) and VARBINARY(max)

    - by John Assymptoth
    TEXT, NTEXT and IMAGE have been deprecated a long time ago and will, eventually, be removed from SQL Server. However, they are not going to be discontinued right away, not even in the next version of SQL Server, so it's not convenient for my enterprise to transform thousands of columns right away, even if it is using SQL Server 2012. What arguments can I use to postpone this migration? I know there are some advantages in using the new types. But I'm strictly looking for reasons not to migrate my data that is already functioning pretty well in the old types.

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  • Using recursion to to trim a binary tree based on a given min and max value

    - by Justin
    As the title says, I have to trim a binary tree based on a given min and max value. Each node stores a value, and a left/right node. I may define private helper methods to solve this problem, but otherwise I may not call any other methods of the class nor create any data structures such as arrays, lists, etc. An example would look like this: overallRoot _____[50]____________________ / \ __________[38] _______________[90] / \ / _[14] [42] [54]_____ / \ \ [8] [20] [72] \ / \ [26] [61] [83] trim(52, 65); should return: overallRoot [54] \ [61] My attempted solution has three methods: public void trim(int min, int max) { rootFinder(overallRoot, min, max); } First recursive method finds the new root perfectly. private void rootFinder(IntTreeNode node, int min, int max) { if (node == null) return; if (overallRoot.data < min) { node = overallRoot = node.right; rootFinder(node, min, max); } else if (overallRoot.data > max) { node = overallRoot = node.left; rootFinder(node, min, max); } else cutter(overallRoot, min, max); } This second method should eliminate any further nodes not within the min/max, but it doesn't work as I would hope. private void cutter(IntTreeNode node, int min, int max) { if (node == null) return; if (node.data <= min) { node.left = null; } if (node.data >= max) { node.right = null; } if (node.data < min) { node = node.right; } if (node.data > max) { node = node.left; } cutter(node.left, min, max); cutter(node.right, min, max); } This returns: overallRoot [54]_____ \ [72] / [61] Any help is appreciated. Feel free to ask for further explanation as needed.

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  • Max Value of a Multiway Tree in OCaml

    - by Trigork
    I'm an IT Student and kind of a newbie to OCaml Recently, studying for an exam I found this exercise. Given: type 'a tree = Tree of 'a * 'a tree list Define a function mtree : 'a tree -'a, that returns the greatest value of all the nodes in a multiway tree, following the usual order relation in OCaml (<=) I've come to do something like this below, which, of course, is not working. let rec mtree (Tree (r, sub)) = let max_node (Tree(a, l1)) (Tree(b, l2)) = if a >= b then (Tree(a, l1)) else (Tree(b, l2)) in List.fold_left max_node r sub;; After reading an answer to this I'm posting the fixed code. let rec mtree (Tree(r,sub)) = let max_node (Tree(a, l1)) (Tree(b, l2)) = if a >= b then a else b in List.fold_left (max_node) r (List.map mtree sub);; The idea is the same, fold the list comparing the nodes making use of my local function to do so and travel through the tree by calling the function itself over the nodes lists of the consecutive levels. Is still not working, though. Now complains about max_node in the fold_left part. Error: This expression has type 'a tree -> 'a tree -> 'a but an expression was expected of type 'a tree -> 'a tree -> 'a tree And here I'm definitely lost because I can't see why does it expects to return an 'a tree

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