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  • how does sql count work?

    - by ria
    I would like to understand how exactly does sql count work. Is it a whole table scan that happens or is it some property of the table that is read. However I feel a table scan would be an overhead in case of huge tables with lots of records.

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  • oracle using index even though there is no filter criteeria specified

    - by Kaushik
    In this query: SELECT WTTEMPLATE.TEMPLATEuID, MAX (WTTRX.VALUEDATE) AS template_last_use_date FROM wttemplate, wttrx WHERE WTTEMPLATE.TEMPLATEID = WTTRX.TEMPLATEID(+) AND WTTEMPLATE.CUSTID = WTTRX.CUSTID GROUP BY WTTEMPLATE.TEMPLATEuID The explain plan shows:index fast full scan using indexes on WTTEMPLATE.TEMPLATEID and (WTTRX.TEMPLATEID,WTTRX.CUSTID). My question is this: I have not specified any filter criteria , so how can it use indexes? It should do full scan...right?

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  • best scala idiom for find & return

    - by IttayD
    This is something I encounter frequently, but I don't know the elegant way of doing. I have a collection of Foo objects. Foo has a method bar() that may return null or a Bar object. I want to scan the collection, calling each object's bar() method and stop on the first one returning an actual reference and return that reference from the scan. Obviously: foos.find(_.bar != null).bar does the trick, but calls #bar twice.

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  • Ubuntu server has slow performance

    - by Rich
    I have a custom built Ubuntu 11.04 server with a 6 disk software RAID 10 primary drive. On it I'm primarily running a PostgreSQL and a few other utilities that stream data from the web. I often find after a few hours of uptime the server starts to lag with all kinds of processes. For example, it may take 10-15 seconds after log-in to get a shell prompt. It might take 5-10 seconds for top to come up. An ls might take a second or two. When I look at top there is almost no CPU usage. There's a fair amount of memory used by the PostgreSQL server but not enough to bleed into swap. I have no idea where to go from here, other than to suspect the RAID10 (I've only ever had software RAID 1's before). Edit: Output from top: top - 11:56:03 up 1:46, 3 users, load average: 0.89, 0.73, 0.72 Tasks: 119 total, 1 running, 118 sleeping, 0 stopped, 0 zombie Cpu(s): 0.2%us, 0.0%sy, 0.0%ni, 93.5%id, 6.2%wa, 0.0%hi, 0.0%si, 0.0%st Mem: 16325596k total, 3478248k used, 12847348k free, 20880k buffers Swap: 19534176k total, 0k used, 19534176k free, 3041992k cached PID USER PR NI VIRT RES SHR S %CPU %MEM TIME+ COMMAND 1747 woodsp 20 0 109m 10m 4888 S 1 0.1 0:42.70 python 357 root 20 0 0 0 0 S 0 0.0 0:00.40 jbd2/sda3-8 1 root 20 0 24324 2284 1344 S 0 0.0 0:00.84 init 2 root 20 0 0 0 0 S 0 0.0 0:00.00 kthreadd 3 root 20 0 0 0 0 S 0 0.0 0:00.24 ksoftirqd/0 6 root RT 0 0 0 0 S 0 0.0 0:00.00 migration/0 7 root RT 0 0 0 0 S 0 0.0 0:00.01 watchdog/0 8 root RT 0 0 0 0 S 0 0.0 0:00.00 migration/1 10 root 20 0 0 0 0 S 0 0.0 0:00.02 ksoftirqd/1 12 root RT 0 0 0 0 S 0 0.0 0:00.01 watchdog/1 13 root RT 0 0 0 0 S 0 0.0 0:00.00 migration/2 14 root 20 0 0 0 0 S 0 0.0 0:00.00 kworker/2:0 15 root 20 0 0 0 0 S 0 0.0 0:00.00 ksoftirqd/2 16 root RT 0 0 0 0 S 0 0.0 0:00.01 watchdog/2 17 root RT 0 0 0 0 S 0 0.0 0:00.00 migration/3 18 root 20 0 0 0 0 S 0 0.0 0:00.00 kworker/3:0 19 root 20 0 0 0 0 S 0 0.0 0:00.02 ksoftirqd/3 20 root RT 0 0 0 0 S 0 0.0 0:00.01 watchdog/3 21 root 0 -20 0 0 0 S 0 0.0 0:00.00 cpuset 22 root 0 -20 0 0 0 S 0 0.0 0:00.00 khelper 23 root 20 0 0 0 0 S 0 0.0 0:00.00 kdevtmpfs 24 root 0 -20 0 0 0 S 0 0.0 0:00.00 netns 26 root 20 0 0 0 0 S 0 0.0 0:00.00 sync_supers df -h rpsharp@ncp-skookum:~$ df -h Filesystem Size Used Avail Use% Mounted on /dev/sda3 1.8T 549G 1.2T 32% / udev 7.8G 4.0K 7.8G 1% /dev tmpfs 3.2G 492K 3.2G 1% /run none 5.0M 0 5.0M 0% /run/lock none 7.8G 0 7.8G 0% /run/shm /dev/sda2 952M 128K 952M 1% /boot/efi /dev/md0 5.5T 562G 4.7T 11% /usr/local free -m psharp@ncp-skookum:~$ free -m total used free shared buffers cached Mem: 15942 3409 12533 0 20 2983 -/+ buffers/cache: 405 15537 Swap: 19076 0 19076 tail -50 /var/log/syslog Jul 3 06:31:32 ncp-skookum rsyslogd: [origin software="rsyslogd" swVersion="5.8.6" x-pid="1070" x-info="http://www.rsyslog.com"] rsyslogd was HUPed Jul 3 06:39:01 ncp-skookum CRON[14211]: (root) CMD ( [ -x /usr/lib/php5/maxlifetime ] && [ -d /var/lib/php5 ] && find /var/lib/php5/ -depth -mindepth 1 -maxdepth 1 -type f -cmin +$(/usr/lib/php5/maxlifetime) ! -execdir fuser -s {} 2>/dev/null \; -delete) Jul 3 06:40:01 ncp-skookum CRON[14223]: (smmsp) CMD (test -x /etc/init.d/sendmail && /usr/share/sendmail/sendmail cron-msp) Jul 3 07:00:01 ncp-skookum CRON[14328]: (woodsp) CMD (/home/woodsp/bin/mail_tweetupdate # email an update) Jul 3 07:00:01 ncp-skookum CRON[14327]: (smmsp) CMD (test -x /etc/init.d/sendmail && /usr/share/sendmail/sendmail cron-msp) Jul 3 07:00:28 ncp-skookum sendmail[14356]: q63E0SoZ014356: from=woodsp, size=2328, class=0, nrcpts=2, msgid=<[email protected]>, relay=woodsp@localhost Jul 3 07:00:29 ncp-skookum sm-mta[14357]: q63E0Si6014357: from=<[email protected]>, size=2569, class=0, nrcpts=2, msgid=<[email protected]>, proto=ESMTP, daemon=MTA-v4, relay=localhost [127.0.0.1] Jul 3 07:00:29 ncp-skookum sendmail[14356]: q63E0SoZ014356: to=Spencer Wood <[email protected]>,Martin Lacayo <[email protected]>, ctladdr=woodsp (1004/1005), delay=00:00:01, xdelay=00:00:01, mailer=relay, pri=62328, relay=[127.0.0.1] [127.0.0.1], dsn=2.0.0, stat=Sent (q63E0Si6014357 Message accepted for delivery) Jul 3 07:00:29 ncp-skookum sm-mta[14359]: STARTTLS=client, relay=mx3.stanford.edu., version=TLSv1/SSLv3, verify=FAIL, cipher=DHE-RSA-AES256-SHA, bits=256/256 Jul 3 07:00:29 ncp-skookum sm-mta[14359]: q63E0Si6014357: to=<[email protected]>,<[email protected]>, ctladdr=<[email protected]> (1004/1005), delay=00:00:01, xdelay=00:00:00, mailer=esmtp, pri=152569, relay=mx3.stanford.edu. [171.67.219.73], dsn=2.0.0, stat=Sent (Ok: queued as 8F3505802AC) Jul 3 07:09:08 ncp-skookum CRON[14396]: (root) CMD ( [ -x /usr/lib/php5/maxlifetime ] && [ -d /var/lib/php5 ] && find /var/lib/php5/ -depth -mindepth 1 -maxdepth 1 -type f -cmin +$(/usr/lib/php5/maxlifetime) ! -execdir fuser -s {} 2>/dev/null \; -delete) Jul 3 07:17:01 ncp-skookum CRON[14438]: (root) CMD ( cd / && run-parts --report /etc/cron.hourly) Jul 3 07:20:01 ncp-skookum CRON[14453]: (smmsp) CMD (test -x /etc/init.d/sendmail && /usr/share/sendmail/sendmail cron-msp) Jul 3 07:39:01 ncp-skookum CRON[14551]: (root) CMD ( [ -x /usr/lib/php5/maxlifetime ] && [ -d /var/lib/php5 ] && find /var/lib/php5/ -depth -mindepth 1 -maxdepth 1 -type f -cmin +$(/usr/lib/php5/maxlifetime) ! -execdir fuser -s {} 2>/dev/null \; -delete) Jul 3 07:40:01 ncp-skookum CRON[14562]: (smmsp) CMD (test -x /etc/init.d/sendmail && /usr/share/sendmail/sendmail cron-msp) Jul 3 08:00:01 ncp-skookum CRON[14668]: (smmsp) CMD (test -x /etc/init.d/sendmail && /usr/share/sendmail/sendmail cron-msp) Jul 3 08:09:01 ncp-skookum CRON[14724]: (root) CMD ( [ -x /usr/lib/php5/maxlifetime ] && [ -d /var/lib/php5 ] && find /var/lib/php5/ -depth -mindepth 1 -maxdepth 1 -type f -cmin +$(/usr/lib/php5/maxlifetime) ! -execdir fuser -s {} 2>/dev/null \; -delete) Jul 3 08:17:01 ncp-skookum CRON[14766]: (root) CMD ( cd / && run-parts --report /etc/cron.hourly) Jul 3 08:20:01 ncp-skookum CRON[14781]: (smmsp) CMD (test -x /etc/init.d/sendmail && /usr/share/sendmail/sendmail cron-msp) Jul 3 08:39:01 ncp-skookum CRON[14881]: (root) CMD ( [ -x /usr/lib/php5/maxlifetime ] && [ -d /var/lib/php5 ] && find /var/lib/php5/ -depth -mindepth 1 -maxdepth 1 -type f -cmin +$(/usr/lib/php5/maxlifetime) ! -execdir fuser -s {} 2>/dev/null \; -delete) Jul 3 08:40:01 ncp-skookum CRON[14892]: (smmsp) CMD (test -x /etc/init.d/sendmail && /usr/share/sendmail/sendmail cron-msp) Output of hdparm -t /dev/sd{a,b,c,d,e,f} This looks suspicious? /dev/sda: Timing buffered disk reads: 2 MB in 4.84 seconds = 423.39 kB/sec /dev/sdb: Timing buffered disk reads: 420 MB in 3.01 seconds = 139.74 MB/sec /dev/sdc: Timing buffered disk reads: 390 MB in 3.00 seconds = 129.87 MB/sec /dev/sdd: Timing buffered disk reads: 416 MB in 3.00 seconds = 138.51 MB/sec /dev/sde: Timing buffered disk reads: 422 MB in 3.00 seconds = 140.50 MB/sec /dev/sdf: Timing buffered disk reads: 416 MB in 3.01 seconds = 138.26 MB/sec

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  • Why am I unable to telnet to a local port that has a listening service?

    - by Skip Huffman
    I suspect this is either a very simple question, or a very complex one. I have a headless server running ubuntu 10.04 that I can ssh into. I have full root access to the system. I am trying to set up an ssh tunnel to allow me to vnc to the system (but that isn't my question. I have vnc running on port 5903, here is the netstat output for that: Proto Recv-Q Send-Q Local Address Foreign Address State PID/Program name tcp 0 0 0.0.0.0:5903 0.0.0.0:* LISTEN 7173/Xtightvnc tcp 0 0 0.0.0.0:22 0.0.0.0:* LISTEN 465/sshd But when I try to telnet to that port, from within the same system and login, I get unable to connect errors # telnet localhost 5903 Trying ::1... Trying 127.0.0.1... telnet: Unable to connect to remote host: Connection timed out I am able to telnet to port 22 (as a verification) ~# telnet localhost 22 Trying ::1... Connected to localhost. Escape character is '^]'. SSH-2.0-OpenSSH_5.3p1 Debian-3ubuntu7 I have tried to open up any possible ports using ufw (probably clumsy fashion) # ufw status numbered Status: active To Action From -- ------ ---- [ 1] 5903 ALLOW IN Anywhere [ 2] 22 ALLOW IN Anywhere What else might be blocking this connection locally? Thank you, Edit: The only reference to port 5903 in iptable -L -n is this: Chain ufw-user-input (1 references) target prot opt source destination ACCEPT tcp -- 0.0.0.0/0 0.0.0.0/0 tcp dpt:5903 ACCEPT udp -- 0.0.0.0/0 0.0.0.0/0 udp dpt:5903 ACCEPT tcp -- 0.0.0.0/0 0.0.0.0/0 tcp dpt:22 ACCEPT udp -- 0.0.0.0/0 0.0.0.0/0 udp dpt:22 ACCEPT tcp -- 0.0.0.0/0 0.0.0.0/0 tcp dpt:8080 ACCEPT udp -- 0.0.0.0/0 0.0.0.0/0 udp dpt:8080 I can post the whole output if that will be useful. hosts.allow and hosts.deny both contain only comments. Re-Edit: Some other questions pointed me to nmap, so I ran a portscan through that utility: # nmap -v -sT localhost -p1-65535 Starting Nmap 5.00 ( http://nmap.org ) at 2011-11-09 09:58 PST NSE: Loaded 0 scripts for scanning. Warning: Hostname localhost resolves to 2 IPs. Using 127.0.0.1. Initiating Connect Scan at 09:58 Scanning localhost (127.0.0.1) [65535 ports] Discovered open port 22/tcp on 127.0.0.1 Connect Scan Timing: About 18.56% done; ETC: 10:01 (0:02:16 remaining) Connect Scan Timing: About 44.35% done; ETC: 10:00 (0:01:17 remaining) Completed Connect Scan at 10:00, 112.36s elapsed (65535 total ports) Host localhost (127.0.0.1) is up (0.00s latency). Interesting ports on localhost (127.0.0.1): Not shown: 65533 filtered ports PORT STATE SERVICE 22/tcp open ssh 80/tcp closed http Read data files from: /usr/share/nmap Nmap done: 1 IP address (1 host up) scanned in 112.43 seconds Raw packets sent: 0 (0B) | Rcvd: 0 (0B) I think this shows that 5903 is blocked somehow. Which I pretty much knew. The question remains what is blocking it and how to modify. Re-re-edit: To check Paul Lathrop's suggested answer, I first verified my ip address with ifconfig: eth0 Link encap:Ethernet HWaddr 02:16:3e:42:28:8f inet addr:10.0.10.3 Bcast:10.0.10.255 Mask:255.255.255.0 Then tried to telnet to 5903 from that address: # telnet 10.0.10.3 5903 Trying 10.0.10.3... telnet: Unable to connect to remote host: Connection timed out No luck. Re-re-re-re-edit: Ok, I think I have isolated it a bit to vncserver, not the firewall, darn it. I shut off vncserver and had netcat listen on port 5903. My vnc client then was able to establish a connnection and sit and wait for a response. Looks like I should be chasing a vnc problem. At least that is progress Thanks for the help

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  • When is a SQL function not a function?

    - by Rob Farley
    Should SQL Server even have functions? (Oh yeah – this is a T-SQL Tuesday post, hosted this month by Brad Schulz) Functions serve an important part of programming, in almost any language. A function is a piece of code that is designed to return something, as opposed to a piece of code which isn’t designed to return anything (which is known as a procedure). SQL Server is no different. You can call stored procedures, even from within other stored procedures, and you can call functions and use these in other queries. Stored procedures might query something, and therefore ‘return data’, but a function in SQL is considered to have the type of the thing returned, and can be used accordingly in queries. Consider the internal GETDATE() function. SELECT GETDATE(), SomeDatetimeColumn FROM dbo.SomeTable; There’s no logical difference between the field that is being returned by the function and the field that’s being returned by the table column. Both are the datetime field – if you didn’t have inside knowledge, you wouldn’t necessarily be able to tell which was which. And so as developers, we find ourselves wanting to create functions that return all kinds of things – functions which look up values based on codes, functions which do string manipulation, and so on. But it’s rubbish. Ok, it’s not all rubbish, but it mostly is. And this isn’t even considering the SARGability impact. It’s far more significant than that. (When I say the SARGability aspect, I mean “because you’re unlikely to have an index on the result of some function that’s applied to a column, so try to invert the function and query the column in an unchanged manner”) I’m going to consider the three main types of user-defined functions in SQL Server: Scalar Inline Table-Valued Multi-statement Table-Valued I could also look at user-defined CLR functions, including aggregate functions, but not today. I figure that most people don’t tend to get around to doing CLR functions, and I’m going to focus on the T-SQL-based user-defined functions. Most people split these types of function up into two types. So do I. Except that most people pick them based on ‘scalar or table-valued’. I’d rather go with ‘inline or not’. If it’s not inline, it’s rubbish. It really is. Let’s start by considering the two kinds of table-valued function, and compare them. These functions are going to return the sales for a particular salesperson in a particular year, from the AdventureWorks database. CREATE FUNCTION dbo.FetchSales_inline(@salespersonid int, @orderyear int) RETURNS TABLE AS  RETURN (     SELECT e.LoginID as EmployeeLogin, o.OrderDate, o.SalesOrderID     FROM Sales.SalesOrderHeader AS o     LEFT JOIN HumanResources.Employee AS e     ON e.EmployeeID = o.SalesPersonID     WHERE o.SalesPersonID = @salespersonid     AND o.OrderDate >= DATEADD(year,@orderyear-2000,'20000101')     AND o.OrderDate < DATEADD(year,@orderyear-2000+1,'20000101') ) ; GO CREATE FUNCTION dbo.FetchSales_multi(@salespersonid int, @orderyear int) RETURNS @results TABLE (     EmployeeLogin nvarchar(512),     OrderDate datetime,     SalesOrderID int     ) AS BEGIN     INSERT @results (EmployeeLogin, OrderDate, SalesOrderID)     SELECT e.LoginID, o.OrderDate, o.SalesOrderID     FROM Sales.SalesOrderHeader AS o     LEFT JOIN HumanResources.Employee AS e     ON e.EmployeeID = o.SalesPersonID     WHERE o.SalesPersonID = @salespersonid     AND o.OrderDate >= DATEADD(year,@orderyear-2000,'20000101')     AND o.OrderDate < DATEADD(year,@orderyear-2000+1,'20000101')     ;     RETURN END ; GO You’ll notice that I’m being nice and responsible with the use of the DATEADD function, so that I have SARGability on the OrderDate filter. Regular readers will be hoping I’ll show what’s going on in the execution plans here. Here I’ve run two SELECT * queries with the “Show Actual Execution Plan” option turned on. Notice that the ‘Query cost’ of the multi-statement version is just 2% of the ‘Batch cost’. But also notice there’s trickery going on. And it’s nothing to do with that extra index that I have on the OrderDate column. Trickery. Look at it – clearly, the first plan is showing us what’s going on inside the function, but the second one isn’t. The second one is blindly running the function, and then scanning the results. There’s a Sequence operator which is calling the TVF operator, and then calling a Table Scan to get the results of that function for the SELECT operator. But surely it still has to do all the work that the first one is doing... To see what’s actually going on, let’s look at the Estimated plan. Now, we see the same plans (almost) that we saw in the Actuals, but we have an extra one – the one that was used for the TVF. Here’s where we see the inner workings of it. You’ll probably recognise the right-hand side of the TVF’s plan as looking very similar to the first plan – but it’s now being called by a stack of other operators, including an INSERT statement to be able to populate the table variable that the multi-statement TVF requires. And the cost of the TVF is 57% of the batch! But it gets worse. Let’s consider what happens if we don’t need all the columns. We’ll leave out the EmployeeLogin column. Here, we see that the inline function call has been simplified down. It doesn’t need the Employee table. The join is redundant and has been eliminated from the plan, making it even cheaper. But the multi-statement plan runs the whole thing as before, only removing the extra column when the Table Scan is performed. A multi-statement function is a lot more powerful than an inline one. An inline function can only be the result of a single sub-query. It’s essentially the same as a parameterised view, because views demonstrate this same behaviour of extracting the definition of the view and using it in the outer query. A multi-statement function is clearly more powerful because it can contain far more complex logic. But a multi-statement function isn’t really a function at all. It’s a stored procedure. It’s wrapped up like a function, but behaves like a stored procedure. It would be completely unreasonable to expect that a stored procedure could be simplified down to recognise that not all the columns might be needed, but yet this is part of the pain associated with this procedural function situation. The biggest clue that a multi-statement function is more like a stored procedure than a function is the “BEGIN” and “END” statements that surround the code. If you try to create a multi-statement function without these statements, you’ll get an error – they are very much required. When I used to present on this kind of thing, I even used to call it “The Dangers of BEGIN and END”, and yes, I’ve written about this type of thing before in a similarly-named post over at my old blog. Now how about scalar functions... Suppose we wanted a scalar function to return the count of these. CREATE FUNCTION dbo.FetchSales_scalar(@salespersonid int, @orderyear int) RETURNS int AS BEGIN     RETURN (         SELECT COUNT(*)         FROM Sales.SalesOrderHeader AS o         LEFT JOIN HumanResources.Employee AS e         ON e.EmployeeID = o.SalesPersonID         WHERE o.SalesPersonID = @salespersonid         AND o.OrderDate >= DATEADD(year,@orderyear-2000,'20000101')         AND o.OrderDate < DATEADD(year,@orderyear-2000+1,'20000101')     ); END ; GO Notice the evil words? They’re required. Try to remove them, you just get an error. That’s right – any scalar function is procedural, despite the fact that you wrap up a sub-query inside that RETURN statement. It’s as ugly as anything. Hopefully this will change in future versions. Let’s have a look at how this is reflected in an execution plan. Here’s a query, its Actual plan, and its Estimated plan: SELECT e.LoginID, y.year, dbo.FetchSales_scalar(p.SalesPersonID, y.year) AS NumSales FROM (VALUES (2001),(2002),(2003),(2004)) AS y (year) CROSS JOIN Sales.SalesPerson AS p LEFT JOIN HumanResources.Employee AS e ON e.EmployeeID = p.SalesPersonID; We see here that the cost of the scalar function is about twice that of the outer query. Nicely, the query optimizer has worked out that it doesn’t need the Employee table, but that’s a bit of a red herring here. There’s actually something way more significant going on. If I look at the properties of that UDF operator, it tells me that the Estimated Subtree Cost is 0.337999. If I just run the query SELECT dbo.FetchSales_scalar(281,2003); we see that the UDF cost is still unchanged. You see, this 0.0337999 is the cost of running the scalar function ONCE. But when we ran that query with the CROSS JOIN in it, we returned quite a few rows. 68 in fact. Could’ve been a lot more, if we’d had more salespeople or more years. And so we come to the biggest problem. This procedure (I don’t want to call it a function) is getting called 68 times – each one between twice as expensive as the outer query. And because it’s calling it in a separate context, there is even more overhead that I haven’t considered here. The cheek of it, to say that the Compute Scalar operator here costs 0%! I know a number of IT projects that could’ve used that kind of costing method, but that’s another story that I’m not going to go into here. Let’s look at a better way. Suppose our scalar function had been implemented as an inline one. Then it could have been expanded out like a sub-query. It could’ve run something like this: SELECT e.LoginID, y.year, (SELECT COUNT(*)     FROM Sales.SalesOrderHeader AS o     LEFT JOIN HumanResources.Employee AS e     ON e.EmployeeID = o.SalesPersonID     WHERE o.SalesPersonID = p.SalesPersonID     AND o.OrderDate >= DATEADD(year,y.year-2000,'20000101')     AND o.OrderDate < DATEADD(year,y.year-2000+1,'20000101')     ) AS NumSales FROM (VALUES (2001),(2002),(2003),(2004)) AS y (year) CROSS JOIN Sales.SalesPerson AS p LEFT JOIN HumanResources.Employee AS e ON e.EmployeeID = p.SalesPersonID; Don’t worry too much about the Scan of the SalesOrderHeader underneath a Nested Loop. If you remember from plenty of other posts on the matter, execution plans don’t push the data through. That Scan only runs once. The Index Spool sucks the data out of it and populates a structure that is used to feed the Stream Aggregate. The Index Spool operator gets called 68 times, but the Scan only once (the Number of Executions property demonstrates this). Here, the Query Optimizer has a full picture of what’s being asked, and can make the appropriate decision about how it accesses the data. It can simplify it down properly. To get this kind of behaviour from a function, we need it to be inline. But without inline scalar functions, we need to make our function be table-valued. Luckily, that’s ok. CREATE FUNCTION dbo.FetchSales_inline2(@salespersonid int, @orderyear int) RETURNS table AS RETURN (SELECT COUNT(*) as NumSales     FROM Sales.SalesOrderHeader AS o     LEFT JOIN HumanResources.Employee AS e     ON e.EmployeeID = o.SalesPersonID     WHERE o.SalesPersonID = @salespersonid     AND o.OrderDate >= DATEADD(year,@orderyear-2000,'20000101')     AND o.OrderDate < DATEADD(year,@orderyear-2000+1,'20000101') ); GO But we can’t use this as a scalar. Instead, we need to use it with the APPLY operator. SELECT e.LoginID, y.year, n.NumSales FROM (VALUES (2001),(2002),(2003),(2004)) AS y (year) CROSS JOIN Sales.SalesPerson AS p LEFT JOIN HumanResources.Employee AS e ON e.EmployeeID = p.SalesPersonID OUTER APPLY dbo.FetchSales_inline2(p.SalesPersonID, y.year) AS n; And now, we get the plan that we want for this query. All we’ve done is tell the function that it’s returning a table instead of a single value, and removed the BEGIN and END statements. We’ve had to name the column being returned, but what we’ve gained is an actual inline simplifiable function. And if we wanted it to return multiple columns, it could do that too. I really consider this function to be superior to the scalar function in every way. It does need to be handled differently in the outer query, but in many ways it’s a more elegant method there too. The function calls can be put amongst the FROM clause, where they can then be used in the WHERE or GROUP BY clauses without fear of calling the function multiple times (another horrible side effect of functions). So please. If you see BEGIN and END in a function, remember it’s not really a function, it’s a procedure. And then fix it. @rob_farley

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  • Install server 2012 on HP ML110 G7 with B110i controller, no disks found

    - by Molotch
    I'm trying to install Server 2012 on a HP ML110 G7 with a B110i controller and four non hot-swappable SATA drives. I just can't get the Server 2012 boot disk PE environment to find any disks. I have downloaded the latest SPP (Service Pack for Proliant 2012.10) and flashed the BIOS. I have tried two different HP drivers for B110i and Windows X64, 6.18.0.64 and 6.18.2.64 to no avail. I have tried setting the controller to both AHCI and legacy mode in the BIOS, no difference. HP:s SmartStart disc for G7 servers only support installation of up to Windows Server 2008R2. HP:s installation instructions for Server 2012 Essentials says boot from the windows disk and use the storage drivers found on the SPP (I can't find any storage drivers on the SPP disk).

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  • Need HP recovery partition info

    - by ggambett
    I'm configuring a new HP Pavillion DV4 with a 320 GB disk. I made the recovery DVDs, then did a couple other things (including deleting the recovery partition), and finally decided to restore the system. Unfortunately, the recovery process fails; the three DVDs are read (the recovery program says "Reformatting the Windows partition" and "Copying files required to restore the hard drive") but after it finishes reading the 3rd, and the progress bar reaches 100%, it fails with error 0xe0f00013 - Googling it didn't return anything at all. I'm afraid this may be because I deleted the partitions. So, I'm kindly asking for one of the following, in order of preference, from a HP Pavillion DV4 with a 320 GB hard disk or a similar enough one : 1) A dump of the MBR 2) The type and size of all the partitions in a "new" system so I can try to make a partition table resembling the original one. BTW, I thought the recovery DVDs were supposed to work even if the entire disk was wiped - isn't that the case? Thanks!

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  • Can't boot up computer windows 8 installation

    - by danny ramirez
    I wanted to install Windows 8 with a volume partition and when the Windows 8 was installing it rebooted and it gave me an error: The digital signature for this file couldn't be verified. File :windows \system 32\winload.exe error code 0xc0000428. I have tried bootec commands and they didn't seem to fix it. Also my Windows 7 got deleted and I only have to boot with the Windows 8 error, so I can't do anything not even boot to safe mode. I have tried to install Windows 8 from disk later on and it won't let me because it keeps rebooting and starting the installation again, so I took off the disk before it rebooted and it takes me to that error again. Remember that's my only boot option so I'm stuck in the installation disk.

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  • Trouble upgrading OSX, because HD doesn't use GUID Partition Table Scheme

    - by Erik Vold
    So I have a intel-based macbook with osx 10.5 and I'm trying to upgrade to 10.6, but when I run the upgrade 'install' I quickly get to a page where I am supposed to 'Select the disk where you want to install Mac OS X' and there is only the one hard drive, so it is auto selected, and below that I see a warning message and the only button available is the 'Go Back' button. The warning message says: "Macintosh HD" can't be used because it doesn't use the GUID Partition Table scheme. Use Disk Utility to change the partition scheme. Select the disk, choose the Partition tab, select the Volume Scheme and then click Options. So I followed the above instructions, and I got to the last step, where I'm supposed to click the 'Options' button, the problem is that I cannot click that button, it is disabled.. So what am I supposed to do?

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  • Windows Server 2003 - Handling hundreds of simultaneous downloads

    - by Paul Hinett
    At the moment I have a single server with 4 1TB hard disks, daily I haver over 150 MP3 music files uploaded (around 80mb each). At busy periods there is over 300 people streaming / downloading these mixes all at once, 75% of the activity is on the most recently uploaded stuff which is all on a single hard disk. My read speads on the hard disk are very low due to such high activity of 200+ reads all happening at the same time on a single hard disk (ran some tests with HDTach). What would be a logical solution to solve this, a couple of ideas I had are: Load balance with another server Install faster hard disks (what are best these days? SCSI / SATA) Spread the most accessed files over the 4 drives so it is sharing the load between all 4 disks, instead of all the most accessed (most recent) all on the most recently installed drive. Obviouslly load balance is the most expensive option, but would it dramatically help? Some help on this situation would be great!

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  • WinSat command line closes too fast

    - by Rob Cowell
    I'm trying to do some analysis under Windows 7 as to why I can't get a Windows Experience Index (WEI) rating due to disk issues. To this end, I'm trying to run winsat from the command line with :- winsat disk -seq -read -drive c and winsat disk -ran -write -n 2 but the command window is closing too quickly to be able to read the results. I've tried opening a seperate cmd window to run it in but it still insists on launching its own window to run in, closing straight away. Any idea how I can see the output?

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  • Cannot access Ubuntu 10/04 after reinstalling Windows

    - by meyosef
    I had Windows Vista Home Premium on HP pavilion desktop. I partitioned the disk for Ubuntu and a swap disk partition and then I installed Ubuntu 10.04. When I would start my computer, I would be given a choice to start Windows or Ubuntu. Today I reinstalled Windows, and now the menu has disappeared. I don't see in windows the disk partition that I allocated to Ubuntu, so it means Ubuntu still exist but I can't load it. What can I do ?

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  • MacBook Air 2010 Not letting me use Boot Camp

    - by SeniorShizzle
    Hi, I'm trying to install Windows 7 on my MBA, 2010. I open up Boot Camp and attempt to use it, but after the first screen it comes up with an error saying it can't install unless theirs 10GB of free disk space. Right now my hdd has 21.5 GB free. Disk Utility won't let me partition my drive either, so a manual install is out of the question. I need help! My original thoughts were that my disk needed to be defraged to get 10GB of contiguous space. Is this a possibility? This is kind of pissing me off because I really need to get Windows up on this Mac (as much as it dreads me to do it). Thanks!

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  • many partitions on a single filegroup?¿ does it make sense?

    - by river0
    Hi, I'm designing a datawarehouse solution and I'm a newbie in disk configuration issues, let me explain you. Our storage is spread over 6 storage enlosures having each of them 5 raid-1 disk arrays, and having 2 LUNS defined per each disk array, which makes a total 48 LUNS (this is following Microsoft fast track recommendations for datawarehouse architectures). I would like to partition my data, on other projects I have worked before, we always followed a 1 partition - 1 filegroup rule. On the microsoft fast track recomendations it is advised to create a filegroup and then for that filegroup a data file per each lun... but I pretend to have a week level partitioning... if I apply that rule I think that I'll get too many files and a complex layout. I'm thinking of just creating just one filegroup (with the 48 lun data files), but still create the partitions since I want to keep soem of the benefits of partitions like partition switching... Is this scenario not recommended? What would you suggest?

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  • Device cannot be added on software-raid-1 array on Ubuntu 12.04

    - by George Pligor
    Unfortunately all tutorials I have found online until now on how to setup software-raid-1 are outdated on ubuntu 12.40 My target is to setup it on a system with a secondary disk drive that is already running. Format is not an option! I am trying to follow and adapt from 11.10 to 12.04 the following tutorial: http://www.howtoforge.com/how-to-set-up-software-raid1-on-a-running-lvm-system-incl-grub2-configuration-ubuntu-11.10-p2 On the above tutorial there is a successful command which creates a raid-1 array by setting the first disk drive with the installed system as missing: mdadm --create /dev/md0 --level=1 --raid-disks=2 missing /dev/sdb1 But when the time comes to add the first main drive with the installed system on the raid-array with this command: mdadm --add /dev/md0 /dev/sda1 I receive an error message. The error message says that the device /dev/sda is (which makes sense) busy! Note: hardware raid solution is not available since the system is a laptop with two disk drives! Thank you

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  • How to recover bitlocker encrypted partition that is now 'unallocated'/'free space'?

    - by Atishay Jain
    My hard drive had 5 partitions(including 1(some 4-5GB) bit locker encrypted one). When I used disk mgmt I could view 2 partitions(24.4GB and 8.94GB) in green colour labeled Empty space. So, I wanted to merge them and I used minitool partition wizard for the purpose. I don't know, what that software did, but all I was left with 2 partitions and lots of green free space. I recovered 2 partitions using EaseUS partition master, but the bitlocker encrypted partition cannot be searched by it(and also minitool partition recovery). Now, the disk mgmt shows 2 free space partitions of 28.36GB and 8.94GB respectively. Here is a screenshot http://s14.postimage.org/4tvij041t/Screen_Shot003.jpg Please, tell me a way to recover the bitlocker encrypted partition that is showing as a free space in disk management. P.S. - It contains very important data.

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  • Cannot add VMDK to VM that was cloned with FlexClone

    - by Daniel Lucas
    I have a virtual machine called VM-A with two VMDK's on volume VOL-A. Call these VMDK-1 (system) and VMDK-2 (data). I want to clone VMDK-2 and attach that clone to VM-A as a new disk, but I'm getting an error. Here are my steps: I use the following command to clone : clone start /vol/VOL-A/VMDK-2 /vol/VOL-A/VMDK-3 Run clone status which shows successful and I can see the new file in the volume In vCenter I edit the settings of VM-A and try to add VMDK-3, but get the following error: Failed to add disk scsi0:4, Failed to power on scsi0:4 I've tried adding this cloned disk to other VMs and get the same error. What could be the issue? My specs are below. NetApp FAS 2040 Data ONTAP 8.0.1 vSphere ESXi 4.1 vCenter Server 4.1 Thanks, Daniel

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  • Can't Shut Down computer with windows 8 Release Preview

    - by Bram Winda
    I downloaded Release Preview and installed it. I have a problem when I shut down the computer. After I press windows key + I Power Shut Down, a box pops up titled "Winlogon.exe - No Disk. Then under that is the message "There is no disk in the drive. Please insert a disk into Drive G". After hitting the retry button a couple of different times the screen finally goes dark. When I was going to turn it back on, I noticed the Power Button was still list. The only way to do a complete Shutdown is to hold in the power button directly. How do I fix this problem?

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  • Improving terminal server performance for a specfic app

    - by Matt
    We have a windows 2003 terminal server running 2X application load balancign that is hosting a client's application that is accessed by around 50 users. Each user has there own database. The database is a file based database. The application is developed under Delphi so I think the database may be BDE based. As you can imagine, there is probably quite a lot of disk i/o. Here are some of the perfmon settings. Logged in users (average) 20 - 25 CPU Utilization (average) 80 - 100% Disk Queue Length (average) 1.6 % Disk time (average) 111 Page faults/sec (average) 1400 The application takes on average about a minute to load up. As usual, the budget is tight. Is there basic windows performance tuning tips that people can recommend to improve things before we fork out on more RAM etc. Server is a 2.8GHz Xeon with 3GB of RAM.

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  • Do different operating systems have different read and write speeds?

    - by Ivan
    If I have two different operating systems, such as Windows 8 and Ubuntu, running on the same hardware, will the two operating systems have different read and write speeds? My guess is that there would be minimal difference between operating systems and read and write speeds to the hard disk since the major limited factor is seeking; however, different operating systems may use different file systems in order to attempt to reduce seek time in the hard disk. Likewise, I'm sure that modern operating systems will not actually write directly to the hard disk, and instead will just have it in memory and marked with a dirty bit. Are there any studies that show differences in read and write speeds between OSs? Or would the file system being used by the OS matter more than the OS itself?

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  • ubuntu automount: only mounting drives as root?

    - by glisignoli
    I'm sharing the /mount dir with smb so users on my network can access use drives added to my linux box. Users are able to read files but not write, modify or delete files or directories. I'm using ubuntu 10.04 server edition with halevt installed for usb auto mounting. Afaik halevt is automounting the drives to /media/ but the drives are showing up as: drwxrwxr-x 1 root root 20480 2010-12-29 20:40 disk drwxrwxr-x 1 root root 24576 2010-12-21 17:20 Sparta mount gives me: /dev/sda1 on /boot type ext2 (rw) /dev/sdb1 on /media/disk type fuseblk (rw,nosuid,nodev,sync,allow_other,blksize=4096,default_permissions) /dev/sdc1 on /media/Sparta type fuseblk (rw,nosuid,nodev,sync,allow_other,blksize=4096,default_permissions) When I umount the drives, the folders /media/disk and /media/Sparta are both removed. I tried changing the permissions with chown to nobody:nogroup but it doesn't work (which I assume is because they are ntfs drives).

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  • Complete refresh of the OS - windows 7.

    - by Vael Victus
    So, the girlfriend's got a virus, and it's pretty nasty. I don't have the time to mess around with it, and she's not so bad with computers, but this is out of her league. (considering her idea of fixing the problem was to download a virus scan that was malware itself) I think I remember hearing that it's possible to put in some restore/recovery disk that comes with the computer to bring Windows back to the way it was originally right from the factory. The problem is, I bought that computer off the shelf - literally - and I don't think I received a disk, or if I had, it's probably gone now. Do you guys have any idea what I could do? I'm all for reformatting, it's just a toy PC for her. I just don't have a disk to re-install windows 7 with.

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  • How can you make a Windows USB HDD Modify All for All Users

    - by David Allan Finch
    Hi, I use a USB HDD a lot between lots of different Windows Boxes. What I find after a while is that there get to be lots of different Permission on the files in some cases stopping me looking at files or removing them. They want Admin rights or even sometimes you need to put the disk back into the original machine with the original user. This is a right pain. Is there away of making the disk have Modify All for All Users and making this the default for all files on the disk. Thanks

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