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  • Can I autoregister my clients/servers in local DNS?

    - by Christian Wattengård
    Right now I have a W2k12 server at home that I run as a domain controller. This has the extra benefit of registering every "subordinate" computers name in it's DNS so that I don't have to go around remembering IP's all the time. (And it let's me easily run dhcp also on my servers). I need to rework my home network for several odd reasons, and in this new scenario there is no place for a big honking W2k12 server box. I have a RasPI, and I have other smallish linux boxen I can use. (In a worst case scenario I'll use my NUC, but then I'll be forced to use my home cinema's UPnP-client for media... The HORROR!!) Is it possible to set up a DNS-server-"appliance" that somehow autoregisters it's own hostname.. Scenario: Router (N66u) on 172.20.20.1. Runs DHCP on 172.20.20.100-200 range. Server [verdant] of a *nix flavor on 172.20.20.2 Laptop [speedy] of W8 flavor on DHCP assigned Laptop [canary] of W8 flavor on DHCP assigned Desktop [lianyu] of Ubunto flavor on DHCP assigned What I would like is that all of the above servers (except possibly the router) would be available on verdant.starling.lan and canary.starling.lan and so on. This is how it works right now (except the Ubuntu box... I haven't cracked that one yet) because Windows just does this for you.. I would also be able to do this without any manual labor on the server. When I tell my box it's name is smoak it should "immediately" be available as smoak.starling.lan without any extra configuration on my part. How can I do this in a Linux (Ubuntu) environment? (Bonus comment upvote for naming the naming scheme :P )

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  • publickey authentication only works with existing ssh session

    - by aaron
    publickey authentication only works for me if I've already got one ssh session open. I am trying to log into a host running Ubuntu 10.10 desktop with publickey authentication, and it fails when I first log in: [me@my-laptop:~]$ ssh -vv host ... debug1: Next authentication method: publickey debug1: Offering public key: /Users/me/.ssh/id_rsa ... debug2: we did not send a packet, disable method debug1: Next authentication method: password me@hosts's password: And the /var/log/auth.log output: Jan 16 09:57:11 host sshd[1957]: reverse mapping checking getaddrinfo for cpe-70-114-155-20.austin.res.rr.com [70.114.155.20] failed - POSSIBLE BREAK-IN ATTEMPT! Jan 16 09:57:13 host sshd[1957]: pam_sm_authenticate: Called Jan 16 09:57:13 host sshd[1957]: pam_sm_authenticate: username = [astacy] Jan 16 09:57:13 host sshd[1959]: Passphrase file wrapped Jan 16 09:57:15 host sshd[1959]: Error attempting to add filename encryption key to user session keyring; rc = [1] Jan 16 09:57:15 host sshd[1957]: Accepted password for astacy from 70.114.155.20 port 42481 ssh2 Jan 16 09:57:15 host sshd[1957]: pam_unix(sshd:session): session opened for user astacy by (uid=0) Jan 16 09:57:20 host sudo: astacy : TTY=pts/0 ; PWD=/home/astacy ; USER=root ; COMMAND=/usr/bin/tail -f /var/log/auth.log The strange thing is that once I've got this first login session, I run the exact same ssh command, and publickey authentication works: [me@my-laptop:~]$ ssh -vv host ... debug1: Server accepts key: pkalg ssh-rsa blen 277 ... [me@host:~]$ And the /var/log/auth.log output is: Jan 16 09:59:11 host sshd[2061]: reverse mapping checking getaddrinfo for cpe-70-114-155-20.austin.res.rr.com [70.114.155.20] failed - POSSIBLE BREAK-IN ATTEMPT! Jan 16 09:59:11 host sshd[2061]: Accepted publickey for astacy from 70.114.155.20 port 39982 ssh2 Jan 16 09:59:11 host sshd[2061]: pam_unix(sshd:session): session opened for user astacy by (uid=0) What do I need to do to make publickey authentication work on the first login? NOTE: When I installed Ubuntu 10.10, I checked the 'encrypt home folder' option. I'm wondering if this has something to do with the log message "Error attempting to add filename encryption key to user session keyring"

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  • Nmap XML parsing with Powershell

    - by Craig620
    I am trying to parse the XML output from NMAP and isolate just the hostadddress and the vendor from the osmatch. I've actually done that with the following: select-xml -path nmap.xml -xpath "nmaprun/host/address/@addr|nmaprun/host/os/osmatch/osclass/@vendor" | select -expandproperty node Which produces: #text ----- 10.20.30.1 HP 10.20.30.2 Linux 10.20.30.3 HP What I was not expecting is that it would jam it all into a single column.Silly me would like the address in one column, and the vendor in another column. I Would like: #addr #vendor ----- ------- 10.20.30.1 HP 10.20.30.2 Linux 10.20.30.3 HP In the several hours I spent learning xpath today, I also realized that this file has a single address for each host, but multiple OS guesses for each host. I would also like to use only the first osGuess in the output. Tired using: -xpath "(nmaprun/host/os/osmatch/osclass/@vendor)[1]" But that truncates the whole data set to a single line of output, instead of only limiting the only the first osclass element of each host. Changing the parens to surround only the @vendor element like .../(@vendor)[1] and .../(@vendor[1]) but both fail with "Expression must evaluate to a node-set." Thanks in advance

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  • Can I autoregister my servers hostname in my local DNS? [on hold]

    - by Christian Wattengård
    We have evaluated a W2k12 server as a domain controller at work. This has the extra benefit of registering every "subordinate" computers name in it's DNS so that I don't have to go around remembering IP's all the time. (And it let's me easily run dhcp also on my "pop-up" dev-servers). We need to rework our work network for several odd reasons, and in this new scenario there was no money for an extra Windows 2012 license. We have at our disposal several old boxes that run linux quite well. Is it possible to set up a DNS-server-"appliance" that somehow autoregisters it's own hostname.. Scenario: Router (N66u) on 172.20.20.1. Runs DHCP on 172.20.20.100-200 range. Server [verdant] of a *nix flavor on 172.20.20.2 Laptop [speedy] of W8 flavor on DHCP assigned Laptop [canary] of W8 flavor on DHCP assigned Desktop [lianyu] of Ubuntu flavor on DHCP assigned What I would like is that all of the above servers (except possibly the router) would be available on verdant.starling.lan and canary.starling.lan and so on. This is how it works right now (except the Ubuntu box... I haven't cracked that one yet) because Windows just does this for you.. I would also be able to do this without any manual labor on the server. When I tell my box it's name is smoak it should "immediately" be available as smoak.starling.lan without any extra configuration on my part. How can I do this in a Linux (Ubuntu) environment?

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  • Separated virtual networks with same subnet range with 2 interface

    - by Coolpet
    I'm having some problems with routing with the following: I have a server with 2 interfaces. It has 1-1 alias contains the same subnet. the 2 interface is connected to 2 switch, which are separated from each other. Infrastructure: Eth0 192.168.16.2/20 Eth0:eth0 192.168.1.222/20 Eth1 192.168.32.3/20 Eth1:eth1 192.168.1.223/20 I have a PC which has the IP address: 192.168.1.3/24 The problem is the next: If PC is on subnet 1, I can ping it. If PC is on subnet 2, I can't ping it. traceroute shows the route is across 192.168.1.222 ping -I 192.168.1.223 192.168.1.3 is not working on subnet 2. arp entries show the MAC address belonging to the correct interface (eth1 on subnet 2) How can I force the server to look on both interface same ranged subnet for specific IP? It searches only in the first subnet. The routing table has these 2 entries: 192.168.0.0/20 dev eth0 proto kernel scope link src 192.168.1.222 192.168.0.0/20 dev eth1 proto kernel scope link src 192.168.1.223

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  • What cause high CPU usage on the server during file upload

    - by bosiang
    When I try to upload a huge file size (approx 2GB), the server cpu usage goes really high. What should I do to fix this? I just use standard html form and php, for file upload. I'm sorry if I post on the wrong forum. Please point me to the right direction here is the result of "top" command during uploading 4 files (18mb, 38mb, 60mb, 33mb) 1904 apache 20 0 33504 5740 1952 R 28.3 0.2 0:02.19 httpd 1905 apache 20 0 33504 5740 1952 R 28.3 0.2 0:01.99 httpd 1903 apache 20 0 33232 6968 3060 R 28.0 0.2 0:01.98 httpd 1910 apache 20 0 33240 6020 2248 S 11.5 0.2 0:02.85 httpd 2133 root 20 0 2656 1124 896 R 1.6 0.0 0:00.71 top 1 root 20 0 2864 1404 1188 S 0.0 0.0 0:03.99 init the code for chunking, although eventhough I don't use this code (just simple file upload), it still cause that high cpu usage function sendRequest() { //clean the screen //bars.innerHTML = ''; var file = document.getElementById('fileToUpload'); for(var i = 0; i < file.files.length; i++) { var blob = file.files[i]; var originalFileName = blob.name; var filePart = 0 const BYTES_PER_CHUNK = 100 * 1024 * 1024; // 10MB chunk sizes. var realFileSize = blob.size; var start = 0; var end = BYTES_PER_CHUNK; totalChunks = Math.ceil(realFileSize / BYTES_PER_CHUNK); alert(realFileSize); while( start < realFileSize ) { if (blob.webkitSlice) { //for Google Chrome var chunk = blob.webkitSlice(start, end); } else if (blob.mozSlice) { //for Mozilla Firefox var chunk = blob.mozSlice(start, end); } uploadFile(chunk, originalFileName, filePart, totalChunks, i); filePart++; start = end; end = start + BYTES_PER_CHUNK; } } }

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  • Project Euler #15

    - by Aistina
    Hey everyone, Last night I was trying to solve challenge #15 from Project Euler: Starting in the top left corner of a 2×2 grid, there are 6 routes (without backtracking) to the bottom right corner. How many routes are there through a 20×20 grid? I figured this shouldn't be so hard, so I wrote a basic recursive function: const int gridSize = 20; // call with progress(0, 0) static int progress(int x, int y) { int i = 0; if (x < gridSize) i += progress(x + 1, y); if (y < gridSize) i += progress(x, y + 1); if (x == gridSize && y == gridSize) return 1; return i; } I verified that it worked for a smaller grids such as 2×2 or 3×3, and then set it to run for a 20×20 grid. Imagine my surprise when, 5 hours later, the program was still happily crunching the numbers, and only about 80% done (based on examining its current position/route in the grid). Clearly I'm going about this the wrong way. How would you solve this problem? I'm thinking it should be solved using an equation rather than a method like mine, but that's unfortunately not a strong side of mine. Update: I now have a working version. Basically it caches results obtained before when a n×m block still remains to be traversed. Here is the code along with some comments: // the size of our grid static int gridSize = 20; // the amount of paths available for a "NxM" block, e.g. "2x2" => 4 static Dictionary<string, long> pathsByBlock = new Dictionary<string, long>(); // calculate the surface of the block to the finish line static long calcsurface(long x, long y) { return (gridSize - x) * (gridSize - y); } // call using progress (0, 0) static long progress(long x, long y) { // first calculate the surface of the block remaining long surface = calcsurface(x, y); long i = 0; // zero surface means only 1 path remains // (we either go only right, or only down) if (surface == 0) return 1; // create a textual representation of the remaining // block, for use in the dictionary string block = (gridSize - x) + "x" + (gridSize - y); // if a same block has not been processed before if (!pathsByBlock.ContainsKey(block)) { // calculate it in the right direction if (x < gridSize) i += progress(x + 1, y); // and in the down direction if (y < gridSize) i += progress(x, y + 1); // and cache the result! pathsByBlock[block] = i; } // self-explanatory :) return pathsByBlock[block]; } Calling it 20 times, for grids with size 1×1 through 20×20 produces the following output: There are 2 paths in a 1 sized grid 0,0110006 seconds There are 6 paths in a 2 sized grid 0,0030002 seconds There are 20 paths in a 3 sized grid 0 seconds There are 70 paths in a 4 sized grid 0 seconds There are 252 paths in a 5 sized grid 0 seconds There are 924 paths in a 6 sized grid 0 seconds There are 3432 paths in a 7 sized grid 0 seconds There are 12870 paths in a 8 sized grid 0,001 seconds There are 48620 paths in a 9 sized grid 0,0010001 seconds There are 184756 paths in a 10 sized grid 0,001 seconds There are 705432 paths in a 11 sized grid 0 seconds There are 2704156 paths in a 12 sized grid 0 seconds There are 10400600 paths in a 13 sized grid 0,001 seconds There are 40116600 paths in a 14 sized grid 0 seconds There are 155117520 paths in a 15 sized grid 0 seconds There are 601080390 paths in a 16 sized grid 0,0010001 seconds There are 2333606220 paths in a 17 sized grid 0,001 seconds There are 9075135300 paths in a 18 sized grid 0,001 seconds There are 35345263800 paths in a 19 sized grid 0,001 seconds There are 137846528820 paths in a 20 sized grid 0,0010001 seconds 0,0390022 seconds in total I'm accepting danben's answer, because his helped me find this solution the most. But upvotes also to Tim Goodman and Agos :) Bonus update: After reading Eric Lippert's answer, I took another look and rewrote it somewhat. The basic idea is still the same but the caching part has been taken out and put in a separate function, like in Eric's example. The result is some much more elegant looking code. // the size of our grid const int gridSize = 20; // magic. static Func<A1, A2, R> Memoize<A1, A2, R>(this Func<A1, A2, R> f) { // Return a function which is f with caching. var dictionary = new Dictionary<string, R>(); return (A1 a1, A2 a2) => { R r; string key = a1 + "x" + a2; if (!dictionary.TryGetValue(key, out r)) { // not in cache yet r = f(a1, a2); dictionary.Add(key, r); } return r; }; } // calculate the surface of the block to the finish line static long calcsurface(long x, long y) { return (gridSize - x) * (gridSize - y); } // call using progress (0, 0) static Func<long, long, long> progress = ((Func<long, long, long>)((long x, long y) => { // first calculate the surface of the block remaining long surface = calcsurface(x, y); long i = 0; // zero surface means only 1 path remains // (we either go only right, or only down) if (surface == 0) return 1; // calculate it in the right direction if (x < gridSize) i += progress(x + 1, y); // and in the down direction if (y < gridSize) i += progress(x, y + 1); // self-explanatory :) return i; })).Memoize(); By the way, I couldn't think of a better way to use the two arguments as a key for the dictionary. I googled around a bit, and it seems this is a common solution. Oh well.

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  • MySQL: LIMIT then RAND rather than RAND then LIMIT

    - by Larry
    I'm using full text search to pull rows. I order the rows based on score (ORDER BY SCORE) , then of the top 20 rows (LIMIT 20), I want to rand (RAND) the result set. So for any specific search term, I want to randomly show 5 of the top 20 results. My workaround is code based- where I put the top 20 into an array then randomly select 5. Is there sql way to do this?

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  • LIMIT then RAND rather than RAND then LIMIT

    - by Larry
    I'm using full text search to pull rows. I order the rows based on score (ORDER BY SCORE) , then of the top 20 rows (LIMIT 20), I want to rand (RAND) the result set. So for any specific search term, I want to randomly show 5 of the top 20 results. My workaround is code based- where I put the top 20 into an array then randomly select 5. Is there sql way to do this?

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  • Linux buffer cache effect on IO writes?

    - by Patrick LeBoutillier
    I'm copying large files (3 x 30G) between 2 filesystems on a Linux server (kernel 2.6.37, 16 cores, 32G RAM) and I'm getting poor performance. I suspect that the usage of the buffer cache is killing the I/O performance. To try and narrow down the problem I used fio directly on the SAS disk to monitor the performance. Here is the output of 2 fio runs (the first with direct=1, the second one direct=0): Config: [test] rw=write blocksize=32k size=20G filename=/dev/sda # direct=1 Run 1: test: (g=0): rw=write, bs=32K-32K/32K-32K, ioengine=sync, iodepth=1 Starting 1 process Jobs: 1 (f=1): [W] [100.0% done] [0K/205M /s] [0/6K iops] [eta 00m:00s] test: (groupid=0, jobs=1): err= 0: pid=4667 write: io=20,480MB, bw=199MB/s, iops=6,381, runt=102698msec clat (usec): min=104, max=13,388, avg=152.06, stdev=72.43 bw (KB/s) : min=192448, max=213824, per=100.01%, avg=204232.82, stdev=4084.67 cpu : usr=3.37%, sys=16.55%, ctx=655410, majf=0, minf=29 IO depths : 1=100.0%, 2=0.0%, 4=0.0%, 8=0.0%, 16=0.0%, 32=0.0%, >=64=0.0% submit : 0=0.0%, 4=100.0%, 8=0.0%, 16=0.0%, 32=0.0%, 64=0.0%, >=64=0.0% complete : 0=0.0%, 4=100.0%, 8=0.0%, 16=0.0%, 32=0.0%, 64=0.0%, >=64=0.0% issued r/w: total=0/655360, short=0/0 lat (usec): 250=99.50%, 500=0.45%, 750=0.01%, 1000=0.01% lat (msec): 2=0.01%, 4=0.02%, 10=0.01%, 20=0.01% Run status group 0 (all jobs): WRITE: io=20,480MB, aggrb=199MB/s, minb=204MB/s, maxb=204MB/s, mint=102698msec, maxt=102698msec Disk stats (read/write): sda: ios=0/655238, merge=0/0, ticks=0/79552, in_queue=78640, util=76.55% Run 2: test: (g=0): rw=write, bs=32K-32K/32K-32K, ioengine=sync, iodepth=1 Starting 1 process Jobs: 1 (f=1): [W] [100.0% done] [0K/0K /s] [0/0 iops] [eta 00m:00s] test: (groupid=0, jobs=1): err= 0: pid=4733 write: io=20,480MB, bw=91,265KB/s, iops=2,852, runt=229786msec clat (usec): min=16, max=127K, avg=349.53, stdev=4694.98 bw (KB/s) : min=56013, max=1390016, per=101.47%, avg=92607.31, stdev=167453.17 cpu : usr=0.41%, sys=6.93%, ctx=21128, majf=0, minf=33 IO depths : 1=100.0%, 2=0.0%, 4=0.0%, 8=0.0%, 16=0.0%, 32=0.0%, >=64=0.0% submit : 0=0.0%, 4=100.0%, 8=0.0%, 16=0.0%, 32=0.0%, 64=0.0%, >=64=0.0% complete : 0=0.0%, 4=100.0%, 8=0.0%, 16=0.0%, 32=0.0%, 64=0.0%, >=64=0.0% issued r/w: total=0/655360, short=0/0 lat (usec): 20=5.53%, 50=93.89%, 100=0.02%, 250=0.01%, 500=0.01% lat (msec): 2=0.01%, 4=0.01%, 10=0.01%, 20=0.01%, 50=0.12% lat (msec): 100=0.38%, 250=0.04% Run status group 0 (all jobs): WRITE: io=20,480MB, aggrb=91,265KB/s, minb=93,455KB/s, maxb=93,455KB/s, mint=229786msec, maxt=229786msec Disk stats (read/write): sda: ios=8/79811, merge=7/7721388, ticks=9/32418456, in_queue=32471983, util=98.98% I'm not knowledgeable enough with fio to interpret the results, but I don't expect the overall performance using the buffer cache to be 50% less than with O_DIRECT. Can someone help me interpret the fio output? Are there any kernel tunings that could fix/minimize the problem? Thanks a lot,

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  • abstract data type list. . .

    - by aldrin
    A LIST is an ordered collection of items where items may be inserted anywhere in the list. Implement a LIST using an array as follows: struct list { int *items; // pointer to the array int size; // actual size of the array int count; // number of items in the array }; typedef struct list *List; // pointer to the structure Implement the following functions: a) List newList(int size); - will create a new List and return its pointer. Allocate space for the structure, allocate space for the array, then initialize size and count, return the pointer. b) void isEmpty(List list); c) void display(List list); d) int contains(List list, int item); e) void remove(List list, int i) ; f) void insertAfter(List list,int item, int i); g) void addEnd(List list,int item) - add item at the end of the list – simply store the data at position count, then increment count. If the array is full, allocate an array twice as big as the original. count = 5 size = 10 0 1 2 3 4 5 6 7 8 9 5 10 15 20 30 addEnd(list,40) will result to count = 6 size = 10 0 1 2 3 4 5 6 7 8 9 5 10 15 20 30 40 h) void addFront(List list,int item) - shift all elements to the right so that the item can be placed at position 0, then increment count. Bonus: if the array is full, allocate an array twice as big as the original. count = 5 size = 10 0 1 2 3 4 5 6 7 8 9 5 10 15 20 30 addFront(list,40) will result to count = 6 size = 10 0 1 2 3 4 5 6 7 8 9 40 5 10 15 20 30 i) void removeFront(List list) - shift all elements to the left and decrement count; count = 6 size = 10 0 1 2 3 4 5 6 7 8 9 40 5 10 15 20 30 removeFront(list) will result to count = 5 size = 10 0 1 2 3 4 5 6 7 8 9 5 10 15 20 30 j) void remove(List list,int item) - get the index of the item in the list and then shift all elements to the count = 6 size = 10 0 1 2 3 4 5 6 7 8 9 40 5 10 15 20 30 remove(list,10) will result to count = 5 size = 10 0 1 2 3 4 5 6 7 8 9 40 5 15 20 30

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  • Linux buffer cache effect on IO writes?

    - by Patrick LeBoutillier
    Hi, I'm copying large files (3 x 30G) between 2 filesystems on a Linux server (kernel 2.6.37, 16 cores, 32G RAM) and I'm getting poor performance. I suspect that the usage of the buffer cache is killing the I/O performance. To try and narrow down the problem I used fio directly on the SAS disk to monitor the performance. Here is the output of 2 fio runs (the first with direct=1, the second one direct=0): Config: [test] rw=write blocksize=32k size=20G filename=/dev/sda # direct=1 Run 1: test: (g=0): rw=write, bs=32K-32K/32K-32K, ioengine=sync, iodepth=1 Starting 1 process Jobs: 1 (f=1): [W] [100.0% done] [0K/205M /s] [0/6K iops] [eta 00m:00s] test: (groupid=0, jobs=1): err= 0: pid=4667 write: io=20,480MB, bw=199MB/s, iops=6,381, runt=102698msec clat (usec): min=104, max=13,388, avg=152.06, stdev=72.43 bw (KB/s) : min=192448, max=213824, per=100.01%, avg=204232.82, stdev=4084.67 cpu : usr=3.37%, sys=16.55%, ctx=655410, majf=0, minf=29 IO depths : 1=100.0%, 2=0.0%, 4=0.0%, 8=0.0%, 16=0.0%, 32=0.0%, >=64=0.0% submit : 0=0.0%, 4=100.0%, 8=0.0%, 16=0.0%, 32=0.0%, 64=0.0%, >=64=0.0% complete : 0=0.0%, 4=100.0%, 8=0.0%, 16=0.0%, 32=0.0%, 64=0.0%, >=64=0.0% issued r/w: total=0/655360, short=0/0 lat (usec): 250=99.50%, 500=0.45%, 750=0.01%, 1000=0.01% lat (msec): 2=0.01%, 4=0.02%, 10=0.01%, 20=0.01% Run status group 0 (all jobs): WRITE: io=20,480MB, aggrb=199MB/s, minb=204MB/s, maxb=204MB/s, mint=102698msec, maxt=102698msec Disk stats (read/write): sda: ios=0/655238, merge=0/0, ticks=0/79552, in_queue=78640, util=76.55% Run 2: test: (g=0): rw=write, bs=32K-32K/32K-32K, ioengine=sync, iodepth=1 Starting 1 process Jobs: 1 (f=1): [W] [100.0% done] [0K/0K /s] [0/0 iops] [eta 00m:00s] test: (groupid=0, jobs=1): err= 0: pid=4733 write: io=20,480MB, bw=91,265KB/s, iops=2,852, runt=229786msec clat (usec): min=16, max=127K, avg=349.53, stdev=4694.98 bw (KB/s) : min=56013, max=1390016, per=101.47%, avg=92607.31, stdev=167453.17 cpu : usr=0.41%, sys=6.93%, ctx=21128, majf=0, minf=33 IO depths : 1=100.0%, 2=0.0%, 4=0.0%, 8=0.0%, 16=0.0%, 32=0.0%, >=64=0.0% submit : 0=0.0%, 4=100.0%, 8=0.0%, 16=0.0%, 32=0.0%, 64=0.0%, >=64=0.0% complete : 0=0.0%, 4=100.0%, 8=0.0%, 16=0.0%, 32=0.0%, 64=0.0%, >=64=0.0% issued r/w: total=0/655360, short=0/0 lat (usec): 20=5.53%, 50=93.89%, 100=0.02%, 250=0.01%, 500=0.01% lat (msec): 2=0.01%, 4=0.01%, 10=0.01%, 20=0.01%, 50=0.12% lat (msec): 100=0.38%, 250=0.04% Run status group 0 (all jobs): WRITE: io=20,480MB, aggrb=91,265KB/s, minb=93,455KB/s, maxb=93,455KB/s, mint=229786msec, maxt=229786msec Disk stats (read/write): sda: ios=8/79811, merge=7/7721388, ticks=9/32418456, in_queue=32471983, util=98.98% I'm not knowledgeable enough with fio to interpret the results, but I don't expect the overall performance using the buffer cache to be 50% less than with O_DIRECT. Can someone help me interpret the fio output? Are there any kernel tunings that could fix/minimize the problem? Thanks a lot,

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  • How to connect to a Virtualbox guest from the host when network cable unplugged

    - by Greg K
    I'd like to work offline (I'm flying to the US twice this month), to do this I need access to a linux development server. When I work from home I boot a VirtualBox VM and that acts as my dev server for the day (providing Apache, PHP & MySQL to run my server side code). However, I'd like to work with my VM when I'm not connected to a network. I have my Ubuntu VM guest set up with a bridge connection so it can serve HTTP and provide SSH access from inside my local network. I've tried to manually configure my network settings on both Mac OSX (the host) and Ubuntu (the guest) but I can't even ping my own NIC address (127.0.0.1 can, 192.168.21.x I can't) in OS X when I unplug the cable. Manual network settings: $ ifconfig en0 en0: flags=8963<UP,BROADCAST,SMART,RUNNING,PROMISC,SIMPLEX,MULTICAST> mtu 1500 ether 00:xx:xx:xx:xx:xx inet 192.168.21.5 netmask 0xffffff00 broadcast 192.168.21.255 media: autoselect (100baseTX <full-duplex,flow-control>) status: active I can ping localhost fine, as well as my VM (.20) and SSH too. $ ping 192.168.21.5 PING 192.168.21.5 (192.168.21.5): 56 data bytes 64 bytes from 192.168.21.5: icmp_seq=0 ttl=64 time=0.085 ms 64 bytes from 192.168.21.5: icmp_seq=1 ttl=64 time=0.102 ms 64 bytes from 192.168.21.5: icmp_seq=2 ttl=64 time=0.100 ms 64 bytes from 192.168.21.5: icmp_seq=3 ttl=64 time=0.094 ms $ ping 192.168.21.20 PING 192.168.21.20 (192.168.21.20): 56 data bytes 64 bytes from 192.168.21.20: icmp_seq=0 ttl=64 time=0.910 ms 64 bytes from 192.168.21.20: icmp_seq=1 ttl=64 time=1.181 ms 64 bytes from 192.168.21.20: icmp_seq=2 ttl=64 time=1.159 ms 64 bytes from 192.168.21.20: icmp_seq=3 ttl=64 time=1.320 ms Network cable unplugged: $ ifconfig en0 en0: flags=8963<UP,BROADCAST,SMART,RUNNING,PROMISC,SIMPLEX,MULTICAST> mtu 1500 ether 00:xx:xx:xx:xx:xx media: autoselect status: inactive $ ping 192.168.21.5 PING 192.168.21.5 (192.168.21.5): 56 data bytes ping: sendto: No route to host ping: sendto: No route to host Request timeout for icmp_seq 0 ping: sendto: No route to host Request timeout for icmp_seq 1 Does OS X disable the NIC when the network cable is unplugged? Any way to stop it doing this?

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  • invalid effective address calculation!

    - by Zia ur Rahman
    Hay Dear! Please look at the following program, the error is invalid effective address calculation and i have mentioned that line please tell me why its invalid effective address calculation here is the program [org 0x100] jmp start array1: dw 10,15,20,25,30,35,40,45,50,55 array2: dw 15,10,20,35,40,30,55,50,25,45 start: mov bx,0 mov dx,0 loop: mov ax,[array2+bx] cmp ax,[array1+cx]//here is the error invalid effective address calculation jne NextElementOfArray1 NextElementOfArray2: add bx,2 cmp bx,20 je end mov cx,0 jmp loop NextElementOfArray1: add cx,2 cmp cx,20 je NextElementOfArray2 jmp loop end: mov ax,0x4c00 int 0x21

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  • mySQL - How to select a date interval

    - by fabriciols
    Hello, this is my table : ------------------------------------- | user | item | date_time | | 10 | 01 | 10-10-10 20:10:05 | | 10 | 02 | 10-10-10 20:10:10 | | 10 | 03 | 10-10-10 20:10:10 | | 20 | 02 | 10-10-10 20:15:10 | | 20 | 02 | 10-10-10 20:20:10 | | 30 | 10 | 10-10-10 20:01:10 | | 30 | 20 | 10-10-10 20:01:20 | | 30 | 30 | 10-10-10 20:05:20 | ------------------------------------- i want to do a query that return a user that took multiple items in a 1min interval, like this result : ------------------------------------- | user | item | date_time | | 10 | 01 | 10-10-10 20:10:05 | | 10 | 02 | 10-10-10 20:10:10 | | 10 | 03 | 10-10-10 20:10:10 | | 30 | 10 | 10-10-10 20:01:10 | | 30 | 20 | 10-10-10 20:01:20 | ------------------------------------- how i do this ?

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  • How to get the last element by date of each "type" in LINQ or TSQL

    - by Mauro
    Imagine to have a table defined as CREATE TABLE [dbo].[Price]( [ID] [int] NOT NULL, [StartDate] [datetime] NOT NULL, [Price] [int] NOT NULL ) where ID is the identifier of an action having a certain Price. This price can be updated if necessary by adding a new line with the same ID, different Price, and a more recent date. So with a set of a data like ID StartDate Price 1 01/01/2009 10 1 01/01/2010 20 2 01/01/2009 10 2 01/01/2010 20 How to obtain a set like the following? 1 01/01/2010 20 2 01/01/2010 20

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  • Using Qt CSS to set own Q_PROPERTY(QFont)

    - by Kamil Klimek
    Hi there. I'm using Qt 4.6.2 and i have problem with QCSS. I have own Q_PROPERTY(QFont myFont READ myFont SET setMyFont). I want to change it with QCSS but it doesn't work. I've tried using normal font syntax but it doesn't work. I've also tried few other combinations like: qproperty-myFont: font(serif 20 1 0) font(serif 20 bold) QFont(serif 20 1 0) QFont(serif 20 bold) QFont(bold 20px serif) etc.

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  • Abstract Data Type: Any1 can help me this? thanks..

    - by Aga Hibaya
    Objectives: Implement the Abstract Data Type (ADT) List using dynamically allocated arrays and structures. Description A LIST is an ordered collection of items where items may be inserted anywhere in the list. Implement a LIST using an array as follows: struct list { int *items; // pointer to the array int size; // actual size of the array int count; // number of items in the array }; typedef struct list *List; // pointer to the structure Implement the following functions: a) List newList(int size); - will create a new List and return its pointer. Allocate space for the structure, allocate space for the array, then initialize size and count, return the pointer. b) void isEmpty(List list); c) void display(List list); d) int contains(List list, int item); e) void remove(List list, int i) ; f) void insertAfter(List list,int item, int i); g) void addEnd(List list,int item) - add item at the end of the list – simply store the data at position count, then increment count. If the array is full, allocate an array twice as big as the original. count = 5 size = 10 0 1 2 3 4 5 6 7 8 9 5 10 15 20 30 addEnd(list,40) will result to count = 6 size = 10 0 1 2 3 4 5 6 7 8 9 5 10 15 20 30 40 h) void addFront(List list,int item) - shift all elements to the right so that the item can be placed at position 0, then increment count. Bonus: if the array is full, allocate an array twice as big as the original. count = 5 size = 10 0 1 2 3 4 5 6 7 8 9 5 10 15 20 30 addFront(list,40) will result to count = 6 size = 10 0 1 2 3 4 5 6 7 8 9 40 5 10 15 20 30 i) void removeFront(List list) - shift all elements to the left and decrement count; count = 6 size = 10 0 1 2 3 4 5 6 7 8 9 40 5 10 15 20 30 removeFront(list) will result to count = 5 size = 10 0 1 2 3 4 5 6 7 8 9 5 10 15 20 30 j) void remove(List list,int item) - get the index of the item in the list and then shift all elements to the count = 6 size = 10 0 1 2 3 4 5 6 7 8 9 40 5 10 15 20 30 remove(list,10) will result to count = 5 size = 10 0 1 2 3 4 5 6 7 8 9 40 5 15 20 30 Remarks

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  • SSH Socks Proxy wiith iptables REDIRECT

    - by Radium
    I have googled and haven`t found the answer on my question. Help me please. There are two servers: serverA with public IP 12.0.0.10 and an private IP 10.0.0.5 serverB with public IP 20.0.0.11 I have setup SOCKS proxy on serverB to serverA: ssh -D20.0.0.11:2222 [email protected] So when on my local machine in a browser i specify SOCKS proxy 20.0.0.11:2222 (serverB:2222) as external IP while browsing i get 12.0.0.10 (serverA IP). That is ok. As well if i go onto http://10.0.0.5 (serverA private IP) it is also reachable. That is what i need. I want to make servers A private IP to be available through servers B public IP on certain ports but without specifying SOCKS in my browser. I could use ssh port forward but the problem is - i need to forward many ports and do not know which exactly - i know only the range. So when i connect to 20.0.0.11 to any port , for example, from 3000:4000 range, i want that traffic to be redirected to 10.0.0.5 on the same port. That is why i`ve decided maybe SOCKS proxy via SSH and iptables REDIRECT could help me. Client - serverBPublicIP (any port from range 3000:4000) - serverAPublicIP - serverAPrivateIP (the port was requested on serverBPublicIP) On serverB i do: ssh -D20.0.0.11:2222 [email protected] iptables -t nat -A PREROUTING -d 20.0.0.11 -p tcp --dport 3000:4000 -j REDIRECT --to-port 2222 But that does not work - when i telnet on 20.0.0.11:3001 for example i do not see any proxied traffic on the serverA. What should i do else? I have tried tcpsocks like this (in example i am telneting to 20.0.0.11:3001) Client -> 20.0.0.11:3001 -> iptables REDIRECT from 3001 --to-port 1111 -> tcpsocks from 1111 to 2222 -> SOCKS proxy from serverB to serverA on port 2222 -> serverA But i do not know what to do with the traffic on serverA. How to route it to its private IP. Help me please. I know, VPN removes all the hell i am trying to create, but i have no ability to use tun/tap device. It is disabled.

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  • looping through a 2d array in ruby to display it in a table format?

    - by Sean
    Hi How can i represent a 2d array in a table format in the terminal, where it lines up the columns properly just like a table? so it looks like so: 1 2 3 4 5 1 [ Infinity | 40 | 45 | Infinity | Infinity ] 2 [ Infinity | 20 | 50 | 14 | 20 ] 3 [ Infinity | 30 | 40 | Infinity | 40 ] 4 [ Infinity | 28 | Infinity | 6 | 6 ] 5 [ Infinity | 40 | 80 | 12 | 0 ] instead of: [ Infinity,40,45,Infinity,Infinity ] [ Infinity,20,50,14,20 ] [ Infinity,30,40,Infinity,40 ] [ Infinity,28,Infinity,6,6 ] [ Infinity,40,80,12,0 ]

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  • mysql display each day in a month

    - by Jason
    during a month, display the infor each date, order by date, but this infor is empty in some day. how can i still display each day as a row? Product date ----------------- 20 2008-01-01 10 2008-01-02 20 2008-01-03 10 2008-01-05 09 2008-01-08 30 2008-01-09 result: Product date ----------------- 20 2008-01-01 10 2008-01-02 20 2008-01-03 0 2008-01-04 10 2008-01-05 0 2008-01-06 0 2008-01-07 09 2008-01-08 30 2008-01-09

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  • How to parse date from html page using html agility pack ?

    - by Harikrishna
    I have html pages and I am parsing those pages with html agility pack. Now I want to parse some information.In every pages there is trading date(20/02/02) which I want to parse. Like it will be look like a Trading date : 20/02/02. Now Trading date and date(20/02/02) may be in same column(td) or it can be different column like in first column trading date and in second column 20/02/02 then what should I do ?

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  • t-sql get variable value from string with variable name

    - by Markus
    Hi. Is there a way to convert '@my_variable' string into a value of @my_variable? I have a table which stores names of variables. I need to get the value of this variable. Something like this: DECLARE @second_variable AS NVARCHAR(20); DECLARE @first_variable AS NVARCHAR(20); SET @first_variable = '20'; SET @second_variable = SELECT '@first_variable'; --here I want that @second variable be assigned a value of "20".

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  • which images load the android device at the time of installation?

    - by VSC
    I am placing 20 images in hdpi folder, 20 images in mdpi folder and 20 images in ldpi folder, all are same images but different resolutions . I am placing my app in android market. One user install my app in his normal android device through market, My question is total 60 images download the device or only 20 images(mdpi) download the device at the time of install the app.Thanks for reading my question.

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  • AIA - Server Topologien und Deployment

    - by Hans Viehmann
    Es gibt ein neues White Paper, in dem die verschiedenen Varianten von AIA Topologien ausführlich beschrieben sind. Die unterschiedlichen Möglichkeiten ergeben sich vor allem aus dem Betrieb mehrerer AIA Instanzen parallel (z.B. Aufteilung in Dev/Test/Prod, Aufteilung zur Lastverteilung, Zuordnung zu unterschiedlichen Geografien, ...). Weitere Variationsmöglichkeiten bietet die Verteilung der unterschiedlichen Bestandteile des DB-Schemas (AIA Metadaten, Cross-Referencing Daten, AIA Lifecycle Repository, JMS Schema, etc.). Und schließlich können auch verschiedene Topologien der SOA Laufzeitumgebung, wie beispielsweise SOA Cluster, unterstützt werden. Diese Varianten mit ihren typischen Einsatzfeldern werden in dem Dokument ebenso beschrieben, wie deren Umsetzung mit Hilfe des Installers.In einem zweiten Teil befasst sich das White Paper mit dem Deployment der AIA Artefakte in diese Topologien, und zwar insbesondere mit dem Tool-Support durch die "Project Lifecycle Workbench" und dem "AIA Installation Driver". Erstere ist eine neue Komponente in AIA 11gR1, die in der Lage ist, am Ende des Entwicklungszyklus Deployment Pläne zu generieren, während der ebenfalls neue "AIA Installation Driver" dann die Artefakte dann in die Zielumgebung überträgt.Das White Paper ist noch nicht auf oracle.com verfügbar und ist leider so umfangreich, dass es nicht auf auf blogs.oracle.com hochgeladen werden kann. Bei Interesse verschicke ich es aber gern per Mail.

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