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  • Hidden Features of Grails

    - by knorv
    Inspired by the question series "Hidden features of ..", I am curious to hear about your favorite Grails tips or lesser known but useful features you know of. Rules: One feature per answer Give an example and short description of the feature, not just a link to documentation Label the feature using bold title as the first line

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  • Slow filesystem access

    - by danneh3826
    I'm trying to diagnose a slow filesystem issue on a server I look after. It's been ongoing for quite some time, and I've run out of ideas as to what I can try. Here's the thick of it. The server itself is a Dell Poweredge T310. It has 4 SAS hard drives in it, configured at RAID5, and is running Citrix XenServer 5.6. The VM is a (relatively) old Debian 5.0.6 installation. It's given 4 cores, and 4Gb's of RAM. It has 3 volumes. A 10Gb volume (ext3) for the system, 980Gb volume (xfs) for data (~94% full), and another 200Gb volume (xfs) for data (~13% full). Now here's the weird thing. Read/write access to the 980Gb volume is really slow. I might get 5Mb/s out of it if I'm lucky. At first I figured it was actually disk access in the system, or at a hypervisor level, but ruled that out entirely as other VMs on the same host are running perfectly fine (a good couple hundred Mb/s disk r/w access). So then I started to target this particular VM. I started thinking it was XFS, but to prove it I wasn't going to attempt to change the filesystem on the 980Gb drive, with years and years of billions of files on there. So I provisioned the 200Gb drive, and did the same read/write test (basically dd), and got a good couple hundred Mb/s r/w access to it. So that ruled out the VM, the hardware, and the filesystem type. There's also a lot of these in /var/log/kern.log; (sorry, this is quite long) Sep 4 10:16:59 uriel kernel: [32571790.564689] httpd: page allocation failure. order:5, mode:0x4020 Sep 4 10:16:59 uriel kernel: [32571790.564693] Pid: 7318, comm: httpd Not tainted 2.6.32-4-686-bigmem #1 Sep 4 10:16:59 uriel kernel: [32571790.564696] Call Trace: Sep 4 10:16:59 uriel kernel: [32571790.564705] [<c1092a4d>] ? __alloc_pages_nodemask+0x476/0x4e0 Sep 4 10:16:59 uriel kernel: [32571790.564711] [<c1092ac3>] ? __get_free_pages+0xc/0x17 Sep 4 10:16:59 uriel kernel: [32571790.564716] [<c10b632e>] ? __kmalloc+0x30/0x128 Sep 4 10:16:59 uriel kernel: [32571790.564722] [<c11dd774>] ? pskb_expand_head+0x4f/0x157 Sep 4 10:16:59 uriel kernel: [32571790.564727] [<c11ddbbf>] ? __pskb_pull_tail+0x41/0x1fb Sep 4 10:16:59 uriel kernel: [32571790.564732] [<c11e4882>] ? dev_queue_xmit+0xe4/0x38e Sep 4 10:16:59 uriel kernel: [32571790.564738] [<c1205902>] ? ip_finish_output+0x0/0x5c Sep 4 10:16:59 uriel kernel: [32571790.564742] [<c12058c7>] ? ip_finish_output2+0x187/0x1c2 Sep 4 10:16:59 uriel kernel: [32571790.564747] [<c1204dc8>] ? ip_local_out+0x15/0x17 Sep 4 10:16:59 uriel kernel: [32571790.564751] [<c12055a9>] ? ip_queue_xmit+0x31e/0x379 Sep 4 10:16:59 uriel kernel: [32571790.564758] [<c1279a90>] ? _spin_lock_bh+0x8/0x1e Sep 4 10:16:59 uriel kernel: [32571790.564767] [<eda15a8d>] ? __nf_ct_refresh_acct+0x66/0xa4 [nf_conntrack] Sep 4 10:16:59 uriel kernel: [32571790.564773] [<c103bf42>] ? _local_bh_enable_ip+0x16/0x6e Sep 4 10:16:59 uriel kernel: [32571790.564779] [<c1214593>] ? tcp_transmit_skb+0x595/0x5cc Sep 4 10:16:59 uriel kernel: [32571790.564785] [<c1005c4f>] ? xen_restore_fl_direct_end+0x0/0x1 Sep 4 10:16:59 uriel kernel: [32571790.564791] [<c12165ea>] ? tcp_write_xmit+0x7a3/0x874 Sep 4 10:16:59 uriel kernel: [32571790.564796] [<c121203a>] ? tcp_ack+0x1611/0x1802 Sep 4 10:16:59 uriel kernel: [32571790.564801] [<c10055ec>] ? xen_force_evtchn_callback+0xc/0x10 Sep 4 10:16:59 uriel kernel: [32571790.564806] [<c121392f>] ? tcp_established_options+0x1d/0x8b Sep 4 10:16:59 uriel kernel: [32571790.564811] [<c1213be4>] ? tcp_current_mss+0x38/0x53 Sep 4 10:16:59 uriel kernel: [32571790.564816] [<c1216701>] ? __tcp_push_pending_frames+0x1e/0x50 Sep 4 10:16:59 uriel kernel: [32571790.564821] [<c1212246>] ? tcp_data_snd_check+0x1b/0xd2 Sep 4 10:16:59 uriel kernel: [32571790.564825] [<c1212de3>] ? tcp_rcv_established+0x5d0/0x626 Sep 4 10:16:59 uriel kernel: [32571790.564831] [<c121902c>] ? tcp_v4_do_rcv+0x15f/0x2cf Sep 4 10:16:59 uriel kernel: [32571790.564835] [<c1219561>] ? tcp_v4_rcv+0x3c5/0x5c0 Sep 4 10:16:59 uriel kernel: [32571790.564841] [<c120197e>] ? ip_local_deliver_finish+0x10c/0x18c Sep 4 10:16:59 uriel kernel: [32571790.564846] [<c12015a4>] ? ip_rcv_finish+0x2c4/0x2d8 Sep 4 10:16:59 uriel kernel: [32571790.564852] [<c11e3b71>] ? netif_receive_skb+0x3bb/0x3d6 Sep 4 10:16:59 uriel kernel: [32571790.564864] [<ed823efc>] ? xennet_poll+0x9b8/0xafc [xen_netfront] Sep 4 10:16:59 uriel kernel: [32571790.564869] [<c11e40ee>] ? net_rx_action+0x96/0x194 Sep 4 10:16:59 uriel kernel: [32571790.564874] [<c103bd4c>] ? __do_softirq+0xaa/0x151 Sep 4 10:16:59 uriel kernel: [32571790.564878] [<c103be24>] ? do_softirq+0x31/0x3c Sep 4 10:16:59 uriel kernel: [32571790.564883] [<c103befa>] ? irq_exit+0x26/0x58 Sep 4 10:16:59 uriel kernel: [32571790.564890] [<c118ff9f>] ? xen_evtchn_do_upcall+0x12c/0x13e Sep 4 10:16:59 uriel kernel: [32571790.564896] [<c1008c3f>] ? xen_do_upcall+0x7/0xc Sep 4 10:16:59 uriel kernel: [32571790.564899] Mem-Info: Sep 4 10:16:59 uriel kernel: [32571790.564902] DMA per-cpu: Sep 4 10:16:59 uriel kernel: [32571790.564905] CPU 0: hi: 0, btch: 1 usd: 0 Sep 4 10:16:59 uriel kernel: [32571790.564908] CPU 1: hi: 0, btch: 1 usd: 0 Sep 4 10:16:59 uriel kernel: [32571790.564911] CPU 2: hi: 0, btch: 1 usd: 0 Sep 4 10:16:59 uriel kernel: [32571790.564914] CPU 3: hi: 0, btch: 1 usd: 0 Sep 4 10:16:59 uriel kernel: [32571790.564916] Normal per-cpu: Sep 4 10:16:59 uriel kernel: [32571790.564919] CPU 0: hi: 186, btch: 31 usd: 175 Sep 4 10:16:59 uriel kernel: [32571790.564922] CPU 1: hi: 186, btch: 31 usd: 165 Sep 4 10:16:59 uriel kernel: [32571790.564925] CPU 2: hi: 186, btch: 31 usd: 30 Sep 4 10:16:59 uriel kernel: [32571790.564928] CPU 3: hi: 186, btch: 31 usd: 140 Sep 4 10:16:59 uriel kernel: [32571790.564931] HighMem per-cpu: Sep 4 10:16:59 uriel kernel: [32571790.564933] CPU 0: hi: 186, btch: 31 usd: 159 Sep 4 10:16:59 uriel kernel: [32571790.564936] CPU 1: hi: 186, btch: 31 usd: 22 Sep 4 10:16:59 uriel kernel: [32571790.564939] CPU 2: hi: 186, btch: 31 usd: 24 Sep 4 10:16:59 uriel kernel: [32571790.564942] CPU 3: hi: 186, btch: 31 usd: 13 Sep 4 10:16:59 uriel kernel: [32571790.564947] active_anon:485974 inactive_anon:121138 isolated_anon:0 Sep 4 10:16:59 uriel kernel: [32571790.564948] active_file:75215 inactive_file:79510 isolated_file:0 Sep 4 10:16:59 uriel kernel: [32571790.564949] unevictable:0 dirty:516 writeback:15 unstable:0 Sep 4 10:16:59 uriel kernel: [32571790.564950] free:230770 slab_reclaimable:36661 slab_unreclaimable:21249 Sep 4 10:16:59 uriel kernel: [32571790.564952] mapped:20016 shmem:29450 pagetables:5600 bounce:0 Sep 4 10:16:59 uriel kernel: [32571790.564958] DMA free:2884kB min:72kB low:88kB high:108kB active_anon:0kB inactive_anon:0kB active_file:5692kB inactive_file:724kB unevictable:0kB isolated(anon):0kB isolated(file):0kB present:15872kB mlocked:0kB dirty:8kB writeback:0kB mapped:0kB shmem:0kB slab_reclaimable:5112kB slab_unreclaimable:156kB kernel_stack:56kB pagetables:0kB unstable:0kB bounce:0kB writeback_tmp:0kB pages_scanned:0 all_unreclaimable? no Sep 4 10:16:59 uriel kernel: [32571790.564964] lowmem_reserve[]: 0 698 4143 4143 Sep 4 10:16:59 uriel kernel: [32571790.564977] Normal free:143468kB min:3344kB low:4180kB high:5016kB active_anon:56kB inactive_anon:2068kB active_file:131812kB inactive_file:131728kB unevictable:0kB isolated(anon):0kB isolated(file):0kB present:715256kB mlocked:0kB dirty:156kB writeback:0kB mapped:308kB shmem:4kB slab_reclaimable:141532kB slab_unreclaimable:84840kB kernel_stack:1928kB pagetables:22400kB unstable:0kB bounce:0kB writeback_tmp:0kB pages_scanned:0 all_unreclaimable? no Sep 4 10:16:59 uriel kernel: [32571790.564983] lowmem_reserve[]: 0 0 27559 27559 Sep 4 10:16:59 uriel kernel: [32571790.564995] HighMem free:776728kB min:512kB low:4636kB high:8760kB active_anon:1943840kB inactive_anon:482484kB active_file:163356kB inactive_file:185588kB unevictable:0kB isolated(anon):0kB isolated(file):0kB present:3527556kB mlocked:0kB dirty:1900kB writeback:60kB mapped:79756kB shmem:117796kB slab_reclaimable:0kB slab_unreclaimable:0kB kernel_stack:0kB pagetables:0kB unstable:0kB bounce:0kB writeback_tmp:0kB pages_scanned:0 all_unreclaimable? no Sep 4 10:16:59 uriel kernel: [32571790.565001] lowmem_reserve[]: 0 0 0 0 Sep 4 10:16:59 uriel kernel: [32571790.565011] DMA: 385*4kB 16*8kB 3*16kB 9*32kB 6*64kB 2*128kB 1*256kB 0*512kB 0*1024kB 0*2048kB 0*4096kB = 2900kB Sep 4 10:16:59 uriel kernel: [32571790.565032] Normal: 21505*4kB 6508*8kB 273*16kB 24*32kB 3*64kB 0*128kB 0*256kB 0*512kB 0*1024kB 0*2048kB 0*4096kB = 143412kB Sep 4 10:16:59 uriel kernel: [32571790.565054] HighMem: 949*4kB 8859*8kB 7063*16kB 6186*32kB 4631*64kB 727*128kB 6*256kB 0*512kB 0*1024kB 0*2048kB 0*4096kB = 776604kB Sep 4 10:16:59 uriel kernel: [32571790.565076] 198980 total pagecache pages Sep 4 10:16:59 uriel kernel: [32571790.565079] 14850 pages in swap cache Sep 4 10:16:59 uriel kernel: [32571790.565082] Swap cache stats: add 2556273, delete 2541423, find 82961339/83153719 Sep 4 10:16:59 uriel kernel: [32571790.565085] Free swap = 250592kB Sep 4 10:16:59 uriel kernel: [32571790.565087] Total swap = 385520kB Sep 4 10:16:59 uriel kernel: [32571790.575454] 1073152 pages RAM Sep 4 10:16:59 uriel kernel: [32571790.575458] 888834 pages HighMem Sep 4 10:16:59 uriel kernel: [32571790.575461] 11344 pages reserved Sep 4 10:16:59 uriel kernel: [32571790.575463] 1090481 pages shared Sep 4 10:16:59 uriel kernel: [32571790.575465] 737188 pages non-shared Now, I've no idea what this means. There's plenty of free memory; total used free shared buffers cached Mem: 4247232 3455904 791328 0 5348 736412 -/+ buffers/cache: 2714144 1533088 Swap: 385520 131004 254516 Though now I see the swap is relatively low in size, but would that matter? I've been starting to think about fragmentation, or inode usage on that large partition, but a recent fsck on it showed is as only like 0.5% fragmented. Which leaves me with inode usage, but how much of an effect really would a large inode table or filesystem TOC have? I've love to hear people's opinions on this. It's driving me potty! df -h output; Filesystem Size Used Avail Use% Mounted on /dev/xvda1 9.5G 6.6G 2.4G 74% / tmpfs 2.1G 0 2.1G 0% /lib/init/rw udev 10M 520K 9.5M 6% /dev tmpfs 2.1G 0 2.1G 0% /dev/shm /dev/xvdb 980G 921G 59G 94% /data

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  • Pruning data for better viewing on loglog graph - Matlab

    - by Geodesic
    Hi Guys, just wondering if anyone has any ideas about an issue I'm having. I have a fair amount of data that needs to be displayed on one graph. Two theoretical lines that are bold and solid are displayed on top, then 10 experimental data sets that converge to these lines are graphed, each using a different identifier (eg the + or o or a square etc). These graphs are on a log scale that goes up to 1e6. The first few decades of the graph (< 1e3) look fine, but as all the datasets converge ( 1e3) it's really difficult to see what data is what. There's over 1000 data points points per decade which I can prune linearly to an extent, but if I do this too much the lower end of the graph will suffer in resolution. What I'd like to do is prune logarithmically, strongest at the high end, working back to 0. My question is: how can I get a logarithmically scaled index vector rather than a linear one? My initial assumption was that as my data is lenear I could just use a linear index to prune, which lead to something like this (but for all decades): //%grab indicies per decade ind12 = find(y >= 1e1 & y <= 1e2); indlow = find(y < 1e2); indhigh = find(y > 1e4); ind23 = find(y >+ 1e2 & y <= 1e3); ind34 = find(y >+ 1e3 & y <= 1e4); //%We want ind12 indexes in this decade, find spacing tot23 = round(length(ind23)/length(ind12)); tot34 = round(length(ind34)/length(ind12)); //%grab ones to keep ind23keep = ind23(1):tot23:ind23(end); ind34keep = ind34(1):tot34:ind34(end); indnew = [indlow' ind23keep ind34keep indhigh']; loglog(x(indnew), y(indnew)); But this causes the prune to behave in a jumpy fashion obviously. Each decade has the number of points that I'd like, but as it's a linear distribution, the points tend to be clumped at the high end of the decade on the log scale. Any ideas on how I can do this?

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  • onscroll for div

    - by The Machine
    Does a div element not have , an onscroll event handler ? The behaviour on my page, doesn't seem to indicate the div onscroll eventHandler is recognized. <div id='bd' onscroll='alert('Scroll Called');'></div>. Also, Do div scroll events roll up to window scroll events, as per DOM event bubbling ?

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  • Linq to Sql get SqlCommand when stored procedure execution fails

    - by Tim Mahy
    Hi all, currently I'm assigning a TextWriter to the Log property of my Linq to Sql data context (per request instancing) and write this to my logging when an exception is thrown while executing a stored procedure (is strongly typed mapped in the context, so not executing a custom command) however when using ADO.NET we normally inspect the SqlCommand upon unhandled exception to read out the parameters and log them is it possible to access the SqlCommand that was used for executing a Stored Procedure in L2S so we can reuse that existing logging component? This would be far nicer than the current Log TextWriter solution.... greetings, Tim

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  • Optimal pixel format for drawing on iPhone?

    - by Felixyz
    Pretty simple question: when doing some pretty intense drawing with CoreGraphics on the iPhone, how can I specify the pixel format to get optimal performance? Is the format that I get from the context via UIGraphicsGetCurrentContext per definition the best one? I know that RGB565 is supposed to be the fastest to use in OpenGL. Does that go for CoreGraphics as well? General advice?

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  • What features would you like to see removed from C++?

    - by Justin Ethier
    This question was inspired by what-features-would-you-like-to-see-added-to-c. anBasically, C++ is a great general-purpose language. But perhaps too general and feature-rich... multiple inheritance, operator overloading, manual memory management, templates, smart pointers, virtual destructors, legacy frameworks (think MFC), and I could just go on. Is there any one feature / aspect of C++ that you would like taken away, to make our lives easier as C++ developers? One feature per answer, please.

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  • printing stl containers with gdb 7.0

    - by Andrei
    I have installed GDB 7.0 and python per the following instructions. In the same manual, there is a mention of this file stl-views-1.0.3.gdb. What confuses me, is where should it be placed in order to enable pretty printing of stl containers. Would someone also explain to me all of this work? Thanks

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  • sequence diagram [UML]

    - by Upul
    Hi All, I have a question regarding sequence diagrams. When drawing sequence diagrams, Is it enough to draw one diagram per user case ? or do we need to draw a set of sequence diagrams to cover each user case instance ? (is a sequence of actions a system performs that yields an observable result of value to a particular Use Case Actor)

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  • Is it possible to filter users from a launchd plist?

    - by compriots
    I have some launchd plists that are now placed at the administrator per-user level, particularly inside: - /Library/LaunchAgents OR - /Library/LaunchDaemons I'd like to avoid that some of those services could start only for one user (possibly avoiding to move all the plists to ~/Library/LaunchAgents), which is the simplest way? Thanks in advance.

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  • Using the Google AJAX Search API for SEO Purposes

    - by User
    I am looking at writing a .net application that uses the Google AJAX Search API to determine where our website falls for a given term compared to a competitor. I can find alot about the old SOAP API however for the new AJAX api I cannot find any information on the following: Is this sort of use allowed as the Terms of use are vague Is there a limit to the number of requests per day As you can only get a max of 8 results at a time, is the only way to get the top 100 results to keep requesting the next set and is this an issue? Thanks

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  • I need help on my C++ assignment using Microsoft Visual C++

    - by krayzwytie
    Ok, so I don't want you to do my homework for me, but I'm a little lost with this final assignment and need all the help I can get. Learning about programming is tough enough, but doing it online is next to impossible for me... Now, to get to the program, I am going to paste what I have so far. This includes mostly //comments and what I have written so far. If you can help me figure out where all the errors are and how to complete the assignment, I will really appreciate it. Like I said, I don't want you to do my homework for me (it's my final), but any constructive criticism is welcome. This is my final assignment for this class and it is due tomorrow (Sunday before midnight, Arizona time). This is the assignment: Examine the following situation: Your company, Datamax, Inc., is in the process of automating its payroll systems. Your manager has asked you to create a program that calculates overtime pay for all employees. Your program must take into account the employee’s salary, total hours worked, and hours worked more than 40 in a week, and then provide an output that is useful and easily understood by company management. Compile your program utilizing the following background information and the code outline in Appendix D (included in the code section). Submit your project as an attachment including the code and the output. Company Background: Three employees: Mark, John, and Mary The end user needs to be prompted for three specific pieces of input—name, hours worked, and hourly wage. Calculate overtime if input is greater than 40 hours per week. Provide six test plans to verify the logic within the program. Plan 1 must display the proper information for employee #1 with overtime pay. Plan 2 must display the proper information for employee #1 with no overtime pay. Plans 3-6 are duplicates of plan 1 and 2 but for the other employees. Program Requirements: Define a base class to use for the entire program. The class holds the function calls and the variables related to the overtime pay calculations. Define one object per employee. Note there will be three employees. Your program must take the objects created and implement calculations based on total salaries, total hours, and the total number of overtime hours. See the Employee Summary Data section of the sample output. Logic Steps to Complete Your Program: Define your base class. Define your objects from your base class. Prompt for user input, updating your object classes for all three users. Implement your overtime pay calculations. Display overtime or regular time pay calculations. See the sample output below. Implement object calculations by summarizing your employee objects and display the summary information in the example below. And this is the code: // Final_Project.cpp : Defines the entry point for the console application. // #include "stdafx.h" #include <iostream> #include <string> #include <iomanip> using namespace std; // //CLASS DECLARATION SECTION // class CEmployee { public: void ImplementCalculations(string EmployeeName, double hours, double wage); void DisplayEmployInformation(void); void Addsomethingup (CEmployee, CEmployee, CEmployee); string EmployeeName ; int hours ; int overtime_hours ; int iTotal_hours ; int iTotal_OvertimeHours ; float wage ; float basepay ; float overtime_pay ; float overtime_extra ; float iTotal_salaries ; float iIndividualSalary ; }; int main() { system("cls"); cout << "Welcome to the Employee Pay Center"; /* Use this section to define your objects. You will have one object per employee. You have only three employees. The format is your class name and your object name. */ std::cout << "Please enter Employee's Name: "; std::cin >> EmployeeName; std::cout << "Please enter Total Hours for (EmployeeName): "; std::cin >> hours; std::cout << "Please enter Base Pay for(EmployeeName): "; std::cin >> basepay; /* Here you will prompt for the first employee’s information. Prompt the employee name, hours worked, and the hourly wage. For each piece of information, you will update the appropriate class member defined above. Example of Prompts Enter the employee name = Enter the hours worked = Enter his or her hourly wage = */ /* Here you will prompt for the second employee’s information. Prompt the employee name, hours worked, and the hourly wage. For each piece of information, you will update the appropriate class member defined above. Enter the employee name = Enter the hours worked = Enter his or her hourly wage = */ /* Here you will prompt for the third employee’s information. Prompt the employee name, hours worked, and the hourly wage. For each piece of information, you will update the appropriate class member defined above. Enter the employee name = Enter the hours worked = Enter his or her hourly wage = */ /* Here you will implement a function call to implement the employ calcuations for each object defined above. You will do this for each of the three employees or objects. The format for this step is the following: [(object name.function name(objectname.name, objectname.hours, objectname.wage)] ; */ /* This section you will send all three objects to a function that will add up the the following information: - Total Employee Salaries - Total Employee Hours - Total Overtime Hours The format for this function is the following: - Define a new object. - Implement function call [objectname.functionname(object name 1, object name 2, object name 3)] /* } //End of Main Function void CEmployee::ImplementCalculations (string EmployeeName, double hours, double wage){ //Initialize overtime variables overtime_hours=0; overtime_pay=0; overtime_extra=0; if (hours > 40) { /* This section is for the basic calculations for calculating overtime pay. - base pay = 40 hours times the hourly wage - overtime hours = hours worked – 40 - overtime pay = hourly wage * 1.5 - overtime extra pay over 40 = overtime hours * overtime pay - salary = overtime money over 40 hours + your base pay */ /* Implement function call to output the employee information. Function is defined below. */ } // if (hours > 40) else { /* Here you are going to calculate the hours less than 40 hours. - Your base pay is = your hours worked times your wage - Salary = your base pay */ /* Implement function call to output the employee information. Function is defined below. */ } // End of the else } //End of Primary Function void CEmployee::DisplayEmployInformation(); { // This function displays all the employee output information. /* This is your cout statements to display the employee information: Employee Name ............. = Base Pay .................. = Hours in Overtime ......... = Overtime Pay Amount........ = Total Pay ................. = */ } // END OF Display Employee Information void CEmployee::Addsomethingup (CEmployee Employ1, CEmployee Employ2) { // Adds two objects of class Employee passed as // function arguments and saves them as the calling object's data member values. /* Add the total hours for objects 1, 2, and 3. Add the salaries for each object. Add the total overtime hours. */ /* Then display the information below. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%% EMPLOYEE SUMMARY DATA%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%% Total Employee Salaries ..... = 576.43 %%%% Total Employee Hours ........ = 108 %%%% Total Overtime Hours......... = 5 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% */ } // End of function

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  • YAML front matter for Jekyll and nested lists

    - by motleydev
    I have a set of nested yaml lists with something like the following: title: the example image: link.jpg products: - top-level: Product One arbitrary: Value nested-products: - nested: Associated Product sub-arbitrary: Associated Value - top-level: Product Two arbitrary: Value - top-level: Product Three arbitrary: Value I can loop through the products with no problem using for item in page.products and I can use a logic operator to determine if nested products exist - what I CAN'T do is loop through multiple nested-products per iteration of top-level I have tried using for subitem in item and other options - but I can't get it to work - any ideas?

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  • Summarising (permanently) data in a SQL table

    - by Cylindric
    Geetings, Stackers. I have a huge number of data-points in a SQL table, and I want to summarise them in a way reminiscent of RRD. Assuming a table such as ID | ENTITY_ID | SCORE_DATE | SCORE | SOME_OTHER_DATA ----+-----------+------------+-------+----------------- 1 | A00000001 | 01/01/2010 | 100 | some data 2 | A00000002 | 01/01/2010 | 105 | more data 3 | A00000003 | 01/01/2010 | 104 | various text ... | ......... | .......... | ..... | ... ... | A00009999 | 01/01/2010 | 101 | ... | A00000001 | 02/01/2010 | 104 | ... | A00000002 | 02/01/2010 | 119 | ... | A00000003 | 02/01/2010 | 119 | ... | ......... | .......... | ..... | ... | A00009999 | 02/01/2010 | 101 | arbitrary data ... | ......... | .......... | ..... | ... ... | A00000001 | 01/02/2010 | 104 | ... | A00000002 | 01/02/2010 | 119 | ... | A00000003 | 01/01/2010 | 119 | I want to end up with one record per entity, per month: ID | ENTITY_ID | SCORE_DATE | SCORE | ----+-----------+------------+-------+ ... | A00000001 | 01/01/2010 | 100 | ... | A00000002 | 01/01/2010 | 105 | ... | A00000003 | 01/01/2010 | 104 | ... | A00000001 | 01/02/2010 | 100 | ... | A00000002 | 01/02/2010 | 105 | ... | A00000003 | 01/02/2010 | 104 | (I Don't care about the SOME_OTHER_DATA - I'll pick something - either the first or last record probably.) What's an easy way of doing this on a regular basis, so that anything in the last calendar month is summarised in this way? At the moment my plan is kind of: For each EntityID For each month Find average score for all records in given month Update first record with results of previous step Delete all records that aren't the first I can't think of a neat way of doing it though, that doesn't involve lots of updates and iteration. This can either be done in a SQL Stored Procedure, or it can be incorporated into the .Net app that's generating this data, so the solution doesn't really need to be "one big SQL script", but can be :) (SQL-2005)

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  • How to convert Date format

    - by vinay_rockin
    Hi All, I have "Friday, April 02, 2010" as date now I want to display "04/02/2010" if browser language is selected english and "02.04.2010" if browser language is selected as German. All I want, is to display the date as per the browser format. Any Idea how this can be done?

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  • Ruby sleep or delay less than a second?

    - by Joseph Silvashy
    So I'm making a script with ruby that must render frames at 24 frames per second, but I need to wait 1/24th of a second between sending the commands... how can I do that? sleep seems to only wait in increments of 1 second or more. update Well ya you can do sleep 0.1 if you want, but is this the best way to delay in a ruby script?

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