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  • Django: Filtering datetime field by *only* the year value?

    - by unclaimedbaggage
    Hi folks, I'm trying to spit out a django page which lists all entries by the year they were created. So, for example: 2010: Note 4 Note 5 Note 6 2009: Note 1 Note 2 Note 3 It's proving more difficult than I would have expected. The model from which the data comes is below: class Note(models.Model): business = models.ForeignKey(Business) note = models.TextField() created = models.DateTimeField(auto_now_add=True) updated = models.DateTimeField(auto_now=True) class Meta: db_table = 'client_note' @property def note_year(self): return self.created.strftime('%Y') def __unicode__(self): return '%s' % self.note I've tried a few different ways, but seem to run into hurdles down every path. I'm guessing an effective 'group by' method would do the trick (PostGres DB Backend), but I can't seem to find any Django functionality that supports it. I tried getting individual years from the database but I struggled to find a way of filtering datetime fields by just the year value. Finally, I tried adding the note_year @property but because it's derived, I can't filter those values. Any suggestions for an elegant way to do this? I figure it should be pretty straightforward, but I'm having a heckuva time with it. Any ideas much appreciated.

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  • Should I be worried if I don't get any internships by 3rd year's end?

    - by karamba
    I am in some mediocre college in some corner of India. Am about to complete 3rd year C.S. in a month and a half. I have no idea how to go about "finding an internship" as everyone seems to put it. Looking at online advice, I find that a primary way is to "use your contacts". I am sad to say that I don't have many friends(those I have, I am trying to get help from them for all it's worth), and my family can't help me as they have no idea about the software industry. My college has no official facility for aiding students in this, and the few faculty members who had contacts in "whatever" part of the industry have favoured some students that they have personally come to know. (Though I hear that the "internships" they got involve them stocking equipment in some small companies.... still it's something?) I'm getting nervous. I am considering just spending the coming summer refining my skills in C++ and begin learning MySQL and C#, both of which I have zero experience in. Maybe work on my own project... like a library management system. Relative to those in my college, I think I am among the best programmers there, but that isn't saying much as a lot of students can barely write basic code. I have experience in teaching myself C++, and DirectX9 having created a Tetris clone, some basic 3D apps (bouncing balls), and a basic console-based, text-file-database-using library management system (which I plan to improve this summer). Is it alright if I spend my summer so? Will I be able to get a job later on? I know I have to improve my social skills to get anywhere in life, and I will try, but say I am stuck like this till 4th year's end... will such self studying, online learning help me in landing a decent job? Perhaps after I have learned a bit more, joining some open source project?

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  • Group and sort blog posts by date in Rails

    - by Senthil
    I've searched all over web and have not found the answer. I'm trying to have a very standard archive option for my blog based on date. A request to url blog.com/archive/2009 shows all posts in 2009, blog.com/archive/2009/11 shows all posts in November 2009 etc. I found two different of code but not very helpful to me. def display_by_date year = params[:year] month = params[:month] day = params[:day] day = '0'+day if day && day.size == 1 @day = day if (year && month && day) render(:template => "blog/#{year}/#{month}/#{date}") elsif year render(:template => "blog/#{year}/list") end end def archive year = params[:year] month = params[:month] day = params[:day] day = '0'+day if day && day.size == 1 if (year && month && day) @posts_by_month = Blog.find(:all, :conditions => ["year is?", year]) else @posts_by_month = Blog.find(:all).group_by { |post| post.created_at.strftime("%B") } end end Any help is appreciated.

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  • MDX Current Month Problem

    - by schone
    Hi all, I have the following MDX query: WITH MEMBER [Measures].[TY Sales] AS 'SUM([Measures].[Sales])' MEMBER [Measures].[YTD Sales] AS 'SUM(PERIODSTODATE([Time.Fiscals].[2009], [Time.Fiscals].CurrentMember), [Measures].[Sales])' MEMBER [Measures].[LY Sales] AS 'SUM(PARALLELPERIOD([Time.Fiscals].[2009].[8], 1, [Time.Fiscals].CurrentMember),[Measures].[Sales])' MEMBER [Measures].[LYTD Sales] AS 'SUM(PERIODSTODATE([Time.Fiscals].[2009], PARALLELPERIOD([Time.Fiscals].[2009].[8], 1, [Time.Fiscals].CurrentMember)), [Measures].[Sales])' SELECT {[Measures].[TY Sales],[Measures].[YTD Sales],[Measures].[LY Sales],[Measures].[LYTD Sales]} ON COLUMNS, [Time.Fiscals].[2009].[8] ON ROWS FROM [Sales] The above query displays sales for fiscal year 2009 and fiscal month 8 (Feb), I would like the fiscal year and fiscal month dynamic but I'm unsure on how to do this? I'm using Mondrian. Thanks!

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  • Explaining NULL and Empty to your 6-year old?

    - by Atomiton
    I'm thinking in terms of Objects here. I think it's important to simplify ideas. If you can explain this to a 6-year old, you can teach new programmers the difference. I'm thinking that a cookie object would be apropos: public class Cookie { public string flavor {get; set; } public int numberOfCrumbs { get; set; } }

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  • How would you explain your job to a 5-year old?

    - by Canavar
    Sometimes it's difficult to define programming to people. Especially too old or too young people can not understand what I do to earn money. They think that I repair computers, or they want to think that I (as an engineer) build computers at work. :) It's really hard to tell people that you produce something they can't touch. Here is a funny question, how would you explain your job to a 5-year-old?

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  • Most concise way to convert from date format: yyyy[3 digit day of year] to SQL datetime

    - by Seth Reno
    I'm working with an existing database where all dates are stored as integers in the following format: yyyy[3 digit day of year]. For example: 2010-01-01 == 2010001 2010-12-31 == 2010356 I'm using the following SQL to convert to a datetime: DATEADD(d, CAST(SUBSTRING( CAST(NEW_BIZ_OBS_DATE AS VARCHAR), 5, LEN(NEW_BIZ_OBS_DATE) - 4 ) AS INT) - 1, CAST('1/1/' + SUBSTRING(CAST(NEW_BIZ_OBS_DATE AS VARCHAR),1,4) AS DATETIME)) Does anyone have a more concise way to do this?

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  • 2010's Most Popular Articles

    The end of the year is upon us, 2010 is about to be in the books. When closing out a year I like to take a look back at the articles I wrote over the year and see which ones resonated the most with readers. Which ones generated the most reader emails? Which ones were read the most? Such a retrospective analysis highlights what content was of most interest to developers in the trenches, and this data is used to guide article topics in the new year. I ended last year with a "Best Of" article - see 2009's Most Popular Articles - and decided to continue this tradition. Such "Best Of" articles give both regular and new readers a chance to discover (or rediscover) the most favored content from the year. So here it is - a list and synopsis of the 2010's most popular articles on 4GuysFromRolla.com. Read More >

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  • Oracle Excellence Awards 2012

    - by A&C Redaktion
    Spezialisierte Partner können sich ab sofort bis 29. Juni als „Specialized Partner of the Year“ bewerben. Damit honoriert Oracle diejenigen OPN Partner in EMEA, die sich mithilfe der Spezialisierung besonders ausgezeichnet haben, sei es durch einen hohen Mehrwert für deren Endkunden oder innovativen Lösungen und Services. Voraussetzung für eine Bewerbung ist mindestens eine abgeschlossene Spezialisierung in diesen sieben Kategorien: Specialized Partner of the Year: Database Specialized Partner of the Year: Applications Specialized Partner of the Year: Middleware Specialized Partner of the Year: Industry Specialized Partner of the Year: Oracle Accelerate Specialized Partner of the Year: Servers and Storage Systems Specialized Partner of the Year: Oracle on Oracle Der Gewinner eines “Specialized Partner of the Year” EMEA Awards erhält 5.000 US-Dollar MDF und vielfältige Möglichkeiten, sich in Szene zu setzen. Wie auch im letzten Jahr wird der Award wieder auf der Oracle OpenWorld in San Francisco überreicht. Interviews, Videos, Werbung, Berichte im Oracle Magazine und ein kostenloser Konferenzpass sind natürlich inklusive. Wie immer, gilt die Bewerbung für den EMEA-Award gleichzeitig als Bewerbung für den deutschen Partner-Award 2012, der auf dem Oracle Partner Day (im Herbst nach der OpenWorld) verliehen wird. Normal 0 21 false false false DE X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0cm 5.4pt 0cm 5.4pt; mso-para-margin:0cm; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;} Die Nominierung Ihres erfolgreichen Projektes muss diesmal hier auf Englisch eingereicht werden, da eine internationale Jury entscheidet. Beschreiben Sie Ihr Projekt so ausführlich wie möglich, damit sich die Juroren ein gutes Bild Ihrer Leistungen und Services machen können. Wenn Sie hierbei Unterstützung benötigen, fragen Sie einfach Ihren Channel Manager. Denn: Je aussagekräftiger Ihre Unterlagen sind, desto höher ist Ihre Chance zu gewinnen! Alle Informationen zu den diesjährigen Awards finden Sie auf der Oracle Excellence Awards Website. Wir drücken Ihnen die Daumen!

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  • Oracle Excellence Awards 2012

    - by A&C Redaktion
    Spezialisierte Partner können sich ab sofort bis 29. Juni als „Specialized Partner of the Year“ bewerben. Damit honoriert Oracle diejenigen OPN Partner in EMEA, die sich mithilfe der Spezialisierung besonders ausgezeichnet haben, sei es durch einen hohen Mehrwert für deren Endkunden oder innovativen Lösungen und Services. Voraussetzung für eine Bewerbung ist mindestens eine abgeschlossene Spezialisierung in diesen sieben Kategorien: Specialized Partner of the Year: Database Specialized Partner of the Year: Applications Specialized Partner of the Year: Middleware Specialized Partner of the Year: Industry Specialized Partner of the Year: Oracle Accelerate Specialized Partner of the Year: Servers and Storage Systems Specialized Partner of the Year: Oracle on Oracle Der Gewinner eines “Specialized Partner of the Year” EMEA Awards erhält 5.000 US-Dollar MDF und vielfältige Möglichkeiten, sich in Szene zu setzen. Wie auch im letzten Jahr wird der Award wieder auf der Oracle OpenWorld in San Francisco überreicht. Interviews, Videos, Werbung, Berichte im Oracle Magazine und ein kostenloser Konferenzpass sind natürlich inklusive. Wie immer, gilt die Bewerbung für den EMEA-Award gleichzeitig als Bewerbung für den deutschen Partner-Award 2012, der auf dem Oracle Partner Day (im Herbst nach der OpenWorld) verliehen wird. Normal 0 21 false false false DE X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0cm 5.4pt 0cm 5.4pt; mso-para-margin:0cm; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;} Die Nominierung Ihres erfolgreichen Projektes muss diesmal hier auf Englisch eingereicht werden, da eine internationale Jury entscheidet. Beschreiben Sie Ihr Projekt so ausführlich wie möglich, damit sich die Juroren ein gutes Bild Ihrer Leistungen und Services machen können. Wenn Sie hierbei Unterstützung benötigen, fragen Sie einfach Ihren Channel Manager. Denn: Je aussagekräftiger Ihre Unterlagen sind, desto höher ist Ihre Chance zu gewinnen! Alle Informationen zu den diesjährigen Awards finden Sie auf der Oracle Excellence Awards Website. Wir drücken Ihnen die Daumen!

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  • Do you know of a C macro to compute Unix time and date?

    - by Alexis Wilke
    I'm wondering if someone knows/has a C macro to compute a static Unix time from a hard coded date and time as in: time_t t = UNIX_TIMESTAMP(2012, 5, 10, 9, 26, 13); I'm looking into that because I want to have a numeric static timestamp. This will be done hundred of times throughout the software, each time with a different date, and I want to make sure it is fast because it will run hundreds of times every second. Converting dates that many times would definitively slow down things (i.e. calling mktime() is slower than having a static number compiled in place, right?) [made an update to try to render this paragraph clearer, Nov 23, 2012] Update I want to clarify the question with more information about the process being used. As my server receives requests, for each request, it starts a new process. That process is constantly updated with new plugins and quite often such updates require a database update. Those must be run only once. To know whether an update is necessary, I want to use a Unix date (which is better than using a counter because a counter is much more likely to break once in a while.) The plugins will thus receive an update signal and have their on_update() function called. There I want to do something like this: void some_plugin::on_update(time_t last_update) { if(last_update < UNIX_TIMESTAMP(2010, 3, 22, 20, 9, 26)) { ...run update... } if(last_update < UNIX_TIMESTAMP(2012, 5, 10, 9, 26, 13)) { ...run update... } // as many test as required... } As you can see, if I have to compute the unix timestamp each time, this could represent thousands of calls per process and if you receive 100 hits a second x 1000 calls, you wasted 100,000 calls when you could have had the compiler compute those numbers once at compile time. Putting the value in a static variable is of no interest because this code will run once per process run. Note that the last_update variable changes depending on the website being hit (it comes from the database.) Code Okay, I got the code now: // helper (Days in February) #define _SNAP_UNIX_TIMESTAMP_FDAY(year) \ (((year) % 400) == 0 ? 29LL : \ (((year) % 100) == 0 ? 28LL : \ (((year) % 4) == 0 ? 29LL : \ 28LL))) // helper (Days in the year) #define _SNAP_UNIX_TIMESTAMP_YDAY(year, month, day) \ ( \ /* January */ static_cast<qint64>(day) \ /* February */ + ((month) >= 2 ? 31LL : 0LL) \ /* March */ + ((month) >= 3 ? _SNAP_UNIX_TIMESTAMP_FDAY(year) : 0LL) \ /* April */ + ((month) >= 4 ? 31LL : 0LL) \ /* May */ + ((month) >= 5 ? 30LL : 0LL) \ /* June */ + ((month) >= 6 ? 31LL : 0LL) \ /* July */ + ((month) >= 7 ? 30LL : 0LL) \ /* August */ + ((month) >= 8 ? 31LL : 0LL) \ /* September */+ ((month) >= 9 ? 31LL : 0LL) \ /* October */ + ((month) >= 10 ? 30LL : 0LL) \ /* November */ + ((month) >= 11 ? 31LL : 0LL) \ /* December */ + ((month) >= 12 ? 30LL : 0LL) \ ) #define SNAP_UNIX_TIMESTAMP(year, month, day, hour, minute, second) \ ( /* time */ static_cast<qint64>(second) \ + static_cast<qint64>(minute) * 60LL \ + static_cast<qint64>(hour) * 3600LL \ + /* year day (month + day) */ (_SNAP_UNIX_TIMESTAMP_YDAY(year, month, day) - 1) * 86400LL \ + /* year */ (static_cast<qint64>(year) - 1970LL) * 31536000LL \ + ((static_cast<qint64>(year) - 1969LL) / 4LL) * 86400LL \ - ((static_cast<qint64>(year) - 1901LL) / 100LL) * 86400LL \ + ((static_cast<qint64>(year) - 1601LL) / 400LL) * 86400LL ) WARNING: Do not use these macros to dynamically compute a date. It is SLOWER than mktime(). This being said, if you have a hard coded date, then the compiler will compute the time_t value at compile time. Slower to compile, but faster to execute over and over again.

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  • Inserting Records in Ascending Order function- C homework assignment

    - by Aaron McRuer
    Good day, Stack Overflow. I have a homework assignment that I'm working on this weekend that I'm having a bit of a problem with. We have a struct "Record" (which contains information about cars for a dealership) that gets placed in a particular spot in a linked list according to 1) its make and 2) according to its model year. This is done when initially building the list, when a "int insertRecordInAscendingOrder" function is called in Main. In "insertRecordInAscendingOrder", a third function, "createRecord" is called, where the linked list is created. The function then goes to the function "compareCars" to determine what elements get put where. Depending on the value returned by this function, insertRecordInAscendingOrder then places the record where it belongs. The list is then printed out. There's more to the assignment, but I'll cross that bridge when I come to it. Ideally, and for the assignment to be considered correct, the linked list must be ordered as: Chevrolet 2012 25 Chevrolet 2013 10 Ford 2010 5 Ford 2011 3 Ford 2012 15 Honda 2011 9 Honda 2012 3 Honda 2013 12 Toyota 2009 2 Toyota 2011 7 Toyota 2013 20 from the a text file that has the data ordered the following way: Ford 2012 15 Ford 2011 3 Ford 2010 5 Toyota 2011 7 Toyota 2012 20 Toyota 2009 2 Honda 2011 9 Honda 2012 3 Honda 2013 12 Chevrolet 2013 10 Chevrolet 2012 25 Notice that the alphabetical order of the "make" field takes precedence, then, the model year is arranged from oldest to newest. However, the program produces this as the final list: Chevrolet 2012 25 Chevrolet 2013 10 Honda 2011 9 Honda 2012 3 Honda 2013 12 Toyota 2009 2 Toyota 2011 7 Toyota 2012 20 Ford 2010 5 Ford 2011 3 Ford 2012 15 I sat down with a grad student and tried to work out all of this yesterday, but we just couldn't figure out why it was kicking the Ford nodes down to the end of the list. Here's the code. As you'll notice, I included a printList call at each instance of the insertion of a node. This way, you can see just what is happening when the nodes are being put in "order". It is in ANSI C99. All function calls must be made as they are specified, so unfortunately, there's no real way of getting around this problem by creating a more efficient algorithm. #include <stdio.h> #include <stdlib.h> #include <string.h> #define MAX_LINE 50 #define MAX_MAKE 20 typedef struct record { char *make; int year; int stock; struct record *next; } Record; int compareCars(Record *car1, Record *car2); void printList(Record *head); Record* createRecord(char *make, int year, int stock); int insertRecordInAscendingOrder(Record **head, char *make, int year, int stock); int main(int argc, char **argv) { FILE *inFile = NULL; char line[MAX_LINE + 1]; char *make, *yearStr, *stockStr; int year, stock, len; Record* headRecord = NULL; /*Input and file diagnostics*/ if (argc!=2) { printf ("Filename not provided.\n"); return 1; } if((inFile=fopen(argv[1], "r"))==NULL) { printf("Can't open the file\n"); return 2; } /*obtain values for linked list*/ while (fgets(line, MAX_LINE, inFile)) { make = strtok(line, " "); yearStr = strtok(NULL, " "); stockStr = strtok(NULL, " "); year = atoi(yearStr); stock = atoi(stockStr); insertRecordInAscendingOrder(&headRecord,make, year, stock); } printf("The original list in ascending order: \n"); printList(headRecord); } /*use strcmp to compare two makes*/ int compareCars(Record *car1, Record *car2) { int compStrResult; compStrResult = strcmp(car1->make, car2->make); int compYearResult = 0; if(car1->year > car2->year) { compYearResult = 1; } else if(car1->year == car2->year) { compYearResult = 0; } else { compYearResult = -1; } if(compStrResult == 0 ) { if(compYearResult == 1) { return 1; } else if(compYearResult == -1) { return -1; } else { return compStrResult; } } else if(compStrResult == 1) { return 1; } else { return -1; } } int insertRecordInAscendingOrder(Record **head, char *make, int year, int stock) { Record *previous = *head; Record *newRecord = createRecord(make, year, stock); Record *current = *head; int compResult; if(*head == NULL) { *head = newRecord; printf("Head is null, list was empty\n"); printList(*head); return 1; } else if ( compareCars(newRecord, *head)==-1) { *head = newRecord; (*head)->next = current; printf("New record was less than the head, replacing\n"); printList(*head); return 1; } else { printf("standard case, searching and inserting\n"); previous = *head; while ( current != NULL &&(compareCars(newRecord, current)==1)) { printList(*head); previous = current; current = current->next; } printList(*head); previous->next = newRecord; previous->next->next = current; } return 1; } /*creates records from info passed in from main via insertRecordInAscendingOrder.*/ Record* createRecord(char *make, int year, int stock) { printf("CreateRecord\n"); Record *theRecord; int len; if(!make) { return NULL; } theRecord = malloc(sizeof(Record)); if(!theRecord) { printf("Unable to allocate memory for the structure.\n"); return NULL; } theRecord->year = year; theRecord->stock = stock; len = strlen(make); theRecord->make = malloc(len + 1); strncpy(theRecord->make, make, len); theRecord->make[len] = '\0'; theRecord->next=NULL; return theRecord; } /*prints list. lists print.*/ void printList(Record *head) { int i; int j = 50; Record *aRecord; aRecord = head; for(i = 0; i < j; i++) { printf("-"); } printf("\n"); printf("%20s%20s%10s\n", "Make", "Year", "Stock"); for(i = 0; i < j; i++) { printf("-"); } printf("\n"); while(aRecord != NULL) { printf("%20s%20d%10d\n", aRecord->make, aRecord->year, aRecord->stock); aRecord = aRecord->next; } printf("\n"); } The text file you'll need for a command line argument can be saved under any name you like; here are the contents you'll need: Ford 2012 15 Ford 2011 3 Ford 2010 5 Toyota 2011 7 Toyota 2012 20 Toyota 2009 2 Honda 2011 9 Honda 2012 3 Honda 2013 12 Chevrolet 2013 10 Chevrolet 2012 25 Thanks in advance for your help. I shall continue to plow away at it myself.

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  • At what year in history was computers first used to store porn? [closed]

    - by Emil H
    Of course this sounds like a joke question, but it's meant seriously. I remember being told by an old system administrator back in the early nineties about people asking about good FTPs for porn, and that they would as a joke always tell them to connect to 127.0.0.1. They would come back saying that there was a lot of porn at that address, but that oddly enough it seemed like they already had it all. Point being, it seems like it's been around for quite a while. Anyway. Considering that a considerable portion of the internet is devoted to porn these days, it would be interesting to know if someone has any kind of idea as to when and where the phenomena first arose? There must be some mention of this in old hacker folk lore? (Changed to CW to emphasize that this isn't about rep, but about genuine curiousity. :)

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  • Have optical drive connectors changed in the last year?

    - by Zippityboomba
    In early 2010 my motherboard freaked out, and I was able to drop in a new board and connect my two DVD drives that were initially bought in 2005. Tonight my new components arrived to build my next system, and I figured I'd just reuse the old DVD drives. Surprise, the long flat connector cable has no mate on the new motherboard. Reckon I'll just pick up a new DVD drive for $25, but what gives? Thanks!

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  • Like to Upgrade My PC (7 year old) - for animation and hardcore gaming ! - help me [closed]

    - by sri
    I like to buy a new computer for my studies and as well as gaming. My old pc has 1.5 GB RAM with 512MB Graphics card. And it is very old to run Adobe CS5 version and other high end animation software. My budget will be INR 20k-25k. I have 500GB hard disk, keyboard and mouse new. So apart from this, I like to buy : Intel or AMD is good ? My idea is : Corei5 or Corei7 = which is best and economy ? Which mother board. 4 GB RAM with upto 8 GB RAM slot for future upgrade. 1 GB or 2 GB Graphic card - which one ? If I am wrong - please suggest me

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  • Skype Mac API - Use AppleScript or 5 year old API?

    - by Andrew
    I have a x86_64 app that I would like to have optionally read Skype status messages. However, the 5 year old skype mac framework is 32-bit, and if there is a way to have that compile within a 64-bit app, I haven't found it. My question is, basically, how should I go about doing this? I really only need to get and set the USERSTATUS AWAY/ONLINE string. Using AppleScript, a "Should Skype allow this" dialog pops up... every time. This is highly inefficient and downright irritating. Advice? I'm considering writing a 32-bit CLI wrapper, but that seems like overkill.

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