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  • Why my /usr/share/file/magic is not found, in PHP at CentOS?

    - by Vincenzo
    This is what I'm doing: $ php <?php $finfo=finfo_open(FILEINFO_MIME, '/usr/share/file/magic'); This is what I'm getting: PHP Notice: finfo_open(): Warning: description `8-bit ISDN mu-law compressed (CCITT G.721 ADPCM voice data enco' truncated in - on line 2 PHP Notice: finfo_open(): Warning: description `8-bit ISDN mu-law compressed (CCITT G.721 ADPCM voice data enco' truncated in - on line 2 PHP Notice: finfo_open(): Warning: <= not supported in - on line 2 PHP Notice: finfo_open(): Warning: <= not supported in - on line 2 PHP Notice: finfo_open(): Warning: <= not supported in - on line 2 PHP Notice: finfo_open(): Warning: >= not supported in - on line 2 PHP Warning: finfo_open(): Failed to load magic database at '/usr/share/file/magic'. in - on line 2 This is a clean CentOS 5.5 installation, PHP 5.3. The file /usr/share/file/magic exists and is accessible.

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  • Problem in apache2 with mod rewrite when setting rules in .conf files instead of .htaccess

    - by gonzlikes
    Hi, Because of weird security policies of my hosting provider I have to define my rewrite rules in /etc/apache2/conf.d/examplesite.conf instead of writing them on an .htaccess on the www folder of that site. What I'm trying to do is setup a Wordpress Mu server (http://mu.wordpress.org/forums/topic/17349 ) and so far its working on a 50%. The main blog loads perfectly but other sub blogs (located for example at www.example.com/blog2 ) don't. I'm guessing the problem is that the rewrite rules behave differently when declared at .conf files for each virtual host instead of using .htaccess files. Has anybody else had this problem? How can you fix it?

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  • Memory Leaks - Objective-C

    - by reising1
    Can anyone help point out memory leaks? I'm getting a bunch within this method and I'm not sure exactly how to fix it. - (NSMutableArray *)getTop5AndOtherKeysAndValuesFromDictionary:(NSMutableDictionary *)dict { NSLog(@"get top 5"); int sumOfAllValues = 0; NSMutableArray *arr = [[[NSMutableArray alloc] init] retain]; for(NSString *key in dict){ NSString *value = [[dict objectForKey:key] retain]; [arr addObject:value]; sumOfAllValues += [value intValue]; } //sort values NSArray *sorted = [[arr sortedArrayUsingFunction:sort context:NULL] retain]; [arr release]; //top 5 values int sumOfTop5 = 0; NSMutableArray *top5 = [[[NSMutableArray alloc] init] retain]; for(int i = 0; i < 5; i++) { int proposedIndex = [sorted count] - 1 - i; if(proposedIndex >= 0) { [top5 addObject:[sorted objectAtIndex:([sorted count] - i - 1)]]; sumOfTop5 += [[sorted objectAtIndex:([sorted count] - i - 1)] intValue]; } } [sorted release]; //copy of all keys NSMutableArray *copyOfKeys = [[[NSMutableArray alloc] init] retain]; for(NSString *key in dict) { [copyOfKeys addObject:key]; } //copy of top 5 values NSMutableArray *copyOfTop5 = [[[NSMutableArray alloc] init] retain]; for(int i = 0; i < [top5 count]; i++) { [copyOfTop5 addObject:[top5 objectAtIndex:i]]; } //get keys with top 5 values NSMutableArray *outputKeys = [[[NSMutableArray alloc] init] retain]; for(int i = 0; i < [top5 count]; i++) { NSString *targetValue = [top5 objectAtIndex:i]; for(int j = 0; j < [copyOfKeys count]; j++) { NSString *key = [copyOfKeys objectAtIndex:j]; NSString *val = [dict objectForKey:key]; if([val isEqualToString:targetValue]) { [outputKeys addObject:key]; [copyOfKeys removeObjectAtIndex:j]; break; } } } [outputKeys addObject:@"Other"]; [top5 addObject:[[NSString stringWithFormat:@"%d",(sumOfAllValues - sumOfTop5)] retain]]; NSMutableArray *output = [[NSMutableArray alloc] init]; [output addObject:outputKeys]; [output addObject:top5]; NSMutableArray *percents = [[NSMutableArray alloc] init]; int sum = sumOfAllValues; float leftOverSum = sum * 1.0f; int count = [top5 count]; float val1, val2, val3, val4, val5; if(count >= 1) val1 = ([[top5 objectAtIndex:0] intValue] * 1.0f)/sum; else val1 = 0.0f; if(count >=2) val2 = ([[top5 objectAtIndex:1] intValue] * 1.0f)/sum; else val2 = 0.0f; if(count >= 3) val3 = ([[top5 objectAtIndex:2] intValue] * 1.0f)/sum; else val3 = 0.0f; if(count >= 4) val4 = ([[top5 objectAtIndex:3] intValue] * 1.0f)/sum; else val4 = 0.0f; if(count >=5) val5 = ([[top5 objectAtIndex:4] intValue] * 1.0f)/sum; else val5 = 0.0f; if(val1 >= .00001f) { NSMutableArray *a1 = [[NSMutableArray alloc] init]; [a1 addObject:[outputKeys objectAtIndex:0]]; [a1 addObject:[top5 objectAtIndex:0]]; [a1 addObject:[NSString stringWithFormat:@"%.01f",(val1*100)]]; [percents addObject:a1]; leftOverSum -= ([[top5 objectAtIndex:0] intValue] * 1.0f); } if(val2 >= .00001f) { NSMutableArray *a2 = [[NSMutableArray alloc] init]; [a2 addObject:[outputKeys objectAtIndex:1]]; [a2 addObject:[top5 objectAtIndex:1]]; [a2 addObject:[NSString stringWithFormat:@"%.01f",(val2*100)]]; [percents addObject:a2]; leftOverSum -= ([[top5 objectAtIndex:1] intValue] * 1.0f); } if(val3 >= .00001f) { NSMutableArray *a3 = [[NSMutableArray alloc] init]; [a3 addObject:[outputKeys objectAtIndex:2]]; [a3 addObject:[top5 objectAtIndex:2]]; [a3 addObject:[NSString stringWithFormat:@"%.01f",(val3*100)]]; [percents addObject:a3]; leftOverSum -= ([[top5 objectAtIndex:2] intValue] * 1.0f); } if(val4 >= .00001f) { NSMutableArray *a4 = [[NSMutableArray alloc] init]; [a4 addObject:[outputKeys objectAtIndex:3]]; [a4 addObject:[top5 objectAtIndex:3]]; [a4 addObject:[NSString stringWithFormat:@"%.01f",(val4*100)]]; [percents addObject:a4]; leftOverSum -= ([[top5 objectAtIndex:3] intValue] * 1.0f); } if(val5 >= .00001f) { NSMutableArray *a5 = [[NSMutableArray alloc] init]; [a5 addObject:[outputKeys objectAtIndex:4]]; [a5 addObject:[top5 objectAtIndex:4]]; [a5 addObject:[NSString stringWithFormat:@"%.01f",(val5*100)]]; [percents addObject:a5]; leftOverSum -= ([[top5 objectAtIndex:4] intValue] * 1.0f); } float valOther = (leftOverSum/sum); if(valOther >= .00001f) { NSMutableArray *a6 = [[NSMutableArray alloc] init]; [a6 addObject:[outputKeys objectAtIndex:5]]; [a6 addObject:[top5 objectAtIndex:5]]; [a6 addObject:[NSString stringWithFormat:@"%.01f",(valOther*100)]]; [percents addObject:a6]; } [output addObject:percents]; NSLog(@"mu - a"); //[arr release]; NSLog(@"mu - b"); //[copyOfKeys release]; NSLog(@"mu - c"); //[copyOfTop5 release]; NSLog(@"mu - c"); //[outputKeys release]; //[top5 release]; //[percents release]; return output; }

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  • Cleaning strings in R: add punctuation w/o overwriting last character

    - by spearmint
    I'm new to R and unable to find other threads with a similar issue. I'm cleaning data that requires punctuation at the end of each line. I am unable to add, say, a period without overwriting the final character of the line preceding the carriage return + line feed. Sample code: Data1 <- "%trn: dads sheep\r\n*MOT: hunn.\r\n%trn: yes.\r\n*MOT: ana mu\r\n%trn: where is it?" Data2 <- gsub("[^[:punct:]]\r\n\\*", ".\r\n\\*", Data1) The contents of Data2: [1] "%trn: dads shee.\r\n*MOT: hunn.\r\n%trn: yes.\r\n*MOT: ana mu\r\n%trn: where is it?" Notice the "p" of sheep was overwritten with the period. Any thoughts on how I could avoid this?

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  • Console class in java Exception in reading password

    - by satheesh
    Hi, i am trying to use Console class in java. with this code import java.io.Console; public class ConsoleClass { public static void main(String[] args){ Console c=System.console(); char[] pw; pw=c.readPassword("%s","pw :"); for(char ch:pw){ c.format("%c",ch); } c.format("\n"); MyUtility mu =new MyUtility(); while(true){ String name=c.readLine("%s", "input?: "); c.format("output: %s \n",mu.doStuff(name)); } } } class MyUtility{ String doStuff(String arg1){ return "result is " +arg1; } } here i am getting NullPointerException when i tried to run in netbeans but i am not getting any Exception when tried to run in cmd with out netbeans IDE.Why?

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  • Un Service Pack annoncé pour Microsoft Exchange 2010, une beta du SP prévue pour juin

    Mise à jour du 08/04/10 Un Service Pack prévu pour Microsoft Exchange 2010 Améliorera les performances du serveur de messagerie, une beta du SP arrivera en juin Microsoft vient de dévoiler le contenu du prochain Service Pack à venir pour Exchange 2010, le "serveur de communication et de collaboration pour les entreprises, basé sur la messagerie électronique", pour reprendre les propres termes de Redmond. Ce SP1 devrait permettre un meilleur archivage (création de filtres pour mieux automatiser l'archivage ou la suppression des messages), une recherche plus efficace dans les mails (avec l'introduction de la recherche "mu...

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  • Questions about identifying the components in MVC

    - by luiscubal
    I'm currently developing an client-server application in node.js, Express, mustache and MySQL. However, I believe this question should be mostly language and framework agnostic. This is the first time I'm doing a real MVC application and I'm having trouble deciding exactly what means each component. (I've done web applications that could perhaps be called MVC before, but I wouldn't confidently refer to them as such) I have a server.js that ties the whole application together. It does initialization of all other components (including the database connection, and what I think are the "models" and the "views"), receiving HTTP requests and deciding which "views" to use. Does this mean that my server.js file is the controller? Or am I mixing code that doesn't belong there? What components should I break the server.js file into? Some examples of code that's in the server.js file: var connection = mysql.createConnection({ host : 'localhost', user : 'root', password : 'sqlrevenge', database : 'blog' }); //... app.get("/login", function (req, res) { //Function handles a GET request for login forms if (process.env.NODE_ENV == 'DEVELOPMENT') { mu.clearCache(); } session.session_from_request(connection, req, function (err, session) { if (err) { console.log('index.js session error', err); session = null; } login_view.html(res, user_model, post_model, session, mu); //I named my view functions "html" for the case I might want to add other output types (such as a JSON API), or should I opt for completely separate views then? }); }); I have another file that belongs named session.js. It receives a cookies object, reads the stored data to decide if it's a valid user session or not. It also includes a function named login that does change the value of cookies. First, I thought it would be part of the controller, since it kind of dealt with user input and supplied data to the models. Then, I thought that maybe it was a model since it dealt with the application data/database and the data it supplies is used by views. Now, I'm even wondering if it could be considered a View, since it outputs data (cookies are part of HTTP headers, which are output)

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  • Valgrind says "stack allocation," I say "heap allocation"

    - by Joel J. Adamson
    Dear Friends, I am trying to trace a segfault with valgrind. I get the following message from valgrind: ==3683== Conditional jump or move depends on uninitialised value(s) ==3683== at 0x4C277C5: sparse_mat_mat_kron (sparse.c:165) ==3683== by 0x4C2706E: rec_mating (rec.c:176) ==3683== by 0x401C1C: age_dep_iterate (age_dep.c:287) ==3683== by 0x4014CB: main (age_dep.c:92) ==3683== Uninitialised value was created by a stack allocation ==3683== at 0x401848: age_dep_init_params (age_dep.c:131) ==3683== ==3683== Conditional jump or move depends on uninitialised value(s) ==3683== at 0x4C277C7: sparse_mat_mat_kron (sparse.c:165) ==3683== by 0x4C2706E: rec_mating (rec.c:176) ==3683== by 0x401C1C: age_dep_iterate (age_dep.c:287) ==3683== by 0x4014CB: main (age_dep.c:92) ==3683== Uninitialised value was created by a stack allocation ==3683== at 0x401848: age_dep_init_params (age_dep.c:131) However, here's the offending line: /* allocate mating table */ age_dep_data->mtable = malloc (age_dep_data->geno * sizeof (double *)); if (age_dep_data->mtable == NULL) error (ENOMEM, ENOMEM, nullmsg, __LINE__); for (int j = 0; j < age_dep_data->geno; j++) { 131=> age_dep_data->mtable[j] = calloc (age_dep_data->geno, sizeof (double)); if (age_dep_data->mtable[j] == NULL) error (ENOMEM, ENOMEM, nullmsg, __LINE__); } What gives? I thought any call to malloc or calloc allocated heap space; there is no other variable allocated here, right? Is it possible there's another allocation going on (the offending stack allocation) that I'm not seeing? You asked to see the code, here goes: /* Copyright 2010 Joel J. Adamson <[email protected]> $Id: age_dep.c 1010 2010-04-21 19:19:16Z joel $ age_dep.c:main file Joel J. Adamson -- http://www.unc.edu/~adamsonj Servedio Lab University of North Carolina at Chapel Hill CB #3280, Coker Hall Chapel Hill, NC 27599-3280 This file is part of an investigation of age-dependent sexual selection. This code is free software: you can redistribute it and/or modify it under the terms of the GNU General Public License as published by the Free Software Foundation, either version 3 of the License, or (at your option) any later version. This software is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details. You should have received a copy of the GNU General Public License along with haploid. If not, see <http://www.gnu.org/licenses/>. */ #include "age_dep.h" /* global variables */ extern struct argp age_dep_argp; /* global error message variables */ char * nullmsg = "Null pointer: %i"; /* error message for conversions: */ char * errmsg = "Representation error: %s"; /* precision for formatted output: */ const char prec[] = "%-#9.8f "; const size_t age_max = AGEMAX; /* maximum age of males */ static int keep_going_p = 1; int main (int argc, char ** argv) { /* often used counters: */ int i, j; /* read the command line */ struct age_dep_args age_dep_args = { NULL, NULL, NULL }; argp_parse (&age_dep_argp, argc, argv, 0, 0, &age_dep_args); /* set the parameters here: */ /* initialize an age_dep_params structure, set the members */ age_dep_params_t * params = malloc (sizeof (age_dep_params_t)); if (params == NULL) error (ENOMEM, ENOMEM, nullmsg, __LINE__); age_dep_init_params (params, &age_dep_args); /* initialize frequencies: this initializes a list of pointers to initial frqeuencies, terminated by a NULL pointer*/ params->freqs = age_dep_init (&age_dep_args); params->by = 0.0; /* what range of parameters do we want, and with what stepsize? */ /* we should go from 0 to half-of-theta with a step size of about 0.01 */ double from = 0.0; double to = params->theta / 2.0; double stepsz = 0.01; /* did you think I would spell the whole word? */ unsigned int numparts = floor(to / stepsz); do { #pragma omp parallel for private(i) firstprivate(params) \ shared(stepsz, numparts) for (i = 0; i < numparts; i++) { params->by = i * stepsz; int tries = 0; while (keep_going_p) { /* each time through, modify mfreqs and mating table, then go again */ keep_going_p = age_dep_iterate (params, ++tries); if (keep_going_p == ERANGE) error (ERANGE, ERANGE, "Failure to converge\n"); } fprintf (stdout, "%i iterations\n", tries); } /* for i < numparts */ params->freqs = params->freqs->next; } while (params->freqs->next != NULL); return 0; } inline double age_dep_pmate (double age_dep_t, unsigned int genot, double bp, double ba) { /* the probability of mating between these phenotypes */ /* the female preference depends on whether the female has the preference allele, the strength of preference (parameter bp) and the male phenotype (age_dep_t); if the female lacks the preference allele, then this will return 0, which is not quite accurate; it should return 1 */ return bits_isset (genot, CLOCI)? 1.0 - exp (-bp * age_dep_t) + ba: 1.0; } inline double age_dep_trait (int age, unsigned int genot, double by) { /* return the male trait, a function of the trait locus, age, the age-dependent scaling parameter (bx) and the males condition genotype */ double C; double T; /* get the male's condition genotype */ C = (double) bits_popcount (bits_extract (0, CLOCI, genot)); /* get his trait genotype */ T = bits_isset (genot, CLOCI + 1)? 1.0: 0.0; /* return the trait value */ return T * by * exp (age * C); } int age_dep_iterate (age_dep_params_t * data, unsigned int tries) { /* main driver routine */ /* number of bytes for female frequencies */ size_t geno = data->age_dep_data->geno; size_t genosize = geno * sizeof (double); /* female frequencies are equal to male frequencies at birth (before selection) */ double ffreqs[geno]; if (ffreqs == NULL) error (ENOMEM, ENOMEM, nullmsg, __LINE__); /* do not set! Use memcpy (we need to alter male frequencies (selection) without altering female frequencies) */ memmove (ffreqs, data->freqs->freqs[0], genosize); /* for (int i = 0; i < geno; i++) */ /* ffreqs[i] = data->freqs->freqs[0][i]; */ #ifdef PRMTABLE age_dep_pr_mfreqs (data); #endif /* PRMTABLE */ /* natural selection: */ age_dep_ns (data); /* normalized mating table with new frequencies */ age_dep_norm_mtable (ffreqs, data); #ifdef PRMTABLE age_dep_pr_mtable (data); #endif /* PRMTABLE */ double * newfreqs; /* mutate here */ /* i.e. get the new frequency of 0-year-olds using recombination; */ newfreqs = rec_mating (data->age_dep_data); /* return block */ { if (sim_stop_ck (data->freqs->freqs[0], newfreqs, GENO, TOL) == 0) { /* if we have converged, stop the iterations and handle the data */ age_dep_sim_out (data, stdout); return 0; } else if (tries > MAXTRIES) return ERANGE; else { /* advance generations */ for (int j = age_max - 1; j < 0; j--) memmove (data->freqs->freqs[j], data->freqs->freqs[j-1], genosize); /* advance the first age-class */ memmove (data->freqs->freqs[0], newfreqs, genosize); return 1; } } } void age_dep_ns (age_dep_params_t * data) { /* calculate the new frequency of genotypes given additive fitness and selection coefficient s */ size_t geno = data->age_dep_data->geno; double w[geno]; double wbar, dtheta, ttheta, dcond, tcond; double t, cond; /* fitness parameters */ double mu, nu; mu = data->wparams[0]; nu = data->wparams[1]; /* calculate fitness */ for (int j = 0; j < age_max; j++) { int i; for (i = 0; i < geno; i++) { /* calculate male trait: */ t = age_dep_trait(j, i, data->by); /* calculate condition: */ cond = (double) bits_popcount (bits_extract(0, CLOCI, i)); /* trait-based fitness term */ dtheta = data->theta - t; ttheta = (dtheta * dtheta) / (2.0 * nu * nu); /* condition-based fitness term */ dcond = CLOCI - cond; tcond = (dcond * dcond) / (2.0 * mu * mu); /* calculate male fitness */ w[i] = 1 + exp(-tcond) - exp(-ttheta); } /* calculate mean fitness */ /* as long as we calculate wbar before altering any values of freqs[], we're safe */ wbar = gen_mean (data->freqs->freqs[j], w, geno); for (i = 0; i < geno; i++) data->freqs->freqs[j][i] = (data->freqs->freqs[j][i] * w[i]) / wbar; } } void age_dep_norm_mtable (double * ffreqs, age_dep_params_t * params) { /* this function produces a single mating table that forms the input for recombination () */ /* i is female genotype; j is male genotype; k is male age */ int i,j,k; double norm_denom; double trait; size_t geno = params->age_dep_data->geno; for (i = 0; i < geno; i++) { double norm_mtable[geno]; /* initialize the denominator: */ norm_denom = 0.0; /* find the probability of mating and add it to the denominator */ for (j = 0; j < geno; j++) { /* initialize entry: */ norm_mtable[j] = 0.0; for (k = 0; k < age_max; k++) { trait = age_dep_trait (k, j, params->by); norm_mtable[j] += age_dep_pmate (trait, i, params->bp, params->ba) * (params->freqs->freqs)[k][j]; } norm_denom += norm_mtable[j]; } /* now calculate entry (i,j) */ for (j = 0; j < geno; j++) params->age_dep_data->mtable[i][j] = (ffreqs[i] * norm_mtable[j]) / norm_denom; } } My current suspicion is the array newfreqs: I can't memmove, memcpy or assign a stack variable then hope it will persist, can I? rec_mating() returns double *.

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  • Rewriting a for loop in pure NumPy to decrease execution time

    - by Statto
    I recently asked about trying to optimise a Python loop for a scientific application, and received an excellent, smart way of recoding it within NumPy which reduced execution time by a factor of around 100 for me! However, calculation of the B value is actually nested within a few other loops, because it is evaluated at a regular grid of positions. Is there a similarly smart NumPy rewrite to shave time off this procedure? I suspect the performance gain for this part would be less marked, and the disadvantages would presumably be that it would not be possible to report back to the user on the progress of the calculation, that the results could not be written to the output file until the end of the calculation, and possibly that doing this in one enormous step would have memory implications? Is it possible to circumvent any of these? import numpy as np import time def reshape_vector(v): b = np.empty((3,1)) for i in range(3): b[i][0] = v[i] return b def unit_vectors(r): return r / np.sqrt((r*r).sum(0)) def calculate_dipole(mu, r_i, mom_i): relative = mu - r_i r_unit = unit_vectors(relative) A = 1e-7 num = A*(3*np.sum(mom_i*r_unit, 0)*r_unit - mom_i) den = np.sqrt(np.sum(relative*relative, 0))**3 B = np.sum(num/den, 1) return B N = 20000 # number of dipoles r_i = np.random.random((3,N)) # positions of dipoles mom_i = np.random.random((3,N)) # moments of dipoles a = np.random.random((3,3)) # three basis vectors for this crystal n = [10,10,10] # points at which to evaluate sum gamma_mu = 135.5 # a constant t_start = time.clock() for i in range(n[0]): r_frac_x = np.float(i)/np.float(n[0]) r_test_x = r_frac_x * a[0] for j in range(n[1]): r_frac_y = np.float(j)/np.float(n[1]) r_test_y = r_frac_y * a[1] for k in range(n[2]): r_frac_z = np.float(k)/np.float(n[2]) r_test = r_test_x +r_test_y + r_frac_z * a[2] r_test_fast = reshape_vector(r_test) B = calculate_dipole(r_test_fast, r_i, mom_i) omega = gamma_mu*np.sqrt(np.dot(B,B)) # write r_test, B and omega to a file frac_done = np.float(i+1)/(n[0]+1) t_elapsed = (time.clock()-t_start) t_remain = (1-frac_done)*t_elapsed/frac_done print frac_done*100,'% done in',t_elapsed/60.,'minutes...approximately',t_remain/60.,'minutes remaining'

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  • Opening up process and grabbing window title. Where did I go wrong?

    - by user1632018
    In my application I allow for the users to add a program from a open file dialog, and it then adds the item to a listview and saves the items location into the tag. So what I am trying to do is when the program in the listview is selected and the button is pressed, it starts a timer and this timer checks to see if the process is running, and if it isn't launches the process, and once the process is launched it gets the window title of the process and sends it to a textbox on another form. EDIT: The question is if anyone can see why it is not working, by this I mean starting the process, then when it's started closing the form and adding the process window title to a textbox on another form. I have tried to get it working but I can't. I know that the process name it is getting is right I think my problem is to do with my for loop. Basically it isn't doing anything visible right now. I feel like I am very close with my code and im hoping it just needs a couple minor tweaks. Any help would be appreciated. Sorry if my coding practices aren't that great, im pretty new to this. EDIT:I thought I found a solution but it only works now if the process has been started already. It won't start it up. Private Sub Timer1_Tick(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Timer1.Tick Dim s As String = ListView1.SelectedItems(0).Tag Dim myFile As String = Path.GetFileName(s) Dim mu As String = myFile.Replace(".exe", "").Trim() Dim f As Process Dim p As Process() = Process.GetProcessesByName(mu) For Each f In p If p.Length > 0 Then For i As Integer = 0 To p.Length - 1 ProcessID = (p(i).Id) AutoMain.Name.Text = f.MainWindowTitle Timer1.Enabled = False Me.Close() Next Else ProcessID = 0 End If If ProcessID = 0 Then Process.Start(myFile) End If Next End Sub

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  • problem in publishing

    - by girish
    when i am publishing my .net website on my domain it is showing the directory of files when i open my site .. anyone please tell me how to avoid this and make my website open mu home page .

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  • La première bêta de Wordpress 3.0 est disponible, elle est plus fonctionnelle bien qu'encore partiel

    Mise à jour du 09.04.2010 par Katleen La première bêta de Wordpress 3.0 est disponible, elle est plus fonctionnelle bien qu'encore partielle L'infrastructure de blogs Wordpress vient de faire un pas en avant. Son équipe de développeurs vient en effet de publier la version bêta de Wordpress 3.0. Beaucoup de nouveaux thèmes y sont présents, même si les fonctionnalités auxquelles ils sont rattachés ne sont pas encore toutes opérationelles. Première bêta oblige, certains éléments sont encore absents ou incomplets. Au menu des nouveautés, on notera la fusion de Wordpress et de Wordpress MU qui permettra aux posseseurs de plusieurs blogs ou sites de les gérer plus facilement depuis un...

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  • Volume setting isn't remembered after restart/shutdown

    - by Iulian
    This is my first time here and I'm new to linux and also to Ubuntu. I've installed first version 11.10 and there was some problems with the unity dock and also the problem with the volume not being remembered after restart or shutdown. I'm using dualboot with Windows 7. Ubuntu was installed after windows. I have 2 sound cards. One is onboard, on the motherboard, and the other is external, an E-MU 0404 USB 2.0 sound card. The last one is my primary sound card and I've choosed it as default output sound card. I've upgraded to 12.04 hopeing that this was fiex but even in this version the OS doesn't keep the volume where it was last time. The big problem is that sometimes I forget about this problem and start music and it starts at full volume and soon I think I will die of heart attack. Is there a way to make it remember or at least to tell him to start at a specific volume not at 100%?

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  • To mount NAS on a Laptop?

    - by deckoff
    So, I bought a NAS, which I configured successfully in /etc/fstab, on mu Kubuntu 10.10 Thinkpad x40. It works just fine when I am at home. A few days I went out with my laptop and the problem is, that when not at home, both suspend and hibernate functions seem forever to work. I commented out the entry on fstab and the laptop started to work as expected. I played with autofs, but it seems just dies at one moment and I cannot access anything. It works for some time, and then just goes off. Is there any consistent way, to make my laptop access the drive when at home and work OK when away? Probably a script that runs at startup, checks if the mount is there and mounts it if available... or a script that umount the drive at suspend|hibernate and loads it back at startup. Any useful ideas?

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  • WordPress 3.0 est disponible en français : avec cette nouvelle version le CMS libre en PHP a-t-il at

    Mise à jour du 18/06/10 WordPress 3.0 est disponible en français Avec cette version le CMS libre en PHP a-t-il atteint l'âge de maturité ? WordPress a annoncé aujourd'hui la mise à disposition de la dernière évolution de son logiciel de gestion de sites, WordPress 3.0. WordPress 3.0 est une évolution majeure du CMS écrit en PHP. Sa plus grande nouveauté est la fusion avec le projet MU. Désormais, avec la même installation de WordPress, il sera possible de gérer soit un site normal, soit une véritable plate-forme de sites. Avec cette version, WordPress se dote d'une gestion très fine des types personnalisé...

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  • sudo command and how to auto get password

    - by user108988
    I got a problem that: I have a file .sh #!/bin/bash var=$(zenity --forms --title="T?t gi? v? yêu nhé" \ --text="V? mu?n t?t máy sau bao nhiêu phút n?a" \ --separator="," \ --add-entry="V? di?n s? vào dây") case $? in 0) sudo shutdown -h $var ;; 1) exit 0;; -1) echo "An unexpected error has occurred." ;; esac How can sudo command autofill the password from echo command or a file. I read about sudo -S options, but i dont know how it works. Anyone can give an example about it! Thanks guys

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  • 12.04 Login gone wrong

    - by Mark H
    I seem to have a fault in the login procedure somewhere. When I boooted up I found that if I selected Guest I could use the computer. If I selected my administrator account I was taken straight to the terminal. So I opened a guest session, went to user settings, and unlocked my admin account (it accepted the password!), and amended it to show "password None" and automatic login. I now find that when I boot up I am taken straight to the terminal - if I exit terminal I get the login screen - if I select the administrator login I go back to terminal - if i select guest I can use the PC, and if I select mu normal user account I can use the PC. So I cannot login as an administrator - so admin functions such as update are no longer accessible. Sorry to be so long winde but I am stuck. Can anyone suggest anything - I am a beginner with this

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  • Modeling distribution of performance measurements

    - by peterchen
    How would you mathematically model the distribution of repeated real life performance measurements - "Real life" meaning you are not just looping over the code in question, but it is just a short snippet within a large application running in a typical user scenario? My experience shows that you usually have a peak around the average execution time that can be modeled adequately with a Gaussian distribution. In addition, there's a "long tail" containing outliers - often with a multiple of the average time. (The behavior is understandable considering the factors contributing to first execution penalty). My goal is to model aggregate values that reasonably reflect this, and can be calculated from aggregate values (like for the Gaussian, calculate mu and sigma from N, sum of values and sum of squares). In other terms, number of repetitions is unlimited, but memory and calculation requirements should be minimized. A normal Gaussian distribution can't model the long tail appropriately and will have the average biased strongly even by a very small percentage of outliers. I am looking for ideas, especially if this has been attempted/analysed before. I've checked various distributions models, and I think I could work out something, but my statistics is rusty and I might end up with an overblown solution. Oh, a complete shrink-wrapped solution would be fine, too ;) Other aspects / ideas: Sometimes you get "two humps" distributions, which would be acceptable in my scenario with a single mu/sigma covering both, but ideally would be identified separately. Extrapolating this, another approach would be a "floating probability density calculation" that uses only a limited buffer and adjusts automatically to the range (due to the long tail, bins may not be spaced evenly) - haven't found anything, but with some assumptions about the distribution it should be possible in principle. Why (since it was asked) - For a complex process we need to make guarantees such as "only 0.1% of runs exceed a limit of 3 seconds, and the average processing time is 2.8 seconds". The performance of an isolated piece of code can be very different from a normal run-time environment involving varying levels of disk and network access, background services, scheduled events that occur within a day, etc. This can be solved trivially by accumulating all data. However, to accumulate this data in production, the data produced needs to be limited. For analysis of isolated pieces of code, a gaussian deviation plus first run penalty is ok. That doesn't work anymore for the distributions found above. [edit] I've already got very good answers (and finally - maybe - some time to work on this). I'm starting a bounty to look for more input / ideas.

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  • Javascript, autoload text box while typing....

    - by user299676
    I am trying to do this but keep on failing. I have a text box. I have also an array of cities, and i want, while typing to display the cities that the user should be able to select. Like if the user is typing : Mu the drop down should display Mumbai \n MuMu ... etc Like the Tags below is doing ! Does someone have any ideas in how this can be accomplished ?

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  • HARD CODED PATHS IN C#

    - by maya
    Hai I am developing an application in C#.I have a folde called myfolder it contains a file called mypoints.bmp.the folder myfolder is in my project folder it is in D drive .the path is D:\Myproject\myfolder\mypoints.bmp Now in my program whereever i need mypoints.bmp i hardcoded the wholw path.when my project is copied in to a different system under C drive i am not able to run mu project because of the hardcoded path.HOw can i give the path so that there will not be any problem evenif it is loaded in to a different system under a different drive.Please help. THAnks maya

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  • Using Regular Expression in VC++

    - by Benit
    Hi , I am finding Email ids in mu project, where I am preprocessing the input using some Regular Expression. RegExpPhone6.RegComp("[\[\{\(][ -]?[s][h][i][f][t][ -]?[+-][2][ -]?[\]\}\)]"); Here while I am compiling i am getting a warning msg like Warning 39 warning C4129: ')' : unrecognized character escape sequence How can i resolve this ? Why this is occuring and Where will it affect? Kindly help me...

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  • How to check if location services are enabled for a particular app prior to iOS 4.2?

    - by NobleK
    Hello How can I check if the user has allowed location for mu app? Normally I would use authorizationStatus method of the CLLocationManager class, but it is only available in iOS 4.2 and newer. Is it possible to achieve this somehow while still using SDK 4.2, so that the app can still run on devices with older versions of iOS, or do I have to downgrade the SDK? And along the same line, I need a similar alternative for the locationServicesEnabled method prior to iOS 4.0. Thanx

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  • DHCPv6: Provide IPv6 information in your local network

    Even though IPv6 might not be that important within your local network it might be good to get yourself into shape, and be able to provide some details of your infrastructure automatically to your network clients. This is the second article in a series on IPv6 configuration: Configure IPv6 on your Linux system DHCPv6: Provide IPv6 information in your local network Enabling DNS for IPv6 infrastructure Accessing your web server via IPv6 Piece of advice: This is based on my findings on the internet while reading other people's helpful articles and going through a couple of man-pages on my local system. IPv6 addresses for everyone (in your network) Okay, after setting up the configuration of your local system, it might be interesting to enable all your machines in your network to use IPv6. There are two options to solve this kind of requirement... Either you're busy like a bee and you go around to configure each and every system manually, or you're more the lazy and effective type of network administrator and you prefer to work with Dynamic Host Configuration Protocol (DHCP). Obviously, I'm of the second type. Enabling dynamic IPv6 address assignments can be done with a new or an existing instance of a DHCPd. In case of Ubuntu-based installation this might be isc-dhcp-server. The isc-dhcp-server allows address pooling for IP and IPv6 within the same package, you just have to run to independent daemons for each protocol version. First, check whether isc-dhcp-server is already installed and maybe running your machine like so: $ service isc-dhcp-server6 status In case, that the service is unknown, you have to install it like so: $ sudo apt-get install isc-dhcp-server Please bear in mind that there is no designated installation package for IPv6. Okay, next you have to create a separate configuration file for IPv6 address pooling and network parameters called /etc/dhcp/dhcpd6.conf. This file is not automatically provided by the package, compared to IPv4. Again, use your favourite editor and put the following lines: $ sudo nano /etc/dhcp/dhcpd6.conf authoritative;default-lease-time 14400; max-lease-time 86400;log-facility local7;subnet6 2001:db8:bad:a55::/64 {    option dhcp6.name-servers 2001:4860:4860::8888, 2001:4860:4860::8844;    option dhcp6.domain-search "ios.mu";    range6 2001:db8:bad:a55::100 2001:db8:bad:a55::199;    range6 2001:db8:bad:a55::/64 temporary;} Next, save the file and start the daemon as a foreground process to see whether it is going to listen to requests or not, like so: $ sudo /usr/sbin/dhcpd -6 -d -cf /etc/dhcp/dhcpd6.conf eth0 The parameters are explained quickly as -6 we want to run as a DHCPv6 server, -d we are sending log messages to the standard error descriptor (so you should monitor your /var/log/syslog file, too), and we explicitely want to use our newly created configuration file (-cf). You might also use the command switch -t to test the configuration file prior to running the server. In my case, I ended up with a couple of complaints by the server, especially reporting that the necessary lease file wouldn't exist. So, ensure that the lease file for your IPv6 address assignments is present: $ sudo touch /var/lib/dhcp/dhcpd6.leases$ sudo chown dhcpd:dhcpd /var/lib/dhcp/dhcpd6.leases Now, you should be good to go. Stop your foreground process and try to run the DHCPv6 server as a service on your system: $ sudo service isc-dhcp-server6 startisc-dhcp-server6 start/running, process 15883 Check your log file /var/log/syslog for any kind of problems. Refer to the man-pages of isc-dhcp-server and you might check out Chapter 22.6 of Peter Bieringer's IPv6 Howto. The instructions regarding DHCPv6 on the Ubuntu Wiki are not as complete as expected and it might not be as helpful as this article or Peter's HOWTO. But see for yourself. Does the client get an IPv6 address? Running a DHCPv6 server on your local network surely comes in handy but it has to work properly. The following paragraphs describe briefly how to check the IPv6 configuration of your clients, Linux - ifconfig or ip command First, you have enable IPv6 on your Linux by specifying the necessary directives in the /etc/network/interfaces file, like so: $ sudo nano /etc/network/interfaces iface eth1 inet6 dhcp Note: Your network device might be eth0 - please don't just copy my configuration lines. Then, either restart your network subsystem, or enable the device manually using the dhclient command with IPv6 switch, like so: $ sudo dhclient -6 You would either use the ifconfig or (if installed) the ip command to check the configuration of your network device like so: $ sudo ifconfig eth1eth1      Link encap:Ethernet  HWaddr 00:1d:09:5d:8d:98            inet addr:192.168.160.147  Bcast:192.168.160.255  Mask:255.255.255.0          inet6 addr: 2001:db8:bad:a55::193/64 Scope:Global          inet6 addr: fe80::21d:9ff:fe5d:8d98/64 Scope:Link          UP BROADCAST RUNNING MULTICAST  MTU:1500  Metric:1 Looks good, the client has an IPv6 assignment. Now, let's see whether DNS information has been provided, too. $ less /etc/resolv.conf # Dynamic resolv.conf(5) file for glibc resolver(3) generated by resolvconf(8)#     DO NOT EDIT THIS FILE BY HAND -- YOUR CHANGES WILL BE OVERWRITTENnameserver 2001:4860:4860::8888nameserver 2001:4860:4860::8844nameserver 192.168.1.2nameserver 127.0.1.1search ios.mu Nicely done. Windows - netsh Per description on TechNet the netsh is defined as following: "Netsh is a command-line scripting utility that allows you to, either locally or remotely, display or modify the network configuration of a computer that is currently running. Netsh also provides a scripting feature that allows you to run a group of commands in batch mode against a specified computer. Netsh can also save a configuration script in a text file for archival purposes or to help you configure other servers." And even though TechNet states that it applies to Windows Server (only), it is also available on Windows client operating systems, like Vista, Windows 7 and Windows 8. In order to get or even set information related to IPv6 protocol, we have to switch the netsh interface context prior to our queries. Open a command prompt in Windows and run the following statements: C:\Users\joki>netshnetsh>interface ipv6netsh interface ipv6>show interfaces Select the device index from the Idx column to get more details about the IPv6 address and DNS server information (here: I'm going to use my WiFi device with device index 11), like so: netsh interface ipv6>show address 11 Okay, address information has been provided. Now, let's check the details about DNS and resolving host names: netsh interface ipv6> show dnsservers 11 Okay, that looks good already. Our Windows client has a valid IPv6 address lease with lifetime information and details about the configured DNS servers. Talking about DNS server... Your clients should be able to connect to your network servers via IPv6 using hostnames instead of IPv6 addresses. Please read on about how to enable a local named with IPv6.

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