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  • I can I separate multiple logical pages in a text file I create in Perl?

    - by Micah
    So far, I've been successful with generating output to individual files by opening a file for output as part of outer loop and closing it after all output is written. I had used a counting variable ($x) and appended .txt onto it to create a filename, and had written it to the same directory as my perl script. I want to step the code up a bit, prompt for a file name from the user, open that file once and only once, and write my output one "printed letter" per page. Is this possible in plain text? From what I understand, chr(12) is an ascii line feed character and will get me close to what I want, but is there a better way? Thanks in advance, guys. :) sub PersonalizeLetters{ print "\n\n Beginning finalization of letters..."; print "\n\n I need a filename to save these letters to."; print "\n Filename > "; $OutFileName = <stdin>; chomp ($OutFileName); open(OutFile, ">$OutFileName"); for ($x=0; $x<$NumRecords; $x++){ $xIndex = (6 * $x); $clTitle = @ClientAoA[$xIndex]; $clName = @ClientAoA[$xIndex+1]; #I use this 6x multiplier because my records have 6 elements. #For this routine I'm only interested in name and title. #Reset OutLetter array #Midletter has other merged fields that aren't specific to who's receiving the letter. @OutLetter = @MiddleLetter; for ($y=0; $y<=$ifLength; $y++){ #Step through line by line and insert the name. $WorkLine = @OutLetter[$y]; $WorkLine =~ s/\[ClientTitle\]/$clTitle/; $WorkLine =~ s/\[ClientName\]/$clName/; @OutLetter[$y] = $WorkLine; } print OutFile "@OutLetter"; #Will chr(12) work here, or is there something better? print OutFile chr(12); $StatusX = $x+1; print "Writing output $StatusX of $NumRecords... \n\n"; } close(OutFile); }

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  • (PHP) 1)How to genrate Secreate key on User & Client Side ? 3) How to Compare Server side MD5 and Client side Md5 ?

    - by user557994
    /* In Below Code .. My problem is that 1) How to genrate Secreate key on User Side ? 2) How to genrate Secreate key on Client Side ? 3) How to Compare Server side MD5 and Client side Md5 ? Can you solve my problem ? */ $gid = $_GET['id']; if($gid=="") { $filename = "counter.txt"; $fp = fopen( $filename, "r" ) or die("Couldn't Generate Whiteboard"); while ( ! feof( $fp ) ) { $countfile = fgets( $fp); $countfile++; } fclose( $fp ); $fp = fopen( $filename, "w" ) or die("Couldn't generate whiteboard"); fwrite( $fp, $countfile ); fclose( $fp ); $doc = new DOMDocument('1.0', 'UTF-8'); $ele = $doc-createElement( 'root' ); $ele-nodeValue = $uvar; $doc-appendChild( $ele ); $test = $doc-save("$countfile.xml"); genkey($id); echo ""; $uvar=$_POST['msgval']; exit; } else { if($uvar == "") { $xdoc = new DOMDocument( '1.0', 'UTF-8' ); $xdoc-Load("$gid.xml"); $candidate = $xdoc-getElementsByTagName('root')-item(0); $newElement = $xdoc -createElement('root'); $txtNode = $xdoc -createTextNode ($root); $newElement - appendChild($txtNode); $candidate - appendChild($newElement); $msg = $candidate-nodeValue; } } function genkey($id) { $encrypt_key = "GJHsahakst1468464a"; $key = MD5("$id","$$encrypt_key"); return $key; } ? function sendRequest() { var uvar = document.getElementById('txtHint').value; var xmlhttp = new XMLHttpRequest(); xmlhttp.onreadystatechange = function() { if(xmlhttp.readyState == 4 && xmlhttp.status==200) { document.getElementById('txtHint').value = ""; } } xmlhttp.open("POST","post.php?id=",true); xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded"); xmlhttp.send("umsg="+uvar); return; } Msg " /

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  • How to connect two files and use the radio button?

    - by Stupefy101
    I have here a set of form from the index.php to upload a zip file, select an option then perform a converter process. <form action="" method="post" accept-charset="utf-8"> <p class="buttons"><input type="file" value="" name="zip_file"/></p> </form> <form action="index.php" method="post" accept-charset="utf-8" name="form1"> <h3><input type="radio" name="option" value="option1"/> Option1 </h3> <h3><input type="radio" name="option" value="option2"/> Option2 </h3> <h3><input type="radio" name="option" value="option3"/> Option3 </h3> <p class="buttons"><input type="submit" value="Convert"/></p> </form> In the other hand, this is my code for the upload.php that will extract the Zip file. <?php if($_FILES["zip_file"]["name"]) { $filename = $_FILES["zip_file"]["name"]; $source = $_FILES["zip_file"]["tmp_name"]; $type = $_FILES["zip_file"]["type"]; $name = explode(".", $filename); $accepted_types = array('application/zip', 'application/x-zip-compressed', 'multipart/x-zip', 'application/x-compressed'); foreach($accepted_types as $mime_type) { if($mime_type == $type) { $okay = true; break; } } $continue = strtolower($name[1]) == 'zip' ? true : false; if(!$continue) { $message = "The file you are trying to upload is not a .zip file. Please try again."; } $target_path = "C:xampp/htdocs/themer/".$filename; // change this to the correct site path if(move_uploaded_file($source, $target_path)) { $zip = new ZipArchive(); $x = $zip->open($target_path); if ($x === true) { $zip->extractTo("C:xampp/htdocs/themer/"); // change this to the correct site path $zip->close(); unlink($target_path); } $message = "Your .zip file was uploaded and unpacked."; } else { $message = "There was a problem with the upload. Please try again."; } } ?> How can i connect both files that will perform the extracting process? And how to include the codes for radio button after submission? Please Help.

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  • SVN always getting errors when I commit (Subclipse)

    - by jax
    I have setup svn on my server and Subclipse at home. I am the only developer and am mainly using it for the backup and versioning features. Everytime I commit my changes I get eighter: Out Of date errors or Tree conflicts Sometimes I even delete files and they don't delete on svn, in a directory hierarchy only the very last item will delete so I have to delete each folder one at a time. How do I avoid these errors in the future? Update: Another problem I am having is that sometimes eclipse seems to sync with the server so that when I refactor a filename it goes off to the server and does something and makes me wait, which is annoying. And for clarity, this is a typical operation: I might change a filename, move a file to a different folder then change the contents of a file. I select the 'Team menu' and click 'commit'. Then I get all these errors above.

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  • Image URL has the contentType "text/html"

    - by user1503025
    I want to implement a method to download Image from website to laptop. public static void DownloadRemoteImageFile(string uri, string fileName) { HttpWebRequest request = (HttpWebRequest)WebRequest.Create(uri); HttpWebResponse response = (HttpWebResponse)request.GetResponse(); if ((response.StatusCode == HttpStatusCode.OK || response.StatusCode == HttpStatusCode.Moved || response.StatusCode == HttpStatusCode.Redirect) && response.ContentType.StartsWith("image", StringComparison.OrdinalIgnoreCase)) { //if the remote file was found, download it using (Stream inputStream = response.GetResponseStream()) using (Stream outputStream = File.OpenWrite(fileName)) { byte[] buffer = new byte[4096]; int bytesRead; do { bytesRead = inputStream.Read(buffer, 0, buffer.Length); outputStream.Write(buffer, 0, bytesRead); } while (bytesRead != 0); } } } But the ContentType of request or response is not "image/jpg" or "image/png". They're always "text/html". I think that's why after I save them to local, they has incorrect content and I cannot view them. Can anyone has a solution here? Thanks

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  • Removing response.End() includes the html of the page.

    - by vinay_rockin
    HttpResponse response = Context.Http.Response; response.Headers.Clear(); response.Clear(); response.ContentType = Component.OverrideMimeType ? Component.MimeType : "application/download"; response.AppendHeader("Content-Disposition", String.Format("attachment; filename=\"{0}\"", HttpUtility.HtmlEncode(Path.GetFileName(file.FileName)))); response.OutputStream.Write(file.Contents, 0, file.Contents.Length); response.Flush(); // response.End(); if I use response.End() it throws exception and if I comment response.End() it includes the html of the page on which the download link resides. how want to download file with introducing this exta html. Any Idea how to fix this?

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  • Allow users to pull temporary data then delete table?

    - by JM4
    I don't know the best way to title this question but am trying to accomplish the following goal: When a client logs into their profile, they are presented with a link to download data from an existing database in CSV format. The process works, however, I would like for this data to be 'fresh' each time they click the link so my plan was - once a user has clicked the link and downloaded the CSV file, the database table would 'erase' all of its data and start fresh (be empty) until the next set of data populated it. My EXISTING CSV creation code: <?php $host = 'localhost'; $user = 'username'; $pass = 'password'; $db = 'database'; $table = 'tablename'; $file = 'export'; $link = mysql_connect($host, $user, $pass) or die("Can not connect." . mysql_error()); mysql_select_db($db) or die("Can not connect."); $result = mysql_query("SHOW COLUMNS FROM ".$table.""); $i = 0; if (mysql_num_rows($result) > 0) { while ($row = mysql_fetch_assoc($result)) { $csv_output .= $row['Field'].", "; $i++; } } $csv_output .= "\n"; $values = mysql_query("SELECT * FROM ".$table.""); while ($rowr = mysql_fetch_row($values)) { for ($j=0;$j<$i;$j++) { $csv_output .= '"'.$rowr[$j].'",'; } $csv_output .= "\n"; } $filename = $file."_".date("Y-m-d",time()); header("Content-type: application/vnd.ms-excel"); header("Content-disposition: csv" . date("Y-m-d") . ".csv"); header( "Content-disposition: filename=".$filename.".csv"); print $csv_output; exit; ?> any ideas?

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  • How do you handle exceptions from API/library in your code?

    - by 5YrsLaterDBA
    I have the following lines of code: FileInfo dbFile = new FileInfo(fileName); dbFileSize = (long)dbFile.Length / 1024;//KB There are 8 possible exceptions from new FileInfo(fileName) and dbFile.Length calls. I cannot just ignore them. I have to catch them. What you are going to do with those 8 exceptions? Catch them separately (too many lines)? Catch only ONE by catching the super Exception excepton? or ...

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  • How to optimize this script

    - by marks34
    I have written the following script. It opens a file, reads each line from it splitting by new line character and deleting first character in line. If line exists it's being added to array. Next each element of array is splitted by whitespace, sorted alphabetically and joined again. Every line is printed because script is fired from console and writes everything to file using standard output. I'd like to optimize this code to be more pythonic. Any ideas ? import sys def main(): filename = sys.argv[1] file = open(filename) arr = [] for line in file: line = line[1:].replace("\n", "") if line: arr.append(line) for line in arr: lines = line.split(" ") lines.sort(key=str.lower) line = ''.join(lines) print line if __name__ == '__main__': main()

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  • C: theory on how to extract files from an archived file

    - by donok
    In C I have created a program which can archive multiple files into an archive file via the command line. e.g. $echo 'file1/2' > file1/2.txt $./archive file1.txt file2.txt arhivedfile $cat archivedfile file1 file2 How do I create a process so that in my archivedfile I have: header file1 end header file2 end They are all stored in the archive file one after another after another. I know that perhaps a header file is needed(containing filename, size of filename, start and end of file) for extracting these files back out into their original form, but how would I go about doing this. I am stuck on where and how to start. Please could someone help me on some logic as to how to approach extracting files back out of an archived file.

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  • From Dictionary To File Python

    - by user3600560
    I am basically trying to write this information from my dictionary to this file. I have this dictionary named files = {} and it is for a filing system I am making. Anyhow it is always being update with new items, and I want those items to be uploaded to the file. Then if you exit the program the files are loaded back to the dictionary files = {}. Here is the code I have so far: file = {} for i in files: g = open(i, 'r') g.read(i) g.close() EDIT I want the contents of the dictionary to be written to a file. The items inside the dictionary are all stored like this: files[filename] = {filedate:filetext} where filename is the file's name, filedate is the date that the file was made on, and the filetext is the files contents.

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  • Paperclip generating wrong URLs in Heroku

    - by Tony
    Paperclip is generating wrong URLs in Heroku. I have an Audio model which has a mp3 field as follows: class Audio < ActiveRecord::Base has_attached_file :mp3, :storage => :s3, :s3_credentials => S3_CREDENTIALS, :bucket => S3_CREDENTIALS[:bucket], :path => ":rails_root/public/system/:attachment/:id/:style/:filename", :url => "/system/:attachment/:id/:style/:filename" I am calling audio.mp3.url from a controller, and it returns http://s3.amazonaws.com/MyApp/audios/mp3s//original/96a9ae89302fdf8462ee05eb829f2e17578b144e20120908-2-11f61zr.mp3?1347135050 instead of http://s3.amazonaws.com/MyApp/audios/mp3s/000/000/004/original/96a9ae89302fdf8462ee05eb829f2e17578b144e20120908-2-11f61zr.mp3?1347135050 (which works) Why is it missing the '000/000/004' part of the route? The same model is generating the right URL when used in a view. Any help? I am using paperclip 3.2.0 and Rails 3.1.8. Any help?

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  • Image upload not functioning correctly

    - by PurpleSmurf
    I'm still trying to get to grips with PHP, and I'm trying to make a form that uploads a picture to a database. I don't have permissions for move_uploaded_file so I'm using the copy() as an alternative. Everywhere I've seen experiencing similar problems have all been with move_uploaded_file so I'm rather stuck. Trying to copy the image from the desktop doesn't seem to be working, it's not throwing up any more PHP errors but is displaying the error message for if something goes wrong. The form sends data to two tables in the database but I'm mainly concerned with the upload not working. There's over 200 lines so I'll post a snippet of the upload code, thank you in advance: function getExtension($str) { $i = strrpos($str,"."); if (!$i) { return ""; } $l = strlen($str) - $i; $ext = substr($str,$i+1,$l); return $ext; } //This variable is used as a flag. The value is initialized with 0 (meaning no error found) and it will be changed to 1 if an errro occures. If the error occures the file will not be uploaded. $errors=0; //checks if the form has been submitted if(isset($_POST['submitted'])) { //reads the name of the file the user submitted for uploading $image=$_FILES['image']['name']; //if it is not empty if ($image) { //get the original name of the file from the clients machine $filename = stripslashes($_FILES['image']['name']); //get the extension of the file in a lower case format $extension = getExtension($filename); $extension = strtolower($extension); //if it is unknown extension, class as an error and do not upload the file, otherwise continue if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif")) { //print error message echo '<h1>Unknown extension!</h1>'; $errors=1; } else { //get the size of the image in bytes //$_FILES['image']['tmp_name'] is the temporary filename of the file in which the uploaded file was stored on the server $size=filesize($_FILES['image']['tmp_name']); //give an unique name, for example the time in unix time format $image_name=time().'.'.$extension; //the new name will be containing the full path where will be stored (images folder) $newname="/home/k0929907/www/uploads/".$image_name; //verify if the image has been uploaded, and print error instead $copied = copy($_FILES['image']['tmp_name'], $newname); if (!$copied) { echo '<h1>Copy unsuccessfull!</h1>'; $errors=1; } } } }

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  • Scanner class is skipping lines

    - by user2403304
    I'm new to programing and I'm having a problem with my scanner class. This code is in a loop and when the loop comes around the second, third whatever time I have it set to it skips the first title input. I need help please why is it skipping my title scanner input in the beginning? System.out.println("Title:"); list[i].title=keyboard.nextLine(); System.out.println("Author:"); list[i].author=keyboard.nextLine(); System.out.println("Album:"); list[i].album=keyboard.nextLine(); System.out.println("Filename:"); list[i].filename=keyboard.nextLine();

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  • How to select random image of specific size using Django / Python?

    - by Jonathan
    I've been using this little snippet to select random images. However I would like to change it to select only images of a certain size. I'm running into trouble checking against image size. If I use get_image_dimensions() I need to use a conditional statement, which then requires that I allow exceptions. So, I guess I need some pointers on just limiting by image dimensions. Thanks. import os import random import posixpath from django import template from django.conf import settings register = template.Library() def is_image_file(filename): """Does `filename` appear to be an image file?""" img_types = [".jpg", ".jpeg", ".png", ".gif"] ext = os.path.splitext(filename)[1] return ext in img_types @register.simple_tag def random_image(path): """ Select a random image file from the provided directory and return its href. `path` should be relative to MEDIA_ROOT. Usage: <img src='{% random_image "images/whatever/" %}'> """ fullpath = os.path.join(settings.MEDIA_ROOT, path) filenames = [f for f in os.listdir(fullpath) if is_image_file(f)] pick = random.choice(filenames) return posixpath.join(settings.MEDIA_URL, path, pick)

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  • Read video information(date created)?

    - by Lynx
    In window, i can get the date created of the video from properties(right click). I have a few idea on this but i dont know how to do it. 1. Get the video information directly from video(like in windows), 2. By extracting the video name to get the date created(The video's name is in date format, which is the time it created). And i also using taglib-sharp to get the video duration and resolution, but i cant find any sample code on how to get the video creation date. Note: video name in date format - example, 20121119_125550.avi Edit Found this code and so far its working string fileName = Server.MapPath("//video//20121119_125550.avi"); FileInfo fileInfo = new FileInfo(fileName); DateTime creationTime = fileInfo.CreationTime; Output: 2012/11/19 12:55:50 For the file's name, i will add another string in name. For example User1-20121119_125550.avi.avi, so it will get complicated after that.

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  • C++ Regarding cin.ignore()

    - by user1578897
    i would hope someone can modify my code as its so buggy. sometime its work, sometime it dont.. so let me explain more.. Text file data is as below Line3D, [70, -120, -3], [-29, 1, 268] Line3D, [25, -69, -33], [-2, -41, 58] To read the above line.. i use the following char buffer[30]; cout << "Please enter filename: "; cin.ignore(); getline(cin,filename); readFile.open(filename.c_str()); //if successfully open if(readFile.is_open()) { //record counter set to 0 numberOfRecords = 0; while(readFile.good()) { //input stream get line by line readFile.getline(buffer,20,','); if(strstr(buffer,"Point3D")) { Point3D point3d_tmp; readFile>>point3d_tmp; // and so on... Then i did a overload on the ifstream for Line3d ifstream& operator>>(ifstream &input,Line3D &line3d) { int x1,y1,z1,x2,y2,z2; //get x1 input.ignore(2); input>>x1; //get y1 input.ignore(); input>>y1; //get z1 input.ignore(); input>>z1; //get x2 input.ignore(4); input>>x2; //get y2 input.ignore(); input>>y2; //get z2 input.ignore(); input>>z2; input.ignore(2); Point3D pt1(x1,y1,z1); Point3D pt2(x2,y2,z2); line3d.setPt1(pt1); line3d.setPt2(pt2); line3d.setLength(); } But the issue is the record work sometime and sometime it dont.. what i mean is if at this point //i add a cout cout << x1 << y1 << z1; cout << x2 << y2 << z2; //its works! Point3D pt1(x1,y1,z1); Point3D pt2(x2,y2,z2); line3d.setPt1(pt1); line3d.setPt2(pt2); line3d.setLength(); but if i take away the cout it dont work. how do i change my cin.ignore() so the data can be handle properly , consider number range is -999 to 999

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  • Why doesn't this work?

    - by Kyle W
    take = raw_input('Please enter the string of numbers that compose code\n\n\t') y = str(take) l = [] for i in xrange(0, len(y), 3):         l.append(str(y[i:i+3])) b = len(l) a = 0 while(a!=b):         c = l[a].replace('444', ' ')         c = l[a].replace('111', 'a')         c = l[a].replace('112', 'b')         c = l[a].replace('113', 'c')         c = l[a].replace('114', 'd')         c = l[a].replace('115', 'e')         etc...         a = a + 1 filename = 'decmes.txt' file = open(filename, 'w') file.write(c) file.close() I can enter anything, just 111 for example and it gives me back the same thing I put in. Maybe it's something dumb, but I can't figure it out.

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  • find string from the file in somewhere

    - by lightmanhk
    I want to find a string from some file in subdirectory. Like we are in bundle/. and in bundle/ there are multiple subdirectories and multiple txt files I want to do something like find . -type f -exec grep "\<F8\>" {} \; want to get the file where it contain string < F8 this command does work, find the string, but never return filename I hope anyone can give me a better solution to this, like display filename along with the line containing that string

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  • OverRiding Help

    - by user445714
    Few questions on over riding. I am interiting a method openRead from another class, the method is to be overridden so that the scanner class uses the provided delimiter pattern, which is referenced. I need to make use of the scanner class useDelmiter method method from another class [code] public boolean openRead() throws FileNotFoundException { sc = new Scanner(new File(fileName)); if (fileName != null) { return true; } else { return false; } } [/code] delimiter [code] protected final String DELIMITERS = "[\s[^'a-zA-Z]]"; [/code] I'm at a loss to how i over ride this using the constant delimiter.

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  • Java - ClassNotFoundException I have the class included, but why do I get an exception?

    - by aloh
    public static Reservation[] openBinaryFile( String fileName ) { Reservation [] objArray = null; try { ObjectInputStream inStream = new ObjectInputStream( new FileInputStream( fileName ) ); objArray = (Reservation[])inStream.readObject(); inStream.close(); } catch( ClassNotFoundException e ) { System.out.println( "Exception: ClassNotFoundException." ); } I have the class included, but why do I get an exception? The class is in the same package as the others. Why am I getting this exception?

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  • PHP export to excel

    - by user1865240
    I'm having a trouble that I can't export japanese texts to excel (xls). I used the following codes: header('Content-type: application/ms-excel;charset=UTF-8'); header('Content-Disposition: attachment; filename='.$filename); header("Pragma: no-cache"); echo $contents; But in the excel file, the text changed to funny characters like this: é™?定ç‰? ã?¨ã??ã?¯ã??ã?£ã?†ã?ªã?¢å??犬ã?®ã?Œæ??ã? ’è??ã??ã?Ÿã?†ã?£ã??ã??ã??ã?? ï?? Currently, I'm using hostingmanager and I tried on the different server using the same codes and there's no problem. What could be the problem. Because of the PHP version?? Please help me. Thank you, Aino

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  • speeding up parsing of files

    - by user248237
    the following function parses a CSV file into a list of dictionaries, where each element in the list is a dictionary where the values are indexed by the header of the file (assumed to be the first line.) this function is very very slow, taking ~6 seconds for a file that's relatively small (less than 30,000 lines.) how can I speed it up? def csv2dictlist_raw(filename, delimiter='\t'): f = open(filename) header_line = f.readline().strip() header_fields = header_line.split(delimiter) dictlist = [] # convert data to list of dictionaries for line in f: values = map(tryEval, line.strip().split(delimiter)) dictline = dict(zip(header_fields, values)) dictlist.append(dictline) return (dictlist, header_fields) thanks.

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  • Allow users to pull temporary data then delete table data (headers remain)?

    - by JM4
    I don't know the best way to title this question but am trying to accomplish the following goal: When a client logs into their profile, they are presented with a link to download data from an existing database in CSV format. The process works, however, I would like for this data to be 'fresh' each time they click the link so my plan was - once a user has clicked the link and downloaded the CSV file, the database table would 'erase' all of its data and start fresh (be empty) until the next set of data populated it. My EXISTING CSV creation code: <?php $host = 'localhost'; $user = 'username'; $pass = 'password'; $db = 'database'; $table = 'tablename'; $file = 'export'; $link = mysql_connect($host, $user, $pass) or die("Can not connect." . mysql_error()); mysql_select_db($db) or die("Can not connect."); $result = mysql_query("SHOW COLUMNS FROM ".$table.""); $i = 0; if (mysql_num_rows($result) > 0) { while ($row = mysql_fetch_assoc($result)) { $csv_output .= $row['Field'].", "; $i++; } } $csv_output .= "\n"; $values = mysql_query("SELECT * FROM ".$table.""); while ($rowr = mysql_fetch_row($values)) { for ($j=0;$j<$i;$j++) { $csv_output .= '"'.$rowr[$j].'",'; } $csv_output .= "\n"; } $filename = $file."_".date("Y-m-d",time()); header("Content-type: application/vnd.ms-excel"); header("Content-disposition: csv" . date("Y-m-d") . ".csv"); header( "Content-disposition: filename=".$filename.".csv"); print $csv_output; exit; ?> any ideas?

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  • C# windows service location

    - by user743246
    By using following code[C#], we can know the path where googlechrome is installed. But, this code first starts chrome.exe, then takes its path. My question is without starting chrome.exe, how can I know the path? ProcessStartInfo startInfo = new ProcessStartInfo(); startInfo.FileName = "chrome.exe"; string argmnt = @"-g"; startInfo.Arguments = argmnt; String path = null; try { Process p = Process.Start(startInfo); path = p.MainModule.FileName; p.Kill(); } catch (Exception e) { return String.Empty; }

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