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  • How to compare two file structures in PHP?

    - by OM The Eternity
    I have a function which gives me the complete file structure upto n-level, function getDirectory($path = '.', $ignore = '') { $dirTree = array (); $dirTreeTemp = array (); $ignore[] = '.'; $ignore[] = '..'; $dh = @opendir($path); while (false !== ($file = readdir($dh))) { if (!in_array($file, $ignore)) { if (!is_dir("$path/$file")) { //display of file and directory name with their modification time $stat = stat("$path/$file"); $statdir = stat("$path"); $dirTree["$path"][] = $file. " === ". date('Y-m-d H:i:s', $stat['mtime']) . " Directory == ".$path."===". date('Y-m-d H:i:s', $statdir['mtime']) ; } else { $dirTreeTemp = getDirectory("$path/$file", $ignore); if (is_array($dirTreeTemp))$dirTree = array_merge($dirTree, $dirTreeTemp); } } } closedir($dh); return $dirTree; } $ignore = array('.htaccess', 'error_log', 'cgi-bin', 'php.ini', '.ftpquota'); //function call $dirTree = getDirectory('.', $ignore); //file structure array print print_r($dirTree); Now here my requirement is , I have two sites The Development/Test Site- where i do testing of all the changes The Production Site- where I finally post all the changes as per test in development site Now, for example, I have tested an image upload in the Development/test site, and i found it appropriate to publish on Production site then i will completely transfer the Development/Test DB detail to Production DB, but now I want to compare the files structure as well to transfer the corresponding image file to Production folder. There could be the situation when I update the image by editing the image and upload it with same name, now in this case the image file would be already present there, which will restrict the use of "file_exist" logic, so for these type of situations....HOW CAN I COMPARE THE TWO FILE STRUCTURE TO GET THE SYNCHRONIZATION DONE AS PER REQUIREMENT??

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  • Modifying File while in use using Java

    - by Marquinio
    Hi all, I have this recurrent Java JAR program tasks that tries to modify a file every 60seconds. Problem is that if user is viewing the file than Java program will not be able to modify the file. I get the typical IOException. Anyone knows if there is a way in Java to modify a file currently in use? Or anyone knows what would be the best way to solve this problem? I was thinking of using the File canRead(), canWrite() methods to check if file is in use. If file is in use then I'm thinking of making a backup copy of data that could not be written. Then after 60 seconds add some logic to check if backup file is empty or not. If backup file is not empty then add its contents to main file. If empty then just add new data to main file. Of course, the first thing I will always do is check if file is in use. Thanks for all your ideas.

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  • File mkdirs() method not working in android/java

    - by Leif Andersen
    I've been pulling out my hair on this for a while now. The following method is supposed to download a file, and save it to the location specified on the hard drive. private static void saveImage(Context context, boolean backgroundUpdate, URL url, File file) { if (!Tools.checkNetworkState(context, backgroundUpdate)) return; // Get the image try { // Make the file file.getParentFile().mkdirs(); // Set up the connection URLConnection uCon = url.openConnection(); InputStream is = uCon.getInputStream(); BufferedInputStream bis = new BufferedInputStream(is); // Download the data ByteArrayBuffer baf = new ByteArrayBuffer(50); int current = 0; while ((current = bis.read()) != -1) { baf.append((byte) current); } // Write the bits to the file OutputStream os = new FileOutputStream(file); os.write(baf.toByteArray()); os.close(); } catch (Exception e) { // Any exception is probably a newtork faiilure, bail return; } } Also, if the file doesn't exist, it is supposed to make the directory for the file. (And if there is another file already in that spot, it should just not do anything). However, for some reason, the mkdirs() method never makes the directory. I've tried everything from explicit parentheses, to explicitly making the parent file class, and nothing seems to work. I'm fairly certain that the drive is writable, as it's only called after that has already been determined, also that is true after running through it while debugging. So the method fails because the parent directories aren't made. Can anyone tell me if there is anything wrong with the way I'm calling it? Also, if it helps, here is the source for the file I'm calling it in: https://github.com/LeifAndersen/NetCatch/blob/master/src/net/leifandersen/mobile/android/netcatch/services/RSSService.java Thank you

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  • Django rewrites URL as IP address in browser - why?

    - by Mitch
    I am using django, nginx and apache. When I access my site with a URL (e.g., http://www.foo.com/) what appears in my browser address is the IP address with admin appended (e.g., http://123.45.67.890/admin/). When I access the site by IP, it is redirected as expected by django's urls.py (e.g., http://123.45.67.890/ - http://123.45.67.890/accounts/login/?next=/) I would like to have the name URL act the same way as the IP. That is, if the URL goes to a new view, the host in the browser address should remain the same and not change to the IP address. Where should I be looking to fix this? My files: ; cpa.com (apache) NameVirtualHost *:8080 <VirtualHost *:8080> AddOutputFilterByType DEFLATE text/html text/plain text/xml text/css text/javascript application/javascript application/x-javascript BrowserMatch ^Mozilla/4 gzip-only-text/html BrowserMatch ^Mozilla/4\.0[678] no-gzip BrowserMatch \bMSIE !no-gzip !gzip-only-text/htm DocumentRoot /path/to/root ServerName www.foo.com <IfModule mod_rpaf.c> RPAFenable On RPAFsethostname On RPAFproxy_ips 127.0.0.1 </IfModule> <Directory /public/static> AllowOverride None AddHandler mod_python .py PythonHandler mod_python.publisher </Directory> Alias / /dj <Location /> SetHandler python-program PythonPath "['/usr/lib/python2.5/site-packages/django', '/usr/lib/python2.5/site-packages/django/forms'] + sys.path" PythonHandler django.core.handlers.modpython SetEnv DJANGO_SETTINGS_MODULE dj.settings PythonDebug On </Location> </VirtualHost> ; ; ports.conf (apache) Listen 127.0.0.1:8080 ; ; cpa.conf (nginx) server { listen 80; server_name www.foo.com; location /static { root /var/public; index index.html; } location /cpa/js { root /var/public/js; } location /cpa/css { root /var/public/css; } location /djmedia { alias "/usr/lib/python2.5/site-packages/django/contrib/admin/media/"; } location / { include /etc/nginx/proxy.conf; proxy_set_header X-Forwarded-For $proxy_add_x_forwarded_for; proxy_pass http://127.0.0.1:8080; } } ; ; proxy.conf (nginx) proxy_redirect off; proxy_set_header Host $host; proxy_set_header X-Real-IP $remote_addr; proxy_set_header X-Forwarded-For $proxy_add_x_forwarded_for; client_max_body_size 10m; client_body_buffer_size 128k; proxy_connect_timeout 90; proxy_send_timeout 90; proxy_read_timeout 500; proxy_buffers 32 4k;

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  • How Do I Disable URL Pre-Pending in the FireFox 3 Title Bar When Opeing A New Window With JavaScript

    - by N Rahl
    For (understandable) security reasons, Firefox does not allow JavaScript to open a new window without the address/location bar AND without pre-pending the page's URL to the title in the title bar. For example, when you set: <title>My Site</title> in the header, and open the page using location=no FireFox changes the header to read: http://www.mysite.com - My Site - Mozilla Firefox. I would like it to simply say: My Site Everything I've read suggests this behaviour can't be altered with scripting, and as such, this is not a scripting question. What I would like to know is, which setting(s) can I change in the browser itself to disable URL pre-pending to the title of new windows? This is for a company Intranet, and I control all of the computers/browsers that connect to the application.

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  • Unset the system immutable bit in Mac OS X

    - by skylarking
    In theory I believe you can unlock and remove the system immutable bit with: chflags noschg /Path/To/File But how can you do this when you've set the bit as root? I have a file that is locked, and even running this command as root will not work as the operation is not permitted. I tried logging in as Single-User mode to no avail. I seem to remember that even though you are in as root you are in at level '1'. And to be able to remove the system-immutable flag you need to be logged in at level '0'. Does this have something to do with this issue?

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  • How do I remove a URL from Google without having to have a Google E-mail Account

    - by PP
    Really simple question. I do not want a Google account. I just want Google to stop making requests every 2 minutes for a URL it should never have known about (apparently Google harvests URLs from search requests as well as private e-mails, not just from actual web pages). But when I search Google help for removing URLs it appears I have to use their "webmaster tools" which require logging into a GMail account! How do I tell Google not to index my URL without becoming a customer? Note: I already return 404 for the URLs in question using a rewrite rule - this appears to make zero difference to the crawler which continually attempts to fetch the page every 2 minutes.

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  • Nginx fastcgi problems with django (double slashes in url?)

    - by wizard
    I'm deploying my first django app. I'm familiar with nginx and fastcgi from deploying php-fpm. I can't get python to recognize the urls. I'm also at a loss on how to debug this further. I'd welcome solutions to this problem and tips on debugging fastcgi problems. Currently I get a 404 page regardless of the url and for some reason a double slash For http://www.site.com/admin/ Page not found (404) Request Method: GET Request URL: http://www.site.com/admin// My urls.py from the debug output - which work in the dev server. Using the URLconf defined in ahrlty.urls, Django tried these URL patterns, in this order: ^listings/ ^admin/ ^accounts/login/$ ^accounts/logout/$ my nginx config server { listen 80; server_name beta.ahrlty.com; access_log /home/ahrlty/ahrlty/logs/access.log; error_log /home/ahrlty/ahrlty/logs/error.log; location /static/ { alias /home/ahrlty/ahrlty/ahrlty/static/; break; } location /media/ { alias /usr/lib/python2.6/dist-packages/django/contrib/admin/media/; break; } location / { include /etc/nginx/fastcgi_params; fastcgi_pass 127.0.0.1:8001; break; } } and my fastcgi_params fastcgi_param QUERY_STRING $query_string; fastcgi_param REQUEST_METHOD $request_method; fastcgi_param CONTENT_TYPE $content_type; fastcgi_param CONTENT_LENGTH $content_length; fastcgi_param SCRIPT_NAME $fastcgi_script_name; fastcgi_param REQUEST_URI $request_uri; fastcgi_param DOCUMENT_URI $document_uri; fastcgi_param DOCUMENT_ROOT $document_root; fastcgi_param SERVER_PROTOCOL $server_protocol; fastcgi_param GATEWAY_INTERFACE CGI/1.1; fastcgi_param SERVER_SOFTWARE nginx/$nginx_version; fastcgi_param REMOTE_ADDR $remote_addr; fastcgi_param REMOTE_PORT $remote_port; fastcgi_param SERVER_ADDR $server_addr; fastcgi_param SERVER_PORT $server_port; fastcgi_param SERVER_NAME $server_name; fastcgi_param PATH_INFO $fastcgi_script_name; # PHP only, required if PHP was built with --enable-force-cgi-redirect fastcgi_param REDIRECT_STATUS 200; And lastly I'm running fastcgi from the commandline with django's manage.py. python manage.py runfcgi method=threaded host=127.0.0.1 port=8080 pidfile=mysite.pid minspare=4 maxspare=30 daemonize=false I'm having a hard time debugging this one. Does anything jump out at anybody? Notes nginx version: nginx/0.7.62 Django svn trunk rev 13013

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  • Is there a direct URL to download Java JDK updates?

    - by Bob Cross
    I have a whole set of machines that are on the other side of a firewall configured to prevent all Javascript from functioning. All of them (Linux 32 and 64 bit configurations) must be updated to Java 6 update 20. This is a problem given Sun/Oracle's URL redirector and download manager: they simply don't appear or don't work. Is there a URL to download the JDK updates and bypass the redirect? Obviously, a yum configuration that would allow for automatic updates would be optimal but I'd be happy to just have the rpm file.

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  • Reliable file copy (move) process - mostly Unix/Linux

    - by mfinni
    Short story : We have a need for a rock-solid reliable file mover process. We have source directories that are often being written to that we need to move files from. The files come in pairs - a big binary, and a small XML index. We get a CTL file that defines these file bundles. There is a process that operates on the files once they are in the destination directory; that gets rid of them when it's done. Would rsync do the best job, or do we need to get more complex? Long story as follows : We have multiple sources to pull from : one set of directories are on a Windows machine (that does have Cygwin and an SSH daemon), and a whole pile of directories are on a set of SFTP servers (Most of these are also Windows.) Our destinations are a list of directories on AIX servers. We used to use a very reliable Perl script on the Windows/Cygwin machine when it was our only source. However, we're working on getting rid of that machine, and there are other sources now, the SFTP servers, that we cannot presently run our own scripts on. For security reasons, we can't run the copy jobs on our AIX servers - they have no access to the source servers. We currently have a homegrown Java program on a Linux machine that uses SFTP to pull from the various new SFTP source directories, copies to a local tmp directory, verifies that everything is present, then copies that to the AIX machines, and then deletes the files from the source. However, we're finding any number of bugs or poorly-handled error checking. None of us are Java experts, so fixing/improving this may be difficult. Concerns for us are: With a remote source (SFTP), will rsync leave alone any file still being written? Some of these files are large. From reading the docs, it seems like rysnc will be very good about not removing the source until the destination is reliably written. Does anyone have experience confirming or disproving this? Additional info We will be concerned about the ingestion process that operates on the files once they are in the destination directory. We don't want it operating on files while we are in the process of copying them; it waits until the small XML index file is present. Our current copy job are supposed to copy the XML file last. Sometimes the network has problems, sometimes the SFTP source servers crap out on us. Sometimes we typo the config files and a destination directory doesn't exist. We never want to lose a file due to this sort of error. We need good logs If you were presented with this, would you just script up some rsync? Or would you build or buy a tool, and if so, what would it be (or what technologies would it use?) I (and others on my team) are decent with Perl.

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  • Posting from ASP.NET WebForms page to another URL

    - by hajan
    Few days ago I had a case when I needed to make FORM POST from my ASP.NET WebForms page to an external site URL. More specifically, I was working on implementing Simple Payment System (like Amazon, PayPal, MoneyBookers). The operator asks to make FORM POST request to a given URL in their website, sending parameters together with the post which are computed on my application level (access keys, secret keys, signature, return-URL… etc). So, since we are not allowed nesting another form inside the <form runat=”server”> … </form>, which is required because other controls in my ASPX code work on server-side, I thought to inject the HTML and create FORM with method=”POST”. After making some proof of concept and testing some scenarios, I’ve concluded that I can do this very fast in two ways: Using jQuery to create form on fly with the needed parameters and make submit() Using HttpContext.Current.Response.Write to write the form on server-side (code-behind) and embed JavaScript code that will do the post Both ways seemed fine. 1. Using jQuery to create FORM html code and Submit it. Let’s say we have ‘PAY NOW’ button in our ASPX code: <asp:Button ID="btnPayNow" runat="server" Text="Pay Now" /> Now, if we want to make this button submit a FORM using POST method to another website, the jQuery way should be as follows: <script src="http://ajax.aspnetcdn.com/ajax/jquery/jquery-1.5.1.js" type="text/javascript"></script> <script type="text/javascript">     $(function () {         $("#btnPayNow").click(function (event) {             event.preventDefault();             //construct htmlForm string             var htmlForm = "<form id='myform' method='POST' action='http://www.microsoft.com'>" +                 "<input type='hidden' id='name' value='hajan' />" +             "</form>";             //Submit the form             $(htmlForm).appendTo("body").submit();         });     }); </script> Yes, as you see, the code fires on btnPayNow click. It removes the default button behavior, then creates htmlForm string. After that using jQuery we append the form to the body and submit it. Inside the form, you can see I have set the htttp://www.microsoft.com URL, so after clicking the button you should be automatically redirected to the Microsoft website (just for test, of course for Payment I’m using Operator's URL). 2. Using HttpContext.Current.Response.Write to write the form on server-side (code-behind) and embed JavaScript code that will do the post The C# code behind should be something like this: public void btnPayNow_Click(object sender, EventArgs e) {     string Url = "http://www.microsoft.com";     string formId = "myForm1";     StringBuilder htmlForm = new StringBuilder();     htmlForm.AppendLine("<html>");     htmlForm.AppendLine(String.Format("<body onload='document.forms[\"{0}\"].submit()'>",formId));     htmlForm.AppendLine(String.Format("<form id='{0}' method='POST' action='{1}'>", formId, Url));     htmlForm.AppendLine("<input type='hidden' id='name' value='hajan' />");     htmlForm.AppendLine("</form>");     htmlForm.AppendLine("</body>");     htmlForm.AppendLine("</html>");     HttpContext.Current.Response.Clear();     HttpContext.Current.Response.Write(htmlForm.ToString());     HttpContext.Current.Response.End();             } So, with this code we create htmlForm string using StringBuilder class and then just write the html to the page using HttpContext.Current.Response.Write. The interesting part here is that we submit the form using JavaScript code: document.forms["myForm1"].submit() This code runs on body load event, which means once the body is loaded the form is automatically submitted. Note: In order to test both solutions, create two applications on your web server and post the form from first to the second website, then get the values in the second website using Request.Form[“input-field-id”] I hope this was useful post for you. Regards, Hajan

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  • Copy mdf file and use it in run time

    - by Anibas
    After I copy mdf file (and his log file) I tries to Insert data. I receive the following message: "An attempt to attach an auto-named database for file [fileName].mdf failed. A database with the same name exists, or specified file cannot be opened, or it is located on UNC share. When I copied the file manual everything worked normally. Is it correct the order File.Copy leaves the file engaged?

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  • Parse a text file into multiple text file

    - by Vijay Kumar Singh
    I want to get multiple file by parsing a input file Through Java. The Input file contains many fasta format of thousands of protein sequence and I want to generate raw format(i.e., without any comma semicolon and without any extra symbol like "", "[", "]" etc) of each protein sequence. A fasta sequence starts form "" symbol followed by description of protein and then sequence of protein. For example ? lcl|NC_000001.10_cdsid_XP_003403591.1 [gene=LOC100652771] [protein=hypothetical protein LOC100652771] [protein_id=XP_003403591.1] [location=join(12190..12227,12595..12721,13403..13639)] MSESINFSHNLGQLLSPPRCVVMPGMPFPSIRSPELQKTTADLDHTLVSVPSVAESLHHPEITFLTAFCL PSFTRSRPLPDRQLHHCLALCPSFALPAGDGVCHGPGLQGSCYKGETQESVESRVLPGPRHRH Like above formate the input file contains 1000s of protein sequence. I have to generate thousands of raw file containing only individual protein sequence without any special symbol or gaps. I have developed the code for it in Java but out put is : Cannot open a file followed by cannot find file. Please help me to solve my problem. Regards Vijay Kumar Garg Varanasi Bharat (India) The code is /*Java code to convert FASTA format to a raw format*/ import java.io.*; import java.util.*; import java.util.regex.*; import java.io.FileInputStream; // java package for using regular expression public class Arrayren { public static void main(String args[]) throws IOException { String a[]=new String[1000]; String b[][] =new String[1000][1000]; /*open the id file*/ try { File f = new File ("input.txt"); //opening the text document containing genbank ids FileInputStream fis = new FileInputStream("input.txt"); //Reading the file contents through inputstream BufferedInputStream bis = new BufferedInputStream(fis); // Writing the contents to a buffered stream DataInputStream dis = new DataInputStream(bis); //Method for reading Java Standard data types String inputline; String line; String separator = System.getProperty("line.separator"); // reads a line till next line operator is found int i=0; while ((inputline=dis.readLine()) != null) { i++; a[i]=inputline; a[i]=a[i].replaceAll(separator,""); //replaces unwanted patterns like /n with space a[i]=a[i].trim(); // trims out if any space is available a[i]=a[i]+".txt"; //takes the file name into an array try // to handle run time error /*take the sequence in to an array*/ { BufferedReader in = new BufferedReader (new FileReader(a[i])); String inline = null; int j=0; while((inline=in.readLine()) != null) { j++; b[i][j]=inline; Pattern q=Pattern.compile(">"); //Compiling the regular expression Matcher n=q.matcher(inline); //creates the matcher for the above pattern if(n.find()) { /*appending the comment line*/ b[i][j]=b[i][j].replaceAll(">gi",""); //identify the pattern and replace it with a space b[i][j]=b[i][j].replaceAll("[a-zA-Z]",""); b[i][j]=b[i][j].replaceAll("|",""); b[i][j]=b[i][j].replaceAll("\\d{1,15}",""); b[i][j]=b[i][j].replaceAll(".",""); b[i][j]=b[i][j].replaceAll("_",""); b[i][j]=b[i][j].replaceAll("\\(",""); b[i][j]=b[i][j].replaceAll("\\)",""); } /*printing the sequence in to a text file*/ b[i][j]=b[i][j].replaceAll(separator,""); b[i][j]=b[i][j].trim(); // trims out if any space is available File create = new File(inputline+"R.txt"); try { if(!create.exists()) { create.createNewFile(); // creates a new file } else { System.out.println("file already exists"); } } catch(IOException e) // to catch the exception and print the error if cannot open a file { System.err.println("cannot create a file"); } BufferedWriter outt = new BufferedWriter(new FileWriter(inputline+"R.txt", true)); outt.write(b[i][j]); // printing the contents to a text file outt.close(); // closing the text file System.out.println(b[i][j]); } } catch(Exception e) { System.out.println("cannot open a file"); } } } catch(Exception ex) // catch the exception and prints the error if cannot find file { System.out.println("cannot find file "); } } } If you provide me correct it will be much easier to understand.

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  • Designing a database file format

    - by RoliSoft
    I would like to design my own database engine for educational purposes, for the time being. Designing a binary file format is not hard nor the question, I've done it in the past, but while designing a database file format, I have come across a very important question: How to handle the deletion of an item? So far, I've thought of the following two options: Each item will have a "deleted" bit which is set to 1 upon deletion. Pro: relatively fast. Con: potentially sensitive data will remain in the file. 0x00 out the whole item upon deletion. Pro: potentially sensitive data will be removed from the file. Con: relatively slow. Recreating the whole database. Pro: no empty blocks which makes the follow-up question void. Con: it's a really good idea to overwrite the whole 4 GB database file because a user corrected a typo. I will sell this method to Twitter ASAP! Now let's say you already have a few empty blocks in your database (deleted items). The follow-up question is how to handle the insertion of a new item? Append the item to the end of the file. Pro: fastest possible. Con: file will get huge because of all the empty blocks that remain because deleted items aren't actually deleted. Search for an empty block exactly the size of the one you're inserting. Pro: may get rid of some blocks. Con: you may end up scanning the whole file at each insert only to find out it's very unlikely to come across a perfectly fitting empty block. Find the first empty block which is equal or larger than the item you're inserting. Pro: you probably won't end up scanning the whole file, as you will find an empty block somewhere mid-way; this will keep the file size relatively low. Con: there will still be lots of leftover 0x00 bytes at the end of items which were inserted into bigger empty blocks than they are. Rigth now, I think the first deletion method and the last insertion method are probably the "best" mix, but they would still have their own small issues. Alternatively, the first insertion method and scheduled full database recreation. (Probably not a good idea when working with really large databases. Also, each small update in that method will clone the whole item to the end of the file, thus accelerating file growth at a potentially insane rate.) Unless there is a way of deleting/inserting blocks from/to the middle of the file in a file-system approved way, what's the best way to do this? More importantly, how do databases currently used in production usually handle this?

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  • Changing a url depending on what link chosen (HTML) no asp.

    - by Ozaki
    TLDR I need to change a javascript variable on the same page after clicking a link (can be from a different page) so that the getjson request pulls different data without having to duplicate on html pages. I am using some getJSON requests with Jquery, to make calls to populate my pages. I want to be able to (in plain HTML / javascript) when the user clicks say "link 1" or "link 2" to open the same page (say page.html) but change the get request url to "link 1" or "link 2". Page.html var url = ??; $.getJSON(url, function(data){} link 1 var url = host/link1 <a href="page.html">link1</a> link2 var url = host/link2 <a href="page.html">link2</a> So I call the same page but am able to populate it with different content. Purposely staying away from asp. Was thinking maybe of inserting the content into a div after page load so the url can be set or something along those lines. Any ideas how I might go about this?

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  • Reading data from text file in C

    - by themake
    I have a text file which contains words separated by space. I want to take each word from the file and store it. So i have opened the file but am unsure how to assign the word to a char. FILE *fp; fp = fopen("file.txt", "r"); //then i want char one = the first word in the file char two = the second word in the file

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  • opening and viewing a file in php

    - by Christian Burgos
    how do i open/view for editing an uploaded file in php? i have tried this but it doesn't open the file. $my_file = 'file.txt'; $handle = fopen($my_file, 'r'); $data = fread($handle,filesize($my_file)); i've also tried this but it wont work. $my_file = 'file.txt'; $handle = fopen($my_file, 'w') or die('Cannot open file: '.$my_file); $data = 'This is the data'; fwrite($handle, $data); what i have in mind is like when you want to view an uploaded resume,documents or any other ms office files like .docx,.xls,.pptx and be able to edit them, save and close the said file. edit: latest tried code... <?php // Connects to your Database include "configdb.php"; //Retrieves data from MySQL $data = mysql_query("SELECT * FROM employees") or die(mysql_error()); //Puts it into an array while($info = mysql_fetch_array( $data )) { //Outputs the image and other data //Echo "<img src=localhost/uploadfile/images".$info['photo'] ."> <br>"; Echo "<b>Name:</b> ".$info['name'] . "<br> "; Echo "<b>Email:</b> ".$info['email'] . " <br>"; Echo "<b>Phone:</b> ".$info['phone'] . " <hr>"; //$file=fopen("uploadfile/images/".$info['photo'],"r+"); $file=fopen("Applications/XAMPP/xamppfiles/htdocs/uploadfile/images/file.odt","r") or exit("unable to open file");; } ?> i am getting the error: Warning: fopen(Applications/XAMPP/xamppfiles/htdocs/uploadfile/images/file.odt): failed to open stream: No such file or directory in /Applications/XAMPP/xamppfiles/htdocs/uploadfile/view.php on line 17 unable to open file the file is in that folder, i don't know it wont find it.

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  • Copied a file with winscp; only winscp can see it

    - by nilbus
    I recently copied a 25.5GB file from another machine using WinSCP. I copied it to C:\beth.tar.gz, and WinSCP can still see the file. However no other app (including Explorer) can see the file. What might cause this, and how can I fix it? The details that might or might not matter WinSCP shows the size of the file (C:\beth.tar.gz) correctly as 27,460,124,080 bytes, which matches the filesize on the remote host Neither explorer, cmd (command line prompt w/ dir C:\), the 7Zip archive program, nor any other File Open dialog can see the beth.tar.gz file under C:\ I have configured Explorer to show hidden files I can move the file to other directories using WinSCP If I try to move the file to Users/, UAC prompts me for administrative rights, which I grant, and I get this error: Could not find this item The item is no longer located in C:\ When I try to transfer the file back to the remote host in a new directory, the transfer starts successfully and transfers data The transfer had about 30 minutes remaining when I left it for the night The morning after the file transfer, I was greeted with a message saying that the connection to the server had been lost. I don't think this is relevant, since I did not tell it to disconnect after the file was done transferring, and it likely disconnected after the file transfer finished. I'm using an old version of WinSCP - v4.1.8 from 2008 I can view the file properties in WinSCP: Type of file: 7zip (.gz) Location: C:\ Attributes: none (Ready-only, Hidden, Archive, or Ready for indexing) Security: SYSTEM, my user, and Administrators group have full permissions - everything other than "special permissions" is checked under Allow for all 3 users/groups (my user, Administrators, SYSTEM) What's going on?!

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  • C++, Ifstream opens local file but not file on HTTP Server

    - by fammi
    Hi, I am using ifstream to open a file and then read from it. My program works fine when i give location of the local file on my system. for eg /root/Desktop/abc.xxx works fine But once the location is on the http server the file fails to open. for eg http://192.168.0.10/abc.xxx fails to open. Is there any alternate for ifstream when using a URL address? thanks. part of the code where having problem: bool readTillEof = (endIndex == -1) ? true : false; // Open the file in binary mode and seek to the end to determine file size ifstream file ( fileName.c_str ( ), ios::in|ios::ate|ios::binary ); if ( file.is_open ( ) ) { long size = (long) file.tellg ( ); long numBytesRead; if ( readTillEof ) { numBytesRead = size - startIndex; } else { numBytesRead = endIndex - startIndex + 1; } // Allocate a new buffer ptr to read in the file data BufferSptr buf (new Buffer ( numBytesRead ) ); mpStreamingClientEngine->SetResponseBuffer ( nextRequest, buf ); // Seek to the start index of the byte range // and read the data file.seekg ( startIndex, ios::beg ); file.read ( (char *)buf->GetData(), numBytesRead ); // Pass on the data to the SCE // and signal completion of request mpStreamingClientEngine->HandleDataReceived( nextRequest, numBytesRead); mpStreamingClientEngine->MarkRequestCompleted( nextRequest ); // Close the file file.close ( ); } else { // Report error to the Streaming Client Engine // as unable to open file AHS_ERROR ( ConnectionManager, " Error while opening file \"%s\"\n", fileName.c_str ( ) ); mpStreamingClientEngine->HandleRequestFailed( nextRequest, CONNECTION_FAILED ); } }

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  • How do I open a file in such a way that if the file doesn't exist it will be created and opened automatically?

    - by snakile
    Here's how I open a file for writing+ : if( fopen_s( &f, fileName, "w+" ) !=0 ) { printf("Open file failed\n"); return; } fprintf_s(f, "content"); If the file doesn't exist the open operation fails. What's the right way to fopen if I want to create the file automatically if the file doesn't already exist? EDIT: If the file does exist, I would like fprintf to overwrite the file, not to append to it.

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  • Problem with File uplolad in javascript.

    - by Nikhil
    I have used javascript to upload more than one file. when user clicks on 'add more' javascript appends new object to older div using innerHTML. Now the problem is if I select a file and then click on "add more" then new file button exist but older selected file removes and two blank file buttons display. I want this old file must be selected when user add new file button. If anybody can, Help Plz!!! tnX.

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  • I want the actual file name that is returned by a PHP script

    - by Aymon Fournier
    I am writing a python script that downloads a file given by a URL. Unfortuneatly the URL is in the form of a PHP script i.e. www.website.com/generatefilename.php?file=5233 If you visit the link in a browser, you are prompted to download the actual file and extension. I need to send this link to the downloader, but I can't send the downloader the PHP link. How would I get the full file name in a usable variable?

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  • CakePhp on IIS: How can I Edit URL Rewrite module for SSL Redirects

    - by AdrianB
    I've not dealt much with IIS rewrites, but I was able to import (and edit) the rewrites found throughout the cake structure (.htaccess files). I'll explain my configuration a little, then get to the meat of the problem. So my Cake php framework is working well and made possible by the url rewrite module 2.0 which I have successfully installed and configured for the site. The way cake is set up, the webroot folder (for cake, not iis) is set as the default folder for the site and exists inside the following hierarchy inetpub -wwwroot --cakePhp root ---application ----models ----views ----controllers ----WEBROOT // *** HERE *** ---cake core --SomeOtherSite Folder For this implementation, the url rewrite module uses the following rules (from the web.config file) ... <rewrite> <rules> <rule name="Imported Rule 1" stopProcessing="true"> <match url="^(.*)$" ignoreCase="false" /> <conditions logicalGrouping="MatchAll"> <add input="{REQUEST_FILENAME}" matchType="IsDirectory" negate="true" /> <add input="{REQUEST_FILENAME}" matchType="IsFile" negate="true" /> </conditions> <action type="Rewrite" url="index.php?url={R:1}" appendQueryString="true" /> </rule> <rule name="Imported Rule 2" stopProcessing="true"> <match url="^$" ignoreCase="false" /> <action type="Rewrite" url="/" /> </rule> <rule name="Imported Rule 3" stopProcessing="true"> <match url="(.*)" ignoreCase="false" /> <action type="Rewrite" url="/{R:1}" /> </rule> <rule name="Imported Rule 4" stopProcessing="true"> <match url="^(.*)$" ignoreCase="false" /> <conditions logicalGrouping="MatchAll"> <add input="{REQUEST_FILENAME}" matchType="IsDirectory" negate="true" /> <add input="{REQUEST_FILENAME}" matchType="IsFile" negate="true" /> </conditions> <action type="Rewrite" url="index.php?url={R:1}" appendQueryString="true" /> </rule> </rules> </rewrite> I've Installed my SSL certificate and created a site binding so that if i use the https:// protocol, everything is working fine within the site. I fear that attempts I have made at creating a rewrite are too far off base to understand results. The rules need to switch protocol without affecting the current set of rules which pass along url components to index.php (which is cake's entry point). My goal is this- Create a couple of rewrite rules that will [#1] redirect all user pages (in this general form http://domain.com/users/page/param/param/?querystring=value ) to use SSL and then [#2} direct all other https requests to use http (is this is even necessary?). [e.g. http://domain.com/users/login , http://domain.com/users/profile/uid:12345 , http://domain.com/users/payments?firsttime=true] ] to all use SSL [e.g. https://domain.com/users/login , https://domain.com/users/profile/uid:12345 , https://domain.com/users/payments?firsttime=true] ] Any help would be greatly appreciated.

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