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  • System Error when running PyQt4's loadUi()

    - by user633804
    Hello, I'm pretty new to Qt, Python and their combinations. I'm currently writing a QGIS plugin in python (I used QtCreator 2.1 (Qt Designer 4.7) to generate a .ui-file and am now trying to use it for a Quantum GIS plugin that's written in Python 2.5 (and running in the Quantum GIS Python 2.5 console)). I am running into trouble when loading the ui-file dynamically when the program runs the loadUi() function. What throws me off is that the error occurs outside my script. Does that mean, I'm passing something wrong into it? Where does the error come in? Any hints on what could be wrong? code_dir = os.path.dirname(os.path.abspath(__file__)) self.ui = loadUi(os.path.join(code_dir, "Ui_myfile.ui"), self) This is the Error Code I am getting (minus the first paragraph): File "C:/Dokumente und Einstellungen/name.name/.qgis/python/plugins\myfile\myfile_gui.py", line 42, in __ init __ self.ui = loadUi(os.path.join(code_dir, "Ui_myfile.ui"), self) File "C:\PROGRA~1\QUANTU~1\apps\Python25\lib\site-packages\PyQt4\uic__init__.py", line 112, in loadUi return DynamicUILoader().loadUi(uifile, baseinstance) File "C:\PROGRA~1\QUANTU~1\apps\Python25\lib\site-packages\PyQt4\uic\Loader\loader.py", line 21, in loadUi return self.parse(filename) File "C:\PROGRA~1\QUANTU~1\apps\Python25\lib\site-packages\PyQt4\uic\uiparser.py", line 768, in parse actor(elem) File "C:\PROGRA~1\QUANTU~1\apps\Python25\lib\site-packages\PyQt4\uic\uiparser.py", line 616, in createUserInterface self.traverseWidgetTree(elem) File "C:\PROGRA~1\QUANTU~1\apps\Python25\lib\site-packages\PyQt4\uic\uiparser.py", line 594, in traverseWidgetTree handler(self, child) File "C:\PROGRA~1\QUANTU~1\apps\Python25\lib\site-packages\PyQt4\uic\uiparser.py", line 233, in createWidget topwidget.setCentralWidget(widget) SystemError: error return without exception set

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  • Making a jQuery selection in IE on html added via .load()

    - by Joel Crawford-Smith
    Scenario: I am using jQuery to lazy load some html and change the relative href attributes of all the anchors to absolute links. The loading function adds the html in all browsers. The url rewrite function works on the original DOM in all browsers. But In IE7, IE8 I can't run that same function on the new lazy loaded html in the DOM. //lazy load a part of a file $(document).ready(function() { $('#tab1-cont') .load('/web_Content.htm #tab1-cont'); return false; }); //convert relative links to absolute links $("#tab1-cont a[href^=/]").each(function() { var hrefValue = $(this).attr("href"); $(this) .attr("href", "http://www.web.org" + hrefValue) .css('border', 'solid 1px green'); return false; }); I think my question is: whats the trick to getting IE to make selections on DOM that is lazy loaded with jQuery? This is my first post. Be gentle :-) Thanks, Joel

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  • file cretaed using exec could not be accessed immediately after creation?

    - by Holicreature
    HI I'm using exec in php to execute a command and it will create a .png file in a temp folder.. After creating that i'm trying to open that file and read contents and process them,, but i end up file could not read error.. I think the time taken by the exec to execute and create a file is the cause for the issue.. but i dont know how to fix it? i tried sleep() but it makes my script to run slow <?php error_reporting(E_ALL); extension_loaded('ffmpeg') or die('Error in loading ffmpeg'); //db connection codes $max_width = 120; $max_height = 72; $path ="/path/"; $qry="select id, input_file, output_file from videos where thumbnail='' or thumbnail is null;"; $res=mysql_query($qry); $cnt = 1; while($row = mysql_fetch_array($res,MYSQL_ASSOC)) { $outfile = $row[output_file]; $imgname = $cnt.".png"; $srcfile = "/path/".$outfile; echo "####$srcfile####"; exec("ffmpeg -i ".$srcfile." -r 1 -ss 00:00:05 -f image2 -s 120x72 ".$path.$imgname); $nname = "./temp/".$imgname; echo "nname===== $nname"; $fileo = fopen($nname,"rb"); if($fileo) { $imgData = addslashes(file_get_contents($nname)); .. ... .... } else echo "Could not open<br><br>"; $cnt = $cnt + 1: } ?>

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  • how to set Content-Type automatically when i download the data that i uploaded.

    - by zjm1126
    this is my code : import os from google.appengine.ext import webapp from google.appengine.ext.webapp import template from google.appengine.ext.webapp.util import run_wsgi_app from google.appengine.ext import db #from login import htmlPrefix,get_current_user class MyModel(db.Model): blob = db.BlobProperty() class BaseRequestHandler(webapp.RequestHandler): def render_template(self, filename, template_args=None): if not template_args: template_args = {} path = os.path.join(os.path.dirname(__file__), 'templates', filename) self.response.out.write(template.render(path, template_args)) class upload(BaseRequestHandler): def get(self): self.render_template('index.html',) def post(self): file=self.request.get('file') obj = MyModel() obj.blob = db.Blob(file.encode('utf8')) obj.put() self.response.out.write('upload ok') class download(BaseRequestHandler): def get(self): #id=self.request.get('id') o = MyModel.all().get() #self.response.out.write(''.join('%s: %s <br/>' % (a, getattr(o, a)) for a in dir(o))) self.response.out.write(o) application = webapp.WSGIApplication( [ ('/?', upload), ('/download',download), ], debug=True ) def main(): run_wsgi_app(application) if __name__ == "__main__": main() my index.html is : <form action="/" method="post"> <input type="file" name="file" /> <input type="submit" /> </form> and it show : <__main__.MyModel object at 0x02506830> but ,i don't want to see this , i want to download it , how to change my code to run, thanks updated it is ok now : class upload(BaseRequestHandler): def get(self): self.render_template('index.html',) def post(self): file=self.request.get('file') obj = MyModel() obj.blob = db.Blob(file) obj.put() self.response.out.write('upload ok') class download(BaseRequestHandler): def get(self): #id=self.request.get('id') o = MyModel.all().order('-').get() #self.response.out.write(''.join('%s: %s <br/>' % (a, getattr(o, a)) for a in dir(o))) self.response.headers['Content-Type'] = "image/png" self.response.out.write(o.blob) and new question is : if you upload a 'png' file ,it will show successful , but ,when i upload a rar file ,i will run error , so how to set Content-Type automatically , and what is the Content-Type of the 'rar' file thanks

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  • How to get the list of files in a directory in a shell script?

    - by jrharshath
    Hi, I'm trying to get the contents of a directory using shell script. My script is: for entry in `ls $search_dir`; do echo $entry done where $search_dir is a relative path. However, $search_dir contains many files with whitespaces in their names. In that case, this script does not run as expected. I know I could use for entry in *, but that would only work for my current directory. I know I can change to that directory, use for entry in * then change back, but my particular situation prevents me from doing that. I have two relative paths $search_dir and $work_dir, and I have to work on both simultaneously, reading them creating/deleting files in them etc. So what do I do now? PS: I use bash.

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  • css position when resizing browser

    - by user478636
    When resizing the browser I noticed that all the elements get out of place and the website layout gets distorted. This also occurs on with low-resolution. Is this because I have used position:relative;? How can I make the page elements not move from their position when resizing. body{ background:url(../img/bg-silver.jpg) #F2F2F2; font-family:"Lucida Sans Unicode", "Lucida Grande", sans-serif; font-size:11px; line-height:18px; color:#636363; margin-top:10%; } #containerHolder { background: #eee; padding: 5px; position:relative; } #container { background: #fff; background:rgba(245,245,245,0.8); border: 1px solid #ddd; } #main { margin: 0 0 0 20px; padding: 0 19px 0 0; }

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  • UnauthorizedAccessException when running desktop application from shared folder

    - by Atara
    I created a desktop application using VS 2008. When I run it locally, all works well. I shared my output folder (WITHOUT allowing network users to change my files) and ran my exe from another Vista computer on our intranet. When running the shared exe, I receive "System.UnauthorizedAccessException" when trying to read a file. How can I give permission to allow reading the file? Should I change the code? Should I grant permission to the application\folder on the Vista computer? how? Notes: I do not use ClickOnce. the application should be distributed using xcopy. My application target framework is ".Net Framework 2.0" On the Vista computer, "controlPanel | UninstallOrChangePrograms" it says it has "Microsoft .Net Framework 3.5 SP1" I also tried to map the folder drive, but got the same errors, only now the fileName is "T:\my.ocx" ' ---------------------------------------------------------------------- ' my code: Dim src As String = mcGlobals.cmcFiles.mcGetFileNameOcx() Dim ioStream As New System.IO.FileStream(src, IO.FileMode.Open) ' ---------------------------------------------------------------------- Public Shared Function mcGetFileNameOcx() As String ' ---------------------------------------------------------------------- Dim dirName As String = Application.StartupPath & "\" Dim sFiles() As String = System.IO.Directory.GetFiles(dirName, "*.ocx") Dim i As Integer For i = 0 To UBound(sFiles) Debug.WriteLine(System.IO.Path.GetFullPath(sFiles(i))) ' if found any - return the first: Return System.IO.Path.GetFullPath(sFiles(i)) Next Return "" End Function ' ---------------------------------------------------------------------- ' The Exception I receive: System.UnauthorizedAccessException: Access to the path '\\computerName\sharedFolderName\my.ocx' is denied. at System.IO._Error(Int32 errorCode, String maybeFullPath) at System.IO.FileStream.Init(...) at System.IO.FileStream..ctor(...) at System.IO.FileStream..ctor(String path, FileMode mode) ' ----------------------------------------------------------------------

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  • Workaround for stopping propagation with live()?

    - by bobsoap
    I've run into the problem that has been addressed here without a workaround: I can't use stopPropagation() on dynamically spawned elements. I've tried creating a condition to exclude a click within the dimensions of the spawned element, but that doesn't seem to work at all. Here is what I got: 1) a large background element ("canvas") that is activated to be "sensitive to clicks on it" by a button 2) the canvas, if activated, catches all clicks on it and spawns a small child form ("child") within it 3) the child is positioned relative to the mouse click position. If the mouse click was on the right half of the canvas, the child will be positioned 200 pixels to the left of that spot. (On the right if the click was in the left half) 4) every new click on the canvas removes the existing child (if any) and spawns a new child at the new position (relative to the click) The problem: Since the spawned child element is on top of the canvas, a click on it counts as a click on the canvas. Even if the child is outside of the boundaries of the canvas, clicking on it will trigger the action as described in 4) again . This shouldn't happen. =========== CODE: The button to activate the canvas: $('a#activate').click(function(event){ event.preventDefault(); canvasActive(); }); I referenced the above to show you that the canvas click-catching is happening in a function. Not sure if this is relevant... This is the function that catches clicks on the canvas: function canvasActive() { $('#canvas').click(function(e){ e.preventDefault(); //get click position relative to canvas posClick = { x : Math.round(e.pageX - $(this).offset().left), y : Math.round(e.pageY - $(this).offset().top) }; //calculate child position if(posClick.x <= $canvas.outerWidth(false)/2) { posChild = { x: posClick.x + 200, //if dot is on the left side of canvas y: posClick.y }; } else { posChild = { x: posClick.x - 600, //if dot is on the right y: posClick.y }; } $(this).append(markup); //markup is just the HTML for the child }); } I left out the unimportant stuff. The question is: How can I prevent a click inside of a spawned child from executing the function? I tried getting the child's dimensions and doing something like "if posClick is within this range, don't do anything" - but I can't seem to get it right. Perhaps someone has come across this dilemma before. Any help is appreciated.

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  • Android::Confused about image sizes in a website

    - by Legend
    I was testing my website inside the Android emulator with the Droid Skin (240 dpi). I have the following css: #container { position: relative; width: 854px; height: 480px; background: #000; margin: auto; } #container li { position: relative; list-style: none; width: 201px; height: 110px; padding-left: 10px; padding-top:10px; padding-bottom:10px; overflow: hidden; float: left; z-index: 2; } The display is not what I expected obviously because I am defining everything in px (when I should have been using dip but css does not allow dip). How can I convert my px to something that is suitable for Android? Any suggestions?

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  • SWIG: From Plain C++ to working Wrapper

    - by duckworthd
    Hi everyone. I've been trying to create a SWIG wrapper for this tiny little C++ class for the better part of 3 hours with no success, so I was hoping one of you out there could lend me a small hand. I have the following class: #include <stdio.h> class Example { public: Example(); ~Example(); int test(); }; #include "example.h" Along with the implementation: Example::Example() { printf("Example constructor called\n"); } Example::~Example() { printf("Example destructor called\n"); } int Example::test() { printf("Holy shit, I work!\n"); return 42; } I've read through the introduction page ( www.swig.org/Doc1.3/Java.html ) a few times without gaining a whole lot of insight into the situation. My steps were Create an example.i file Compile original alongside example_wrap.cxx (no linking) link resulting object files together Create a little java test file (see below) javac all .java files there and run Well steps 4 and 5 have created a host of problems for me, starting with the basic ( library 'example' not found due to not being in java's path ) to the weird ( library not found even unless LD_LIBRARY_PATH is set to something, even if it's nothing at all). I've included my little testing code below public class test2 { static { String libpath = System.getProperty("java.library.path"); String currentDir = System.getProperty("user.dir"); System.setProperty("java.library.path", currentDir + ":" + libpath); System.out.println(System.getProperty("java.library.path")); System.loadLibrary("example"); } public static void main(String[] args){ System.out.println("It loads!"); } } Well, if anyone has navigated these murky waters of wrapping, I could not be happier than if you could light the way, particularly if you could provide the example.i and bash commands to go along with it.

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  • Recursive function for copy of multilevel folder is not working.

    - by OM The Eternity
    Recursive function for copy of multilevel folder is not working. I have a code to copy all the mulitilevel folder to new folder. But in between I feel there is problem of proper path recognition, see the code below.. <?php $source = '/var/www/html/pranav_test'; $destination = '/var/www/html/parth'; copy_recursive_dirs($source, $destination); function copy_recursive_dirs($dirsource, $dirdest) { // recursive function to copy // all subdirectories and contents: if(is_dir($dirsource)) { $dir_handle=opendir($dirsource); } if(!is_dir($dirdest)) { mkdir($dirdest, 0777); } while($file=readdir($dir_handle)) {/*echo "<pre>"; print_r($file);*/ if($file!="." && $file!="..") { if(!is_dir($dirsource.DIRECTORY_SEPARATOR.$file)) { copy ($dirsource.DIRECTORY_SEPARATOR.$file, $dirdest.DIRECTORY_SEPARATOR.$dirsource.DIRECTORY_SEPARATOR.$file); } else{ copy_recursive_dirs($dirsource.DIRECTORY_SEPARATOR.$file, $dirdest); } } } closedir($dir_handle); return true; } ?> from the above code the if loop has a copy function as per requirement, but the path applied for destination here is not proper, I have tried with basename function as well.. but it didnt got the expected result.. below is the if loop i am talking about with comment describing the output... if(!is_dir($dirsource.DIRECTORY_SEPARATOR.$file)) { $basefile = basename($dirsource.DIRECTORY_SEPARATOR.$file);//it gives the file name echo "Pranav<br>".$dirdest.DIRECTORY_SEPARATOR.$dirsource.DIRECTORY_SEPARATOR.$file;//it outputs for example "/var/www/html/parth//var/www/html/pranav_test/media/system/js/caption.js" which is not correct.. copy ($dirsource.DIRECTORY_SEPARATOR.$file, $dirdest.DIRECTORY_SEPARATOR.$dirsource.DIRECTORY_SEPARATOR.$file); } as shown above the i cannot get the files and folders copied to expected path.. please guide me to place proper path in the function....

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  • Organization of linking to external libraries in C++

    - by Nicholas Palko
    In a cross-platform (Windows, FreeBSD) C++ project I'm working on, I am making use of two external libraries, Protocol Buffers and ZeroMQ. In both projects, I am tracking the latest development branch, so these libraries are recompiled / replaced often. For a development scenario, where is the best place to keep libprotobuf.{a,lib} and zeromq.{so,dll}? Should I have my build script copy them from their respective project directories into my local project's directory (say MyProjectRoot/lib or MyProjectRoot/bin) before I build my project? This seems preferable to tossing things into /usr/local/lib, as I wouldn't want to replace a system-wide stable version with the latest experimental one. Cmake warns me whenever I specify a relative path for linking, so I would suspect copying is a better solution then relative linking? Is this the best approach? Thanks for your help!

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  • Importing package as a submodule

    - by wecac
    Hi, I have a package 3rd party open source package "foo"; that is in beta phase and I want to tweak it to my requirements. So I don't want to get it installed in /usr/local/lib/python or anywhere in current sys.path as I can't make frequent changes in top level packages. foo/ __init__.py fmod1.py import foo.mod2 fmod2.py pass I want to install the the package "foo" as a sub package of my namespace say "team.my_pkg". So that the "fullname" of the package becomes "team.my_pkg.foo" without changing the code in inner modules that refer "team.my_pkg.foo" as "foo". team/ __init__.py my_pkg/ __init__.py foo/ fmod1.py import foo.mod2 fmod2.py pass One way to do this is to do this in team.my_pkg.init.py: import os.path import sys sys.path.append(os.path.dirname(__file__)) But I think it is very unsafe. I hope there is some way that only fmod1.py and fmod2.py can call "foo" by its short name everything else should use its complete name "team.my_pkg.foo" I mean this should fail outside team/my_pkg/foo: import team.my_pkg import foo But this should succeed outside team/my_pkg/foo: import team.my_pkg.foo

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  • How to draw shadows that don't suck?

    - by mystify
    A CAShapeLayer uses a CGPathRef to draw it's stuff. So I have a star path, and I want a smooth drop shadow with a radius of about 15 units. Probably there is some nice functionality in some new iPhone OS versions, but I need to do it myself for a old aged version of 3.0 (which most people still use). I tried to do some REALLY nasty stuff: I created a for-loop and sequentially created like 15 of those paths, transform-scaling them step by step to become bigger. Then assigning them to a new created CAShapeLayer and decreasing it's alpha a little bit on every iteration. Not only that this scaling is mathematically incorrect and sucks (it should happen relative to the outline!), the shadow is not rounded and looks really ugly. That's why nice soft shadows have a radius. The tips of a star shouldn't appear totally sharp after a shadow size of 15 units. They should be soft like cream. But in my ugly solution they're just as s harp as the star itself, since all I do is scale the star 15 times and decrease it's alpha 15 times. Ugly. I wonder how the big guys do it? If you had an arbitrary path, and that path must throw a shadow, how does the algorithm to do that work? Probably the path would have to be expanded like 30 times, point-by-point relative to the tangent of the outline away from the filled part, and just by 0.5 units to have a nice blending. Before I re-invent the wheel, maybe someone has a handy example or link?

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  • Crash in C++ Code

    - by Ankuj
    I am trying to list all files in a directory recursively. But my code crashes without giving any error. When the given file is directory I recursively call the function or else print its name. I am using dirent.h int list_file(string path) { DIR *dir; struct dirent *ent; char *c_style_path; c_style_path = new char[path.length()]; c_style_path = (char *)path.c_str(); dir = opendir (c_style_path); if (dir != NULL) { /* print all the files and directories within directory */ while ((ent = readdir (dir)) != NULL) { if(ent->d_type == DT_DIR && (strcmp(ent->d_name,".")!=0) && (strcmp(ent->d_name,"..")!=0)) { string tmp = path + "\\" + ent->d_name; list_file(tmp); } else { cout<<ent->d_name<<endl; } } closedir (dir); } else { /* could not open directory */ perror (""); return EXIT_FAILURE; } delete [] c_style_path; return 0; } I am not getting as to what I am doing wrong here. Any clues ?

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  • how to download data which upload to gae ,

    - by zjm1126
    this is my code : import os from google.appengine.ext import webapp from google.appengine.ext.webapp import template from google.appengine.ext.webapp.util import run_wsgi_app from google.appengine.ext import db #from login import htmlPrefix,get_current_user class MyModel(db.Model): blob = db.BlobProperty() class BaseRequestHandler(webapp.RequestHandler): def render_template(self, filename, template_args=None): if not template_args: template_args = {} path = os.path.join(os.path.dirname(__file__), 'templates', filename) self.response.out.write(template.render(path, template_args)) class upload(BaseRequestHandler): def get(self): self.render_template('index.html',) def post(self): file=self.request.get('file') obj = MyModel() obj.blob = db.Blob(file.encode('utf8')) obj.put() self.response.out.write('upload ok') class download(BaseRequestHandler): def get(self): #id=self.request.get('id') o = MyModel.all().get() #self.response.out.write(''.join('%s: %s <br/>' % (a, getattr(o, a)) for a in dir(o))) self.response.out.write(o) application = webapp.WSGIApplication( [ ('/?', upload), ('/download',download), ], debug=True ) def main(): run_wsgi_app(application) if __name__ == "__main__": main() my index.html is : <form action="/" method="post"> <input type="file" name="file" /> <input type="submit" /> </form> and it show : <__main__.MyModel object at 0x02506830> but ,i don't want to see this , i want to download it , how to change my code to run, thanks

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  • Heroku DJango app development on Windows

    - by Cliff
    I'm trying to start a Django app on Heroku using Windows and I'm getting stuck on the following error when I try to pip install psycopg2: Downloading/unpacking psycopg2 Downloading psycopg2-2.4.5.tar.gz (719Kb): 719Kb downloaded Running setup.py egg_info for package psycopg2 Error: pg_config executable not found. Please add the directory containing pg_config to the PATH or specify the full executable path with the option: python setup.py build_ext --pg-config /path/to/pg_config build ... or with the pg_config option in 'setup.cfg'. Complete output from command python setup.py egg_info: running egg_info creating pip-egg-info\psycopg2.egg-info writing pip-egg-info\psycopg2.egg-info\PKG-INFO writing top-level names to pip-egg-info\psycopg2.egg-info\top_level.txt writing dependency_links to pip-egg-info\psycopg2.egg-info\dependency_links.txt writing manifest file 'pip-egg-info\psycopg2.egg-info\SOURCES.txt' warning: manifest_maker: standard file '-c' not found I've googled the error and it seems you need libpq-dev python-dev as dependencies for postgres under Python. I also turned up a link that says you gt into trouble if you don't have the postgres bin folder in your Path so I installed Postgres manually and tried again. This time I get: error: Unable to find vcvarsall.bat I am still a python N00b so I am lost. Could someone point me in a general direction?

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  • Error: 'Sys' is undefined

    - by thegunner
    Hi, When I moved my website over to another server I've noticed now that ajax doesn't appear to be working. When I remote desktop to the server and go to the url on the server i.e. http://myserver/mywebsite ... everything works ok. When I open up the website in visual studio on the server it works as no problem as well. It's only when I connect remotely that the 'javascript' error occurs. To my web.config I've added: <httpHandlers> <remove verb="*" path="*.asmx"/> <add verb="*" path="*.asmx" validate="false" type="System.Web.Script.Services.ScriptHandlerFactory, System.Web.Extensions, Version=3.5.0.0, Culture=neutral, PublicKeyToken=31BF3856AD364E35"/> <add verb="*" path="*_AppService.axd" validate="false" type="System.Web.Script.Services.ScriptHandlerFactory, System.Web.Extensions, Version=3.5.0.0, Culture=neutral, PublicKeyToken=31BF3856AD364E35"/> <add verb="GET,HEAD" path="ScriptResource.axd" type="System.Web.Handlers.ScriptResourceHandler, System.Web.Extensions, Version=3.5.0.0, Culture=neutral, PublicKeyToken=31BF3856AD364E35" validate="false"/> </httpHandlers> <httpModules> <add name="ScriptModule" type="System.Web.Handlers.ScriptModule, System.Web.Extensions, Version=3.5.0.0, Culture=neutral, PublicKeyToken=31BF3856AD364E35"/> </httpModules> I've tried <compilation debug="false"/> and tried emptying web history and still no luck. Any ideas?

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  • Converting a video to flv format by using ffmpeg is working in local but not working in server

    - by kishore
    Hi all, I want to convert a video to flv format. I am using ffmpeg to convert the video. I am using following code. exec("C:/wamp/www/newtip/ffmpeg/ffmpeg -i C:/wamp/www/newtip/ffmpeg/videos/".$name." -ar 22050 -ab 32 -f flv -s 320x240 C:/wamp/www/newtip/ffmpeg/players/".$name_s.".flv"); It is working correctly in the local system. But in the server it is not working correctly. In the server i changed the code as below exec("http://www.mydomain.com/newtip/ffmpeg/ffmpeg -i http://www.mydomain.com/newtip/ffmpeg/videos/".$name." -ar 22050 -ab 32 -f flv -s 320x240 http://www.mydomain.com/newtip/ffmpeg/players/".$name_s.".flv"); In local I have given the Source path as C:/wamp/www/newtip/ But in server i have given the path as http://www.mydomain.com/newtip/ .I think in the server the path is wrong. Can anybody tell me how to give the path in the server?

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  • div style working in Firefox but not IE8

    - by Nathan Spiwak
    I am creating a banner that resizes to fit the window (I got the code of a blog post) and it works fine in Firefox, but it doesn't display at all in IE8. Please help!! <html> <body> <div style="position:relative; width:100%; height:100%; margin:0px; padding:0px; left:0px;right:0px;z-index:1”><img src="https://na6.salesforce.com/servlet/servlet.ImageServer?id=01580000000pT8r&oid=00D80000000aYeL&lastMod=1273785188000" width="100%"></div> <div style="z-index:2; position:relative; margin:0px; padding:0px;"> </div> </body> </html>

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  • What access token should i use?

    - by user548458
    I made a posting scores. It worked normaly sometimes. But, now I cant post any score. If I use user accessToken: string path = meID + "/scores"; var parameters = new Dictionary<string, object> (){ { "score", score.ToString () }, { "access_token", accessToken } }; facebook.post (path, parameters, ( error, obj ) => I get error: {"error":{"message":"(#15) This method must be called with an app access_token.", "type":"OAuthException", "code":15}} If I use an app access token: string path = meID + "/scores"; var parameters = new Dictionary<string, object> () { { "score", score.ToString () }, { "access_token", appAccessToken } }; facebook.post (path, parameters, ( error, obj ) => I get other error: {"error":{"message":"A user access token is required to request this resource.", "type":"OAuthException", "code":102}} Help me please, what am i doing wrong? PS: I worked well recently, but now - dont. Cant explain it...

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  • Why the databinding fails in ListView (WPF) ?

    - by Ashish Ashu
    I have a ListView of which ItemSource is set to my Custom Collection. I have defined a GridView CellTemplate that contains a combo box as below : <ListView MaxWidth="850" Grid.Row="1" SelectedItem="{Binding Path = SelectedCondition}" ItemsSource="{Binding Path = Conditions}" FontWeight="Normal" FontSize="11" Name="listview"> <ListView.View> <GridView> <GridViewColumn Width="175" Header="Type"> <GridViewColumn.CellTemplate> <DataTemplate> <ComboBox Style="{x:Null}" x:Name="TypeCmbox" Height="Auto" Width="150" SelectedValuePath="Key" DisplayMemberPath="Value" SelectedItem="{Binding Path = MyType}" ItemsSource="{Binding Path = MyTypes}" HorizontalAlignment="Center" /> </DataTemplate> </GridViewColumn.CellTemplate> </GridViewColumn> </ListView> My Custom collection is the ObservableCollection. I have a two buttons - Move Up and Move Down on top of the listview control . When user clicks on the Move Up or Move Down button I call MoveUp and MoveDown methods of Observable Collection. But when I Move Up and Move Down the rows then the Selected Index of a combo box is -1. I have ensured that selectedItem is not equal to null when performing Move Up and Move Down commands. Please Help!!

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  • WPF binding comboboxes to parent- child model

    - by PaulB
    I've got a model with a few tiers in it - something along the lines of ... Company Employees Phone numbers So I've got a ListBox showing all the companys in the model. Each ListBoxItem then contains two comboboxes ... one for employees, one for phone numbers. I can successfully get the employee combo to bind correctly and show the right people, but I'd like the phone combo to show the numbers for the selected employee. I'm just setting the DataContext of the ListBox to the model above and using the following data template for each item <DataTemplate x:Key="CompanyBody"> <StackPanel Orientation="Horizontal"> <Label Content="{Binding Path=CompanyName}"></Label> <ComboBox Name="EmployeesCombo" ItemsSource="{Binding Path=Company.Employees}"></ComboBox> <!-- What goes here --> <ComboBox DataContext="???" ItemsSource="??" ></ComboBox> </StackPanel> </DataTemplate> I've tried (naively) <ComboBox ItemsSource="{Binding Path=Company.Employees.PhoneNumbers}" ></ComboBox> and <ComboBox DataContext="EmployeesCombo.SelectedValue" ItemsSource="{Binding Path=PhoneNumbers}" ></ComboBox> and all other manner of combinations ...

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