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  • Excel 2008 Cant Parse HTML

    - by VictorV
    I need to export a gridview to excel, I put the return html code from the gridview to a HtmlTextWriter and put this into the response. The result file work fine in excel, excel can parse the html and the result is readable, work perfect on excel 2003 and 2007, but in some machines with Excel 2008 (MACOS) excel shows only the raw html code and can't process this html code. Any idea to configure excel? This is the code to convert: public static void ToExcel(GridView gridView, string fileName) { HttpResponse response = HttpContext.Current.Response; response.Clear(); response.Buffer = true; fileName = fileName.Replace(".xls", string.Empty) + ".xls"; response.AddHeader("content-disposition", "attachment;filename=" + fileName); response.Charset = ""; response.ContentEncoding = Encoding.Unicode; response.BinaryWrite(Encoding.Unicode.GetPreamble()); response.ContentType = MimeTypes.GetContentType(fileName); StringWriter sw = new StringWriter(); HtmlTextWriter hw = new HtmlTextWriter(sw); gridView.AllowPaging = false; //gridView.DataBind(); //Change the Header Row back to white color gridView.HeaderRow.Style.Add("background-color", "#FFFFFF"); //Apply style to Individual Cells for (int i = 0; i < gridView.HeaderRow.Cells.Count; i++) { gridView.HeaderRow.Cells[i].Style.Add("background-color", "yellow"); } for (int i = 0; i < gridView.Rows.Count; i++) { GridViewRow row = gridView.Rows[i]; //Change Color back to white row.BackColor = System.Drawing.Color.White; //Apply text style to each Row row.Attributes.Add("class", "textmode"); //Apply style to Individual Cells of Alternating Row if (i % 2 != 0) { for (int j = 0; j < row.Cells.Count; j++) { row.Cells[j].Style.Add("background-color", "#C2D69B"); } } } gridView.RenderControl(hw); //style to format numbers to string string style = @"<style> .textmode { mso-number-format:\@; } </style>"; response.Write(style); response.Output.Write(sw.ToString()); response.Flush(); response.End(); }

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  • problem with struts actions and redirections

    - by Casey
    I am trying to update a simple web app that was built with struts2, jsp and standard servlets. I am trying to redirect a url to a specific action but can't seem to get it to work right. For example, the url that is correct is: http://localhost:8080/theapp/lookup/search.action Here is my web.xml: <?xml version="1.0" encoding="ISO-8859-1"?> <!DOCTYPE web-app PUBLIC "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN" "http://java.sun.com/dtd/web-app_2_3.dtd"><web-app> <display-name>theapp</display-name> <filter> <filter-name>struts2</filter-name> <filter-class> org.apache.struts2.dispatcher.FilterDispatcher </filter-class> </filter> <filter-mapping> <filter-name>struts2</filter-name> <url-pattern>/*</url-pattern> </filter-mapping> <listener> <listener-class> org.springframework.web.context.ContextLoaderListener </listener-class> </listener> And here is my struts.xml: <?xml version="1.0" encoding="UTF-8" ?> <!DOCTYPE struts PUBLIC "-//Apache Software Foundation//DTD Struts Configuration 2.0//EN" "http://struts.apache.org/dtds/struts-2.0.dtd"> <default-action-ref name="search" /> <action name="search" method="search" class="com.theapp.SearchAction" > <result>index.jsp</result> <result name="input" >index.jsp</result> <result name="error" type="redirect">site_locator_mobile/error.action</result> </action> The problem here is that if I don't specify the correct url as above, I just get the index.jsp file, but without any properties in index.jsp being processed because the information is contained in the servlet. What I would like to is if someone just entered: http://localhost:8080/theapp/lookup/ than they would be taken to: http://localhost:8080/theapp/lookup/search.action Thanks

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  • utf8 format in xml

    - by hussain
    i want to know how to store this è (this type of symbols) in xml file if i store this symbol in xml file.. the file shows this symbol like ? i was inserted in front of xml file is <?xml version="1.0" encoding="UTF-8"?> but that doest not shows correct thanks and advance

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  • getURL, parsing web-site with german special characters

    - by Kay
    I am using getURL() and htmlParse() - how can I make web-site content with special characters to be displayed properly? library(RCurl); library(XML) script <- getURL("http://www.floraweb.de/pflanzenarten/foto.xsql?suchnr=814") doc <- htmlParse(script, encoding = "UTF-8") xpathSApply(doc, "//div[@id='content']//p", xmlValue)[2] [1] "Bellis perennis L., Gänseblümchen" # should say: [1] "Bellis perennis L., Gänseblümchen" > Sys.getlocale() [1] "LC_COLLATE=German_Austria.1252;LC_CTYPE=German_Austria.1252;LC_MONETARY=German_Austria.1252;LC_NUMERIC=C;LC_TIME=German_Austria.1252"

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  • Outputing UTF-8 string on Mac OS's Terminal

    - by SuperBloup
    I got a programm in haskell outputting utf-8 using the package utf8-string and using only the output functions of this package. I set the encoding of each file I write to this way : hSetEncoding myFile utf8 {- myFile may be stdout -} but when I try to output : alpha = [fromEnum 0x03B1] {- a -} instead of the nice alpha letter I got on Linux (or in a file on windows), I got the following : α The weird thing is even if I try to write the output on a file, I can't read it back with mvim as an utf-8 file. Is there any way to get the correct behaviour

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  • Parsing json in ios

    - by gaps
    i have the follwing json string ,can anybody tell me how to get the value of role_name {"response":"success","user":{"created_at":"2011-11-16T05:48:31Z","ud_id":"1234567890","last_sign_in_ip":"182.72.141.194","updated_at":"2011-11-19T08:58:27Z","account_id":21,"last_name":"dfg","role_name":"Parent","email":"[email protected]","first_name":"abc"},"status":"200"} code for parsing is NSString *responseString = [[NSString alloc] initWithData:responseData encoding:NSUTF8StringEncoding]; NSLog(@"resp--- %@",responseString); NSArray* latestLoans = [(NSDictionary*)[responseString JSONValue] objectForKey:@"user"]; NSDictionary* loan = [latestLoans objectAtIndex:0]; NSString* name = [loan objectForKey:@"last_name"];

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  • C# Windows Service Multiple Config Files

    - by Goober
    Quick Question Is it possible to have more than 1 config file in a windows service? Or is there some way I can merge them at run time? Scenario Currently I have two config files containing the below contents. After I build and install my Windows Service, I can't get my custom XML Parser class to read the content because it keeps pointing to the wrong config file, even though I am using a few lines of code to ensure it gets the right name of the config file I need to access. ALSO When I navigate to the folder in which the service is installed, there is only sign of the normal APP.Config file, and no sign of the custom config file. (I have even set the normal ones properties to "Do Not Copy" and the custom ones properties to "Copy Always"). Config File Determination Code string settingsFile = String.Format("{0}.exe.config", System.AppDomain.CurrentDomain.BaseDirectory + Assembly.GetExecutingAssembly().GetName().Name); CUSTOM CONFIG File <?xml version="1.0" encoding="utf-8" ?> <configuration> <servers> <SV066930> <add name="Name" value = "SV066930" /> <processes> <SimonTest1> <add name="ProcessName" value="notepad.exe" /> <add name="CommandLine" value="C:\\WINDOWS\\system32\\notepad.exe C:\\WINDOWS\\Profiles\\TA2TOF1\\Desktop\\SimonTest1.txt" /> </SimonTest1> </processes> </SV066930> </servers> </configuration> NORMAL APP.CONFIG File <?xml version="1.0" encoding="utf-8" ?> <configuration> <configSections> <section name="dataConfiguration" type="Microsoft.Practices.EnterpriseLibrary.Data.Configuration.DatabaseSettings, Microsoft.Practices.EnterpriseLibrary.Data, Version=4.0.0.0, Culture=neutral, PublicKeyToken=xxxxxxxxxxx" /> </configSections> <connectionStrings> <add name="DB" connectionString="Data Source=etc......" /> </connectionStrings> </configuration>

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  • asp.net forms authentification security issues

    - by Andrew Florko
    Hi there, I have a kind of asp.net forms authentication with the code like that: FormsAuthentication.SetAuthCookie(account.Id.ToString(), true); HttpContext.Current.User = new GenericPrincipal(new GenericIdentity(account.Id.ToString()), null); What kind of additional efforts shall I take to make authentication cookie (that is user id) more securable? (https, encoding for example) Thank you in advance!

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  • What's the default ScaleType of ImageView?

    - by kknight
    What's the default ScaleType of ImageView?If I put an image which is 400 pixels x 400 pixels on a normal screen (320x480) without specifying ScaleType, how will the image be scaled? Thanks. <?xml version="1.0" encoding="utf-8"?> <RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android" android:layout_width="fill_parent" android:layout_height="fill_parent"> <ImageView android:layout_width="fill_parent" android:layout_height="fill_parent" android:src="@drawable/big_image" /> </RelativeLayout>

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  • UnicodeDecodeError when redirecting to file

    - by zedoo
    Hi, I run this snippet twice, in the ubuntu terminal, (encoding set to utf-8) once with ./test.py and then with ./test.py >out.txt: uni = u"\u001A\u0BC3\u1451\U0001D10C" print uni Without redirection it prints garbage. With redirection I get a UnicodeDecodeError. Can someone explain why I get the error only in the second case, or even better give a detailed explanation of what's going on behind the curtain in both cases?

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  • RIM blackberry Record 3GP video

    - by pankaj_shukla
    Hi All, I am writing an application that can record a 3GP video. I have tried both MMAPI and Invoke API. But have following issues. Using MMAPI: 1. When I record to stream, It records video in RIMM streaming format. when I try to play this video player gives error "Unsupported media format.". 2. When I record to a file. It will create a file of size 0. Using Invoke API: 1. In MMS mode it does not allow to record a video more than 30 seconds. 2. In Normal mode size of the file is very large. 3. Once I invoke camera application I do not have any control on application. Here is my source code: _player = javax.microedition.media.Manager .createPlayer("capture://video?encoding=video/3gpp&mode=mms"); // I have tried every encoding returns from System.getProperty("video.encodings") method _player.realize(); _videoControl = (VideoControl) _player.getControl("VideoControl"); _recordControl = (RecordControl) _player.getControl("RecordControl"); _volumeControl = (VolumeControl) _player.getControl("VolumeControl"); String videoPath = System.getProperty("fileconn.dir.videos"); if (videoPath == null) { videoPath = "file:///store/home/user/videos/"; } _recordControl.setRecordLocation(videoPath + "RecordedVideo.3gp"); _player.addPlayerListener(this); Field videoField = (Field) _videoControl.initDisplayMode( VideoControl.USE_GUI_PRIMITIVE, "net.rim.device.api.ui.Field"); _videoControl.setVisible(true); add(videoField); _player.start(); ON start menu item Selection: try { _recordControl.startRecord(); } catch (Exception e) { _player.close(); showAlert(e.getClass() + " " + e.getMessage()); } On stop menuItem selection: try { _recordControl.commit(); } catch (Exception e) { _player.close(); showAlert(e.getClass() + " " + e.getMessage()); } Please let me if I am doing something wrong. Thanks, Pankaj

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  • Saving program working data

    - by gloris
    When the program runs, the intermediate data must save in the user's computer. Now I use .txt files, encoding with AES. But file/code user can break, delete.... Maybe is better decision?

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  • How to assign XML attribute values to drop down list using XSL

    - by Vijay
    Hi, I have a sample xml as; <?xml version="1.0" encoding="iso-8859-9"?> <DropDownControl id="dd1" name="ShowValues" choices="choice1,choice2,choice3,choice4"> </DropDownControl > I need to create a UI representation of this XML using XSL. I want to fill the drop down list with values specified in choices attribute. Does anyone have any idea about this ? Thanks in advance :)

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  • How can i assign xml scripts in to a php variable?

    - by Himel Khan
    I can not assign a xml scripts in to a php variable. My xml text: <?xml version="1.0" encoding="UTF-8" ?><rss version="2.0"><channel> <title>coolajax.net</title> <link>http://www.hotscripts.com/listings/feed</link> <description>Coolajax Scripts Listings Description</description> and I want to assign this text in to $xml_header variable. can anyone help me..

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  • Change HttpContext.Request.InputStream

    - by user320478
    I am getting lot of errors for HttpRequestValidationException in my event log. Is it possible to HTMLEncode all the inputs from override of ProcessRequest on web page. I have tried this but it gives context.Request.InputStream.CanWrite == false always. Is there any way to HTMLEncode all the feilds when request is made? public override void ProcessRequest(HttpContext context) { if (context.Request.InputStream.CanRead) { IEnumerator en = HttpContext.Current.Request.Form.GetEnumerator(); while (en.MoveNext()) { //Response.Write(Server.HtmlEncode(en.Current + " = " + //HttpContext.Current.Request.Form[(string)en.Current])); } long nLen = context.Request.InputStream.Length; if (nLen > 0) { string strInputStream = string.Empty; context.Request.InputStream.Position = 0; byte[] bytes = new byte[nLen]; context.Request.InputStream.Read(bytes, 0, Convert.ToInt32(nLen)); strInputStream = Encoding.Default.GetString(bytes); if (!string.IsNullOrEmpty(strInputStream)) { List<string> stream = strInputStream.Split('&').ToList<string>(); Dictionary<int, string> data = new Dictionary<int, string>(); if (stream != null && stream.Count > 0) { int index = 0; foreach (string str in stream) { if (str.Length > 3 && str.Substring(0, 3) == "txt") { string textBoxData = str; string temp = Server.HtmlEncode(str); //stream[index] = temp; data.Add(index, temp); index++; } } if (data.Count > 0) { List<string> streamNew = stream; foreach (KeyValuePair<int, string> kvp in data) { streamNew[kvp.Key] = kvp.Value; } string newStream = string.Join("", streamNew.ToArray()); byte[] bytesNew = Encoding.Default.GetBytes(newStream); if (context.Request.InputStream.CanWrite) { context.Request.InputStream.Flush(); context.Request.InputStream.Position = 0; context.Request.InputStream.Write(bytesNew, 0, bytesNew.Length); //Request.InputStream.Close(); //Request.InputStream.Dispose(); } } } } } } base.ProcessRequest(context); }

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  • FileZilla is saying there is an error in the W3.org link!

    - by Anonymous the Great
    I just got my free web hosting set up (trigoblocks.comuf.com), I connected via FileZilla FTP and uploaded my files. The error I get is "Parse error: syntax error, unexpected T_STRING in /home/a3639879/public_html/header.php on line 1". Header.php line 1-2: <?xml version="1.0" encoding="UTF-8"?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "DTD/xhtml1-strict.dtd"> I do not get this error while on localhost with XAMPP.

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  • MS Exam 70-536 - How can you constrain the input before you write any code?

    - by Max Gontar
    Hello! In MS Exam 70-536 .Net Foundation, Chapter 3 "Searching, Modifying, and Encoding Text" in Case Scenario 1 related to regex there is a question: How can you constrain the input before you write any code? I thought it's maybe a in-mind design of regex pattern but it will not really constrain the input, will it? I am not so good in psychokinesis yet! Or maybe the is some other way? Thanks for your time!

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  • Attach file to mail using php

    - by ktsixit
    Hi all, I've created a form which contains an upload field file and some other text fields. I'm using php to send the form's data via email and attach the file. This is the code I'm using but it's not working properly. The file is normally attached to the message but the rest of the data is not sent. $body="bla bla bla"; $attachment = $_FILES['cv']['tmp_name']; $attachment_name = $_FILES['cv']['name']; if (is_uploaded_file($attachment)) { $fp = fopen($attachment, "rb"); $data = fread($fp, filesize($attachment)); $data = chunk_split(base64_encode($data)); fclose($fp); } $headers = "From: $email<$email>\n"; $headers .= "Reply-To: <$email>\n"; $headers .= "MIME-Version: 1.0\n"; $headers .= "Content-Type: multipart/related; type=\"multipart/alternative\"; boundary=\"----=MIME_BOUNDRY_main_message\"\n"; $headers .= "X-Sender: $first_name $family_name<$email>\n"; $headers .= "X-Mailer: PHP4\n"; $headers .= "X-Priority: 3\n"; $headers .= "Return-Path: <$email>\n"; $headers .= "This is a multi-part message in MIME format.\n"; $headers .= "------=MIME_BOUNDRY_main_message \n"; $headers .= "Content-Type: multipart/alternative; boundary=\"----=MIME_BOUNDRY_message_parts\"\n"; $message = "------=MIME_BOUNDRY_message_parts\n"; $message .= "Content-Type: text/html; charset=\"utf-8\"\n"; $message .= "Content-Transfer-Encoding: quoted-printable\n"; $message .= "\n"; $message .= "$body\n"; $message .= "\n"; $message .= "------=MIME_BOUNDRY_message_parts--\n"; $message .= "\n"; $message .= "------=MIME_BOUNDRY_main_message\n"; $message .= "Content-Type: application/octet-stream;\n\tname=\"" . $attachment_name . "\"\n"; $message .= "Content-Transfer-Encoding: base64\n"; $message .= "Content-Disposition: attachment;\n\tfilename=\"" . $attachment_name . "\"\n\n"; $message .= $data; //The base64 encoded message $message .= "\n"; $message .= "------=MIME_BOUNDRY_main_message--\n"; $subject = 'bla bla bla'; $to="[email protected]"; mail($to,$subject,$message,$headers); Why isn't the $body data not sent? Can you help me fix it?

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  • Determining the best audio quality.

    - by The Rook
    How can you determine the best audio quality in a list of audio files, with out looking at the audio file's header. What if all of the files came from differnt encoding types and they where all transcoded to the same format and bit rate.

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  • java: converting part of a ByteBuffer to a string

    - by Jason S
    I have a ByteBuffer containing bytes that were derived by String.getBytes(charsetName), where "containing" means that the string comprises the entire sequence of bytes between the ByteBuffer's position() and limit(). What's the best way for me to get the string back? (assuming I know the encoding charset) Is there anything better than the following (which seems a little clunky) byte[] ba = new byte[bbuf.remaining()]; bbuf.get(ba); try { String s = new String(ba, charsetName); } catch (UnsupportedEncodingException e) { /* take appropriate action */ }

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  • Python regular expression

    - by user3692739
    I have this HTTP Request and I want to display only the Authorization section (base64 Value) : any help ? This Request is stored on a variable called hreq I have tried this : reg = re.search(r"Authorization:\sBasic\s(.*)\r", hreq) print reg.group() but doesn't work Here is the request : HTTP Request: Path: /dynaform/custom.js Http-Version: HTTP/1.1 Host: 192.168.1.254 Accept-Language: en-US,en;q=0.5 Accept-Encoding: gzip, deflate Referer: http://domain.com/userRpm/StatusRpm.htm Authorization: Basic YWhtEWa6MDfGcmVlc3R6bGH I want to display the value YWhtEWa6MDfGcmVlc3R6bGH Please I need your help thanks in advance experts

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  • XSLT 2.0 Header Leaks into Transformed XML

    - by user1303797
    First, a thank you in advance. Second, this is my first post so apologies for any errors or wrongdoings. I am a noob w/ xml and xslt, and can't seem to figure this out. When I transform some xml using xslt 2.0, some of the headers from the xslt leaks into the new xml. It doesn't seem to do it in xslt 1.0 (granted the xslt is a little different). Here is the xml: <?xml version="1.0" encoding="ISO-8859-1" ?> <xml_content> <feed_name>feed</feed_name> <feed_info> <entry_1> <id>1</id> <pub_date>1320814800</pub_date> </entry_1> </feed_info> </xml_content> Here is the xslt: <xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns="http://www.w3.org/TR/xhtml1/strict"> <xsl:output method="xml" indent="yes" /> <xsl:template match="xml_content"> <Records> <xsl:for-each select="feed_info/entry_1"> <Record> <ID><xsl:value-of select="id" /></ID> <PublicationDate><xsl:value-of select='xs:dateTime("1970-01-01T00:00:00") + xs:integer(pub_date) * xs:dayTimeDuration("PT1S")'/></PublicationDate> </Record> </xsl:for-each> </Records> </xsl:template> </xsl:stylesheet> Here is the new xml. Look specifically at the first "Records" element. <?xml version="1.0" encoding="UTF-8"?> <Records xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns="http://www.w3.org/TR/xhtml1/strict"> <Record> <ID>1</ID> <PublicationDate>2011-11-09T05:00:00</PublicationDate> </Record> </Records>

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  • Google Map only showing Grey Blocks on load - Debug Cert has been obtained

    - by Tom
    I am attempting to follow the Google Map View under the views tutorial for the Android. I have followed step by step but still only see grey blocks when viewed. First: I created a Virtual Device using "Google API's(Google Inc.) Platform 2.2 API Level 8" Second: When creating my project I selected "Google API's Google Inc. Platform 2.2 API Level 8". Third: I obtained the SDK Debug Certificate Fouth: Began Coding. Main.xml <?xml version="1.0" encoding="utf-8"?> <com.google.android.maps.MapView xmlns:android="http://schemas.android.com/apk/res/android" android:id="@+id/mapview" android:layout_width="fill_parent" android:layout_height="fill_parent" android:clickable="true" android:apiKey="0l4sCTTyRmXTNo7k8DREHvEaLar2UmHGwnhZVHQ" / HelloGoogleMaps.java package com.example.googlemap; import android.app.Activity; import android.os.Bundle; import com.google.android.maps.MapView; import com.google.android.maps.MapActivity; public class HelloGoogleMaps extends MapActivity { @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.main); } @Override protected boolean isRouteDisplayed() { return false; } } HelloGoogleMaps Manifest: <?xml version="1.0" encoding="utf-8"?> <manifest xmlns:android="http://schemas.android.com/apk/res/android" package="com.example.googlemap" android:versionCode="1" android:versionName="1.0"> <application android:icon="@drawable/icon" android:label="@string/app_name"> <uses-library android:name="com.google.android.maps" /> <activity android:name=".HelloGoogleMaps" android:label="@string/app_name"> <intent-filter> <action android:name="android.intent.action.MAIN" /> <category android:name="android.intent.category.LAUNCHER" /> </intent-filter> </activity> </application> <uses-permission android:name="android.permission.INTERNET" /> <uses-permission android:name="android.permission.ACCESS_FINE_LOCATION"/> </manifest> Any thoughts?? Thanks!

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