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  • Updating target workbook - extracting data from source workbook

    - by Allan
    My question is as follows: I have given a workbook to multiple people. They have this workbook in a folder of their choice. The workbook name is the same for all people, but folder locations vary. Let's assume the common file name is MyData-1.xls. Now I have updated the workbook and want to give it to these people. However when they receive the new one (let's call it MyData-2.xls) I want specific parts of their data pulled from their file (MyData-1) and automatically put into the new one provided (MyData-2). The columns and cells to be copied/imported are identical for both workbooks. Let's assume I want to import cell data (values only) from MyData-1.xls, Sheet 1, cells B8 through C25 ... to ... the same location in the MyData-2.xls workbook. How can I specify in code (possibly attached to a macro driven import data now button) that I want this data brought into this new workbook. I have tried it at my own location by opening the two workbooks and using the copy/paste-special with links process. It works really well, but It seems to create a hard link between the two physical workbooks. I changed the name of the source workbook and it still worked. This makes me believe that there is a "hard link" between the tow and that this will not allow me to give the target (MyData-2.xls) workbook to others and have it find their source workbook.

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  • Uploading multiple files using Spring MVC 3.0.2 after HiddenHttpMethodFilter has been enabled

    - by Tiny
    I'm using Spring version 3.0.2. I need to upload multiple files using the multiple="multiple" attribute of a file browser such as, <input type="file" id="myFile" name="myFile" multiple="multiple"/> (and not using multiple file browsers something like the one stated by this answer, it indeed works I tried). Although no versions of Internet Explorer supports this approach unless an appropriate jQuery plugin/widget is used, I don't care about it right now (since most other browsers support this). This works fine with commons fileupload but in addition to using RequestMethod.POST and RequestMethod.GET methods, I also want to use other request methods supported and suggested by Spring like RequestMethod.PUT and RequestMethod.DELETE in their own appropriate places. For this to be so, I have configured Spring with HiddenHttpMethodFilter which goes fine as this question indicates. but it can upload only one file at a time even though multiple files in the file browser are chosen. In the Spring controller class, a method is mapped as follows. @RequestMapping(method={RequestMethod.POST}, value={"admin_side/Temp"}) public String onSubmit(@RequestParam("myFile") List<MultipartFile> files, @ModelAttribute("tempBean") TempBean tempBean, BindingResult error, Map model, HttpServletRequest request, HttpServletResponse response) throws IOException, FileUploadException { for(MultipartFile file:files) { System.out.println(file.getOriginalFilename()); } } Even with the request parameter @RequestParam("myFile") List<MultipartFile> files which is a List of type MultipartFile (it can always have only one file at a time). I could find a strategy which is likely to work with multiple files on this blog. I have gone through it carefully. The solution below the section SOLUTION 2 – USE THE RAW REQUEST says, If however the client insists on using the same form input name such as ‘files[]‘ or ‘files’ and then populating that name with multiple files then a small hack is necessary as follows. As noted above Spring 2.5 throws an exception if it detects the same form input name of type file more than once. CommonsFileUploadSupport – the class which throws that exception is not final and the method which throws that exception is protected so using the wonders of inheritance and subclassing one can simply fix/modify the logic a little bit as follows. The change I’ve made is literally one word representing one method invocation which enables us to have multiple files incoming under the same form input name. It attempts to override the method protected MultipartParsingResult parseFileItems(List fileItems, String encoding) {} of the abstract class CommonsFileUploadSupport by extending the class CommonsMultipartResolver such as, package multipartResolver; import java.io.UnsupportedEncodingException; import java.util.HashMap; import java.util.Iterator; import java.util.List; import java.util.Map; import javax.servlet.ServletContext; import org.apache.commons.fileupload.FileItem; import org.springframework.util.StringUtils; import org.springframework.web.multipart.MultipartException; import org.springframework.web.multipart.MultipartFile; import org.springframework.web.multipart.commons.CommonsMultipartFile; import org.springframework.web.multipart.commons.CommonsMultipartResolver; final public class MultiCommonsMultipartResolver extends CommonsMultipartResolver { public MultiCommonsMultipartResolver() { } public MultiCommonsMultipartResolver(ServletContext servletContext) { super(servletContext); } @Override @SuppressWarnings("unchecked") protected MultipartParsingResult parseFileItems(List fileItems, String encoding) { Map<String, MultipartFile> multipartFiles = new HashMap<String, MultipartFile>(); Map multipartParameters = new HashMap(); // Extract multipart files and multipart parameters. for (Iterator it = fileItems.iterator(); it.hasNext();) { FileItem fileItem = (FileItem) it.next(); if (fileItem.isFormField()) { String value = null; if (encoding != null) { try { value = fileItem.getString(encoding); } catch (UnsupportedEncodingException ex) { if (logger.isWarnEnabled()) { logger.warn("Could not decode multipart item '" + fileItem.getFieldName() + "' with encoding '" + encoding + "': using platform default"); } value = fileItem.getString(); } } else { value = fileItem.getString(); } String[] curParam = (String[]) multipartParameters.get(fileItem.getFieldName()); if (curParam == null) { // simple form field multipartParameters.put(fileItem.getFieldName(), new String[] { value }); } else { // array of simple form fields String[] newParam = StringUtils.addStringToArray(curParam, value); multipartParameters.put(fileItem.getFieldName(), newParam); } } else { // multipart file field CommonsMultipartFile file = new CommonsMultipartFile(fileItem); if (multipartFiles.put(fileItem.getName(), file) != null) { throw new MultipartException("Multiple files for field name [" + file.getName() + "] found - not supported by MultipartResolver"); } if (logger.isDebugEnabled()) { logger.debug("Found multipart file [" + file.getName() + "] of size " + file.getSize() + " bytes with original filename [" + file.getOriginalFilename() + "], stored " + file.getStorageDescription()); } } } return new MultipartParsingResult(multipartFiles, multipartParameters); } } What happens is that the last line in the method parseFileItems() (the return statement) i.e. return new MultipartParsingResult(multipartFiles, multipartParameters); causes a compile-time error because the first parameter multipartFiles is a type of Map implemented by HashMap but in reality, it requires a parameter of type MultiValueMap<String, MultipartFile> It is a constructor of a static class inside the abstract class CommonsFileUploadSupport, public abstract class CommonsFileUploadSupport { protected static class MultipartParsingResult { public MultipartParsingResult(MultiValueMap<String, MultipartFile> mpFiles, Map<String, String[]> mpParams) { } } } The reason might be - this solution is about the Spring version 2.5 and I'm using the Spring version 3.0.2 which might be inappropriate for this version. I however tried to replace the Map with MultiValueMap in various ways such as the one shown in the following segment of code, MultiValueMap<String, MultipartFile>mul=new LinkedMultiValueMap<String, MultipartFile>(); for(Entry<String, MultipartFile>entry:multipartFiles.entrySet()) { mul.add(entry.getKey(), entry.getValue()); } return new MultipartParsingResult(mul, multipartParameters); but no success. I'm not sure how to replace Map with MultiValueMap and even doing so could work either. After doing this, the browser shows the Http response, HTTP Status 400 - type Status report message description The request sent by the client was syntactically incorrect (). Apache Tomcat/6.0.26 I have tried to shorten the question as possible as I could and I haven't included unnecessary code. How could be made it possible to upload multiple files after Spring has been configured with HiddenHttpMethodFilter? That blog indicates that It is a long standing, high priority bug. If there is no solution regarding the version 3.0.2 (3 or higher) then I have to disable Spring support forever and continue to use commons-fileupolad as suggested by the third solution on that blog omitting the PUT, DELETE and other request methods forever. Just curiously waiting for a solution and/or suggestion. Very little changes to the code in the parseFileItems() method inside the class MultiCommonsMultipartResolver might make it to upload multiple files but I couldn't succeed in my attempts (again with the Spring version 3.0.2 (3 or higher)).

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  • What can a company possibly gain by making Android phones hard to root?

    - by Chinmay Kanchi
    As someone who recently got a HTC Hero, I had to jump through several hoops to get root access on the phone to install custom firmware. Now, Android is open-source and fairly easy to build and hack on an emulator. It seems to be against the spirit of open-source to lock down a phone so you can't hack the phone itself. Now, often, there are understandable (though not always justifiable) reasons for locking a device down. For example, it might have proprietary software on it or you might want to retain control of the platform. However, Android by its open-source nature makes such concerns moot. Everyone and their dog has access to the userland code, and HTC is forced by the GPL to release kernel sources for each of their devices. So, I fail to see any motivation for alienating the hackers, when there is no possible benefit (in my mind) to be had from doing this. Any idea why a company would want to do this? Is it just short-sightedness or am I missing possible commercial implications of this?

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  • How can I format Custom Data and display in autocomplete when source is an DB

    - by Andres Scarpone
    so I'm trying to get some info in the auto-complete widget like it's shown in the JQuery UI demo Demo, the only problem is they use a variable that they fill with the data they want to show, I instead want to access the data and the different description and stuff using a Data Base in MySQL, for this I have changed the source to use another php page that looks up the info. here is the code for the Auto-complete, I really don't understand the methods so I haven't changed it from the basic search. This is the JS: $(document).ready((function(){ $( "#completa" ).autocomplete({ source: "buscar.php", minLength: 1, focus: function (event, ui){ $("#completa").val(ui.item.val); return false; }; })); This is what I have in buscar.php: <?php $conec = mysql_connect(localhost, root, admin); if(!$conec) { die(mysql_error()); } else { $bd = mysql_select_db("ve_test",$conec ); if(!$bd) { die(mysql_error()); } } $termino = trim(strip_tags($_GET['term']));//Obtener el termino que envia el autocompletar $qstring = "SELECT name, descripcion FROM VE_table WHERE name LIKE '%".$termino."%'"; $result = mysql_query($qstring);//Solicitud a la Base de Datos while ($row = mysql_fetch_array($result,MYSQL_ASSOC))//Realizar un LOOP sobre los valores obtenidos { $row['value']=htmlentities(stripslashes($row['name'])); $row_set[] = $row;//build an array } echo json_encode($row_set);//Enviar los datos al autocompletar en codificacion JSON, Altamente Necesario. ?

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  • Report Viewer Configuration error - In View Source of webpage

    - by sarada
    I found the following error message when I checked View source of the web page , but the web page works just fine. Our Test lead found the error while performing Assertion tests. Report Viewer Configuration Error The Report Viewer Web Control HTTP Handler has not been registered in the application's web.config file. Add <add verb=" * " path="Reserved.ReportViewerWebControl.axd" type = "Microsoft.Reporting.WebForms.HttpHandler, Microsoft.ReportViewer.WebForms, Version=10.0.0.0, Culture=neutral, PublicKeyToken=b03f5f7f11d50a3a" /> to the system.web/httpHandlers section of the web.config file, or add <add name="ReportViewerWebControlHandler" preCondition="integratedMode" verb="*" path="Reserved.ReportViewerWebControl.axd" type="Microsoft.Reporting.WebForms.HttpHandler, Microsoft.ReportViewer.WebForms, Version=10.0.0.0, Culture=neutral, PublicKeyToken=b03f5f7f11d50a3a" /> to the system.webServer/handlers section for Internet Information Services 7 or later Why is this error message coming up in view source.. Note:There is a div tag around this error message which has style="display:none" I am trying to find out why but everybody has only discussed this error message as one which is thrown in the webpage. The changes suggested to web.config are already present in our config file.

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  • iPhone Network Share Access

    - by user361988
    I am trying to download a data file from a local network share to an iPhone device. I have placed the file on a computer on the network and can view through browsers such as Chrome or Mozilla, from any computer on the local network. However, Safari on a Mac and the iPhone do not find the file! An example of the URL I use is 'file://computer/SharedDocs/file.csv'. Why do Safari and the iPhone fail to find the file?

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  • Rails original release source code

    - by user547057
    Hi, I've been looking to read some ruby code(specifically Rails) but I don't want to start with the current version of Rails since it has a lot of stuff I don't need and even more stuff that I wouldn't probably understand. I want to read only the core of Rails and supposedly the early versions were small and kind of easy to wrap one's head around(even for a neophyte like me). I have tried searching for the original release of rails, but have not been able to find it. The github repo consists of thousands of commits and I don't want to wade through those. What I want is to know whether there is any place I can get a zip or tar file with the original rails source or even the other early versions. Pointers to links will be very much appreciated. Thanks. p.s I'm new to ruby programming but not programming in general(I know a little python and scheme) and I understand blocks, lambdas and OO stuff, so I think I can tackle the rails source code. If anyone knows of other ruby projects that make for good code reading, i'd love to know of those too.

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  • How to jar java source files from different (sub-)directories?

    - by Holger
    Consider the following directory structure: ./source/com/mypackage/../A.java ./extensions/extension1/source/com/mypackage/../T.java ./extensions/extension2/source/com/mypackage/../U.java ... ./extensions/extensionN/source/com/mypackage/../Z.java I want to produce a source jar with the following contents: com/mypackage/../A.java com/mypackage/../T.java com/mypackage/../U.java ... com/mypackage/../Z.java I know I could use a fileset for each source directory. But is there an easy solution using ANT without having to refer to all extensions explicitly?

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  • Compiling GWT 2.6.1 at Java 7 source level

    - by Neeko
    I've recently updated my GWT project to 2.6.1, and started to make use of Java 7 syntax since 2.6 now supports Java 7. However, when I attempt to compile, I'm receiving compiler errors such as [ERROR] Line 42: '<>' operator is not allowed for source level below 1.7 How do I specify the GWT compiler to target 1.7? I was under the impression that it would do that by default, but I guess not. I've attempted cleaning the project, including deleting the gwt-unitCache directory but to no avail. Here is my Ant compile target. <target name="compile" depends="prepare"> <javac includeantruntime="false" debug="on" debuglevel="lines,vars,source" srcdir="${src.dir}" destdir="${build.dir}"> <classpath refid="project.classpath"/> </javac> </target> <target name="gwt-compile" depends="compile"> <java failonerror="true" fork="true" classname="com.google.gwt.dev.Compiler"> <classpath> <!-- src dir is added to ensure the module.xml file(s) are on the classpath --> <pathelement location="${src.dir}"/> <pathelement location="${build.dir}"/> <path refid="project.classpath"/> </classpath> <jvmarg value="-Xmx256M"/> <arg value="${gwt.module.name}"/> </java> </target>

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  • How to create a named temporary file in memory?

    - by conradlee
    I would like to use Python's tempfile module to create a temporary file that I will use for communication between processes (use of pipes is awkward). The documentation I've linked to above shows two functions that almost do what I want: tempfile.NamedTemporaryFile # For creating named tempfiles tempfile.SpooledTemporaryFile # For creating tempfiles in memory but actually I want a tempfile that is both named AND in memory. Any ideas?

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  • please give me a solution

    - by user327832
    here is the code i have written so far but ended up giving me error import java.io.File; import java.io.FileInputStream; import java.io.IOException; import java.io.InputStream; public class Main { public static void main(String[] args) throws Exception { File file = new File("c:\\filea.txt"); InputStream is = new FileInputStream(file); long length = file.length(); System.out.println (length); bytes[] bytes = new bytes[(int) length]; try { int offset = 0; int numRead = 0; while (numRead >= 0) { numRead = is.read(bytes); } } catch (IOException e) { System.out.println ("Could not completely read file " + file.getName()); } is.close(); Object[] see = new Object[(int) length]; see[1] = bytes; System.out.println ((String[])see[1]); } }

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  • Login to website and use cookie to get source for another page

    - by Stu
    I am trying to login to the TV Rage website and get the source code of the My Shows page. I am successfully logging in (I have checked the response from my post request) but then when I try to perform a get request on the My Shows page, I am re-directed to the login page. This is the code I am using to login: private string LoginToTvRage() { string loginUrl = "http://www.tvrage.com/login.php"; string formParams = string.Format("login_name={0}&login_pass={1}", "xxx", "xxxx"); string cookieHeader; WebRequest req = WebRequest.Create(loginUrl); req.ContentType = "application/x-www-form-urlencoded"; req.Method = "POST"; byte[] bytes = Encoding.ASCII.GetBytes(formParams); req.ContentLength = bytes.Length; using (Stream os = req.GetRequestStream()) { os.Write(bytes, 0, bytes.Length); } WebResponse resp = req.GetResponse(); cookieHeader = resp.Headers["Set-cookie"]; String responseStream; using (StreamReader sr = new StreamReader(resp.GetResponseStream())) { responseStream = sr.ReadToEnd(); } return cookieHeader; } I then pass the cookieHeader into this method which should be getting the source of the My Shows page: private string GetSourceForMyShowsPage(string cookieHeader) { string pageSource; string getUrl = "http://www.tvrage.com/mytvrage.php?page=myshows"; WebRequest getRequest = WebRequest.Create(getUrl); getRequest.Headers.Add("Cookie", cookieHeader); WebResponse getResponse = getRequest.GetResponse(); using (StreamReader sr = new StreamReader(getResponse.GetResponseStream())) { pageSource = sr.ReadToEnd(); } return pageSource; } I have been using this previous question as a guide but I'm at a loss as to why my code isn't working.

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  • [NSData dataWithContentsOfFile:path] doesn't work

    - by Felics
    Hello, when I have the fallowing code to read a binary file: NSString* file = [NSString stringWithUTF8String:fileName]; NSString* filePath = resource ? [[NSBundle mainBundle] pathForResource:file ofType:nil] : [[NSSearchPathForDirectoriesInDomains(NSDocumentDirectory, NSUserDomainMask, YES) objectAtIndex:0] stringByAppendingPathComponent: file]; NSData* fileData = [NSData dataWithContentsOfFile:filePath]; Where "fileName" and resource are load function parameters. "resource" is used to know if the file is located in application bundle or in Documents. Sometimes this code works well and sometimes it doesn't. As far I saw this problem is random. I can run the code 10 times in a row and it works fine and after that it gives me nil data without any modification. Does anybody knows what could be the problem? Could it be related with file extension or file name? Thank you. PS: I use this code on iPhone Simulator and the file exists in application bundle.

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  • Find files in a folder using Java

    - by Sii
    Hello, I'm new here so be kind to my stupidity :) What I need to do if Search a folder say C:\example I then need to go through each file and check to see if it matches a few start characters so if files start temp****.txt tempONE.txt tempTWO.txt So if the file starts with temp and has an extension .txt I would like to then put that file name into a File file = new File("C:/example/temp***.txt); so I can then read in the file, the loop then needs to move onto the next file to check to see if it meets as above. I hope this makes sense, thanks for taking the time to view this I do apperciate it :)

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  • Anyone know any good backend user online file manager?

    - by skyhigh
    Hi I'm looking for a backend system where your clients can login and upload files to your server, download files from the server and you can delete the users, create users, etc. I do not know the proper name for this kind of software. Maybe its called online file manager? Any recommendations? My server supports PHP, apache and mysq. Thanks

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