Please help me with the below issue.
I have to read data from excel .
I have created JSPDynpage component and followd below link :
http://help.sap.com/saphelp_sm40/helpdata/en/63/9c0e41a346ef6fe10000000a1550b0/frameset.htm below is my code. I am trying to read excel file using apache poi 3.2 API
try
{
FileUpload fu = (FileUpload)
this.getComponentByName("myfileupload");
// this is the temporary file
if (fu != null) {
// Output to the console to see size and UI.
System.out.println(fu.getSize());
System.out.println(fu.getUI());
// Get file parameters and write it to the console
IFileParam fileParam = fu.getFile();
System.out.println(fileParam);
// Get the temporary file name
File f = fileParam.getFile();
String fileName = fileParam.getFileName();
// Get the selected file name and write ti to the console
ivSelectedFileName = fu.getFile().getSelectedFileName();
File fp = new File(ivSelectedFileName);
myLoc.errorT("#fp#"+fp);
try {
//
**FileInputStream file = new FileInputStream(fp);** --> error at this line
HSSFWorkbook workbook = new HSSFWorkbook(file);
myLoc.errorT("#workbook#"+workbook);
//Get first sheet from the workbook
HSSFSheet sheet = workbook.getSheetAt(0);
myLoc.errorT("#sheet#"+sheet);
//
} catch(Exception ioe) {
myLoc.errorT("#getLocalizedMessage# " + ioe.getLocalizedMessage());
}
Error :
#getLocalizedMessage# C:\Documents and Settings\10608871\Desktop\test.xls (The system cannot find the path specified)
at line FileInputStream file = new FileInputStream(fp);
I have created the PAR file and deploying it on server.
Thanks in Advance,
Aliya Khan.
i resolved the problem, i was passing the worng parameter instead of f
FileInputStream file = new FileInputStream(f);
Related
I'm new to CodeIgniter and having the following issue. When I upload a file, its successfully uploaded on my local folder and the file name saved to the db. The problem is, file name has been changed after uploading.
eg:
file name "screenshot image 123.png" -> "screenshot_image_123.png"
It just convert the space into underscore;this issue occurred only when the file name having space.Other than this issue all the other files uploading successfully. Given below is my code which I used on my controller page:
public function upload_pic()
{
$image_path = realpath(APPPATH . '../uploads');
$config['upload_path'] = $image_path;
$config['allowed_types'] = "gif|jpg|jpeg|png";
$config['file_name'] = $_FILES['file']['name'];
$config['encrypt_name'] = TRUE;
$config['overwrite'] = TRUE;
$this->load->library('upload',$config);
$this->upload->initialize($config);
if(!$this->upload->do_upload('file'))
{
echo $this->upload->display_errors();
}
else
{
$finfo=$this->upload->data();
$data['uploadInfo'] = $finfo;
}
}
Can anyone help me????
Try before saving file name generate some unique
$filename = $_FILES['file']['name'];
$ext = pathinfo($filename, PATHINFO_EXTENSION);
$filename = sha1_file($filename). md5(date('Y-m-d H:i:s:u')) . '.'. $ext; //not only this you can generate any format
$config['file_name'] = $filename; //Use this file name is db too
I'm using the following code on Mac using Mono to unzip a zip file. The zip file contains entries under directories (for example foo/bar.txt). However, in the unzipped directory, instead of creating a directory foo with a file bar.txt, FastZip creates a file foo\bar.txt. How do I get around this?
FastZip fz = new FastZip();
string filePath = #"path\to\myfile.zip";
fz.ExtractZip(filePath, #"path\to\unzip\to", null);
This creates a file foo\bar.txt in path\to\unzip\to.
Apparently cannot use FastZip for this case so I ended up writing my own unzipping mechanism:
string filePath = #"path\to\myfile.zip";
string unzipDir = #"path\to\unzip\to";
using (var zipFile = new ZipFile(filePath))
{
foreach (var zipEntry in zipFile.OfType<ZipEntry>())
{
var unzipPath = Path.Combine(unzipDir, zipEntry.Name);
var directoryPath = Path.GetDirectoryName(unzipPath);
// create directory if needed
if (directoryPath.Length > 0)
{
Directory.CreateDirectory(directoryPath);
}
// unzip the file
var zipStream = zipFile.GetInputStream(zipEntry);
var buffer = new byte[4096];
using (var unzippedFileStream = File.Create(unzipPath))
{
StreamUtils.Copy(zipStream, unzippedFileStream, buffer);
}
}
}
use a forward slash to separate folders when creating the zip
How to take automatically backup of a log file(.txt) when it's size reached a threshold level, say 5MB. The backup file name should be like (log_file_name)_(system_date) and original log file should be cleaned(0 KB).
Please help. Thanks in advance.
Check your log file size using lenght().Then check if its bigger then 5mb call extendLogFile() func.
This is c# code u can easly convert to java
Size check:
if (size > 400 * 100 * 100)
{
extendLogFile(Path);
}
Copy old log file in archive directory and create new log file:
private static void extendLogFile(string lPath)
{
string name = lPath.Substring(0, lPath.LastIndexOf("."));
string UniquName = GenerateUniqueNameUsingDate(); // create a unique name for old log files like '12-04-2013-12-43-00'
string ArchivePath = System.IO.Path.GetDirectoryName(lPath) + "\\Archive";
if (!string.IsNullOrEmpty(ArchivePath) && !System.IO.Directory.Exists(ArchivePath))
{
System.IO.Directory.CreateDirectory(ArchivePath);
}
string newName = ArcivePath + "\\" + UniquName;
if (!File.Exists(newName))
{
File.Copy(lPath, newName + ".txt");
using (FileStream stream = new FileStream(lPath, FileMode.Create))
using (TextWriter writer = new StreamWriter(stream))
{
writer.WriteLine("");
}
}
}
I want to implement Following stuff with my java code in eclipse.
i need to edit the .dict file which is in directory of jar file.
my directory structure is like
C:\Users\bhavik.kama\Desktop\Sphinx\sphinx4-1.0beta6-bin\sphinx4-1.0beta6\modified_jar_dict\*WSJ_8gau_13dCep_16k_40mel_130Hz_6800Hz.jar*\dict\**cmudict04.dict**
Text with bold character is my text file name which i want to edit
and text with italic foramt is my .jar file
now how can i edit this cmudict04.dict file which is reside in WSJ_8gau_13dCep_16k_40mel_130Hz_6800Hz.jar\dict\ directory on runtime with java application.
and i want the jar file with the updated file i have edited.
please can u provide me any help?
thnank you in advance.
I would recommend to use java.util.zip.Using these classes you can read and write the files inside the archive .But modifying the contents is not guaranteed because it may be cached.
Sample tutorial
http://www.javaworld.com/community/node/8362
You can't edit files that are contained in a Jar file and have it saved in the Jar file ... Without, extracting the file first, updating it and creating a new Jar by copying the contents of the old one over to the new one, deleting the old one and renaming the new one in its place...
My suggestion is find a better solution
I had succeded to edit jar file and wrap it back as it is...with the following code
public void run() throws IOException
{
Manifest manifest = new Manifest();
manifest.getMainAttributes().put(Attributes.Name.MANIFEST_VERSION, "1.0");
// JarOutputStream target = new JarOutputStream(new FileOutputStream("E:\\hiren1\\WSJ_8gau_13dCep_16k_40mel_130Hz_6800Hz.jar"), manifest);
// add(new File("E:\\hiren1\\WSJ_8gau_13dCep_16k_40mel_130Hz_6800Hz/"), target);
JarOutputStream target = new JarOutputStream(new FileOutputStream("C:\\Users\\bhavik.kama\\Desktop\\Sphinx\\sphinx4-1.0beta6-bin\\sphinx4-1.0beta6\\modified_jar_dict\\WSJ_8gau_13dCep_16k_40mel_130Hz_6800Hz.jar"), manifest);
add(new File("C:\\Users\\bhavik.kama\\Desktop\\Sphinx\\sphinx4-1.0beta6-bin\\sphinx4-1.0beta6\\modified_jar_dict\\WSJ_8gau_13dCep_16k_40mel_130Hz_6800Hz/"), target);
target.close();
}
private void add(File source, JarOutputStream target) throws IOException
{
BufferedInputStream in = null;
try
{
if (source.isDirectory())
{
//String name = source.getPath().replace("\\", "/");
if(isFirst)
{
firstDir = source.getParent() + "\\";
isFirst = false;
}
String name = source.getPath();
name = name.replace(firstDir,"");
if (!name.isEmpty())
{
if (!name.endsWith("/"))
name += "/";
JarEntry entry = new JarEntry(name);
entry.setTime(source.lastModified());
target.putNextEntry(entry);
target.closeEntry();
}
for (File nestedFile: source.listFiles())
add(nestedFile, target);
return;
}
String name = source.getPath();
name = name.replace(firstDir,"").replace("\\", "/");
//JarEntry entry = new JarEntry(source.getPath().replace("\\", "/"));
JarEntry entry = new JarEntry(name);
//JarEntry entry = new JarEntry(source.getName());
entry.setTime(source.lastModified());
target.putNextEntry(entry);
in = new BufferedInputStream(new FileInputStream(source));
byte[] buffer = new byte[1024];
while (true)
{
int count = in.read(buffer);
if (count == -1)
break;
target.write(buffer, 0, count);
}
target.closeEntry();
}
finally
{
if (in != null)
in.close();
}
}
I am trying to upload a file via a form and then save in in SQL as a blob.
I already have my form working fine, my database is fully able to take the blob and I have a controller that take the file, saves it in a local directory:
[AcceptVerbs(HttpVerbs.Post)]
public ActionResult FileUpload(int id, HttpPostedFileBase uploadFile)
{
//allowed types
string typesNonFormatted = "text/plain,application/msword,application/pdf,image/jpeg,image/png,image/gif";
string[] types = typesNonFormatted.Split(',');
//
//Starting security check
//checking file size
if (uploadFile.ContentLength == 0 && uploadFile.ContentLength > 10000000)
ViewData["StatusMsg"] = "Could not upload: File too big (max size 10mb) or error while transfering the file.";
//checking file type
else if(types.Contains(uploadFile.ContentType) == false)
ViewData["StatusMsg"] = "Could not upload: Illigal file type!<br/> Allowed types: images, Ms Word documents, PDF, plain text files.";
//Passed all security checks
else
{
string filePath = Path.Combine(HttpContext.Server.MapPath("../Uploads"),
Path.GetFileName(uploadFile.FileName)); //generating path
uploadFile.SaveAs(filePath); //saving file to final destination
ViewData["StatusMsg"] = "Uploaded: " + uploadFile.FileName + " (" + Convert.ToDecimal(uploadFile.ContentLength) / 1000 + " kb)";
//saving file to database
//
//MISSING
}
return View("FileUpload", null);
}
Now all I am missing is putting the file in the database. I could not find anything on the subject... I found some way to do it in a regular website but nothing in MVC2.
Any kind of help would be welcome!
Thank you.
This could help: http://byatool.com/mvc/asp-net-mvc-upload-image-to-database-and-show-image-dynamically-using-a-view/
Since you have HttpPostedFileBase in your controllers method, all you need to do is:
int length = uploadFile.ContentLength;
byte[] tempImage = new byte[length];
myDBObject.ContentType = uploadFile.ContentType;
uploadFile.InputStream.Read(tempImage, 0, length);
myDBObject.ActualImage = tempImage ;
HttpPostedFileBase has a InputStream property
Hope this helps.
Alright thanks to kheit, I finaly got it working. Here's the final solution, it might help someone out there.
This script method takes all the file from a directory and upload them to the database:
//upload all file from a directory to the database as blob
public void UploadFilesToDB(long UniqueId)
{
//directory path
string fileUnformatedPath = "../Uploads/" + UniqueId; //setting final path with unique id
//getting all files in directory ( if any)
string[] FileList = System.IO.Directory.GetFiles(HttpContext.Server.MapPath(fileUnformatedPath));
//for each file in direcotry
foreach (var file in FileList)
{
//extracting file from directory
System.IO.FileStream CurFile = System.IO.File.Open(file, System.IO.FileMode.Open);
long fileLenght = CurFile.Length;
//converting file to a byte array (byte[])
byte[] tempFile = new byte[fileLenght];
CurFile.Read(tempFile, 0, Convert.ToInt32(fileLenght));
//creating new attachment
IW_Attachment CurAttachment = new IW_Attachment();
CurAttachment.attachment_blob = tempFile; //setting actual file
string[] filedirlist = CurFile.Name.Split('\\');//setting file name
CurAttachment.attachment_name = filedirlist.ElementAt(filedirlist.Count() - 1);//setting file name
//uploadind attachment to database
SubmissionRepository.CreateAttachment(CurAttachment);
//deleting current file fromd directory
CurFile.Flush();
System.IO.File.Delete(file);
CurFile.Close();
}
//deleting directory , it should be empty by now
System.IO.Directory.Delete(HttpContext.Server.MapPath(fileUnformatedPath));
}
(By the way IW_Attachment is the name of one of my database table)