File Complexity/Size Maintenance; Paths vs Compund Objects

I have found myself using different methods to make designs, but I was wondering what was the best practice for archiving FINISHED products.
In terms of anchor and path complexity: would it be best to save files as paths with width controls, or should I be outlining strokes and flattening as compound objects?
If it turns out that the results would vary with every project, then I can accept saving two versions and deleting whichever happens to be the bigger file.

Well now, no need to drag my beliefs into this, but if you must know...
My understanding is that progress should be reduced to it's simplest form at regular intervals. The extraction of modular devices (commonly used symbols, purposefully arranged actions, etc.) from previous projects is what drives "development". I have no intention of destroying my work. Just making sure that snapshots exist in case my memory/workspace fails me.
That said, let's pretend I have a client with a portable device of limited capabilities, and I have to keep the poor things CPU from lagging by giving it less instructions when redrawing the vector graphics. Then let's throw in a time limit that only gives time for a few adjustments:
How will I ever prepare this file?

Similar Messages

Outofmemory while creating a new object of file with size more than 100 MB

I have created an application which generates a report by getting the data from our archived files (.zip file).By the time, the application is reaching a file with size more than 100 mb, it is running out fo memory while creating the object of that particular file. Can some one help me by tellin if there id way to resolve this issue?
Thanks in advance

If you're getting OutOfMemoryError, the simplest thing to try is to give the VM more memory at startup. For Sun's VM, I believe the default is 64 MB. You can increase this by using the -X args that control the heap size. For example: java -Xms128m -Xmx256m ... etc. ... This says start with 128 MB of heap, and allow it to grow up to 256 MB.
One thing to consider, though, is do you need that much stuff in memory at once? A more intelligent approach might be to unpack the archive to the file system, and then read a file at a time or a line at a time or a whatever at a time is appropriate for the processing you need to do.

How is it posible to get the File name, size and type from a File out the H

How is it posible to get the File name, size and type from a File out the HttpServletRequest. I want to upload a File from a client and save it on a server with the client name. I want to conrole before saving the name, type and size of the file.How is it posible to get the File name, size and type from a File out the HttpServletRequest.
form JSP
<form name="form" method="post" action="procesuploading.jsp" ENCTYPE="multipart/form-data">
File: <input type="file" name="filename"/
Path: <input type="text" readonly="" name="path" value="c:"/
Saveas: <input type="text" name="saveas"/>
<input name="submit" type="submit" value="Upload" />
</form>
proces JSP
<%@ page language="java" %>
<%@ page import="java.util.*" %>
<%@ page import="FileUploadBean" %>
<jsp:useBean id="TheBean" scope="page" class="FileUploadBean" />
<%
TheBean.doUpload(request);
%>
BEAN
import java.io.*;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import javax.servlet.ServletInputStream;
public class FileUploadBean {
public void doUpload(HttpServletRequest request) throws IOException
          String melding = "";
          String filename = request.getParameter("saveas");
          String path = request.getParameter("path");
          PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("test.java")));
          ServletInputStream in = request.getInputStream();
          int i = in.read();
          System.out.println("filename:"+filename);
          System.out.println("path:"+path);
          while (i != -1)
               pw.print((char) i);
               i = in.read();
          pw.close();
}

Thanks it works great.
Here an excample from my code
import org.apache.commons.fileupload.*;
public class FileUploadBean extends Object implements java.io.Serializable{
String foutmelding = "geen";
String path;
String filename;
public boolean doUpload(HttpServletRequest request) throws IOException
     try
     // Create a new file upload handler
     FileUpload upload = new FileUpload();
     // Set upload parameters
     upload.setSizeMax(100000);
     upload.setSizeThreshold(100000000);
     upload.setRepositoryPath("/");
     // Parse the request
     List items = upload.parseRequest(request);
     // Process the uploaded fields
     Iterator iter = items.iterator();
     while (iter.hasNext())
     FileItem item = (FileItem) iter.next();
          if (item.isFormField())
               String stringitem = item.getString();
     else
          String filename = "";
               int temp = item.getName().lastIndexOf("\\");
               filename = item.getName().substring(temp,item.getName().length());
               File bestand = new File(path+filename);
               if(item.getSize() > SizeMax && SizeMax != -1){foutmelding = "bestand is te groot.";return false;}
               if(bestand.exists()){foutmelding ="bestand bestaat al";return false;}
               FileOutputStream fOut = new FileOutputStream(bestand);
               BufferedOutputStream bOut = new BufferedOutputStream(fOut);
               int bytesRead =0;
               byte[] data = item.get();
               bOut.write(data, 0 , data.length);
               bOut.close();
     catch(Exception e)
          System.out.println("er is een foutontstaan bij het opslaan de een bestand "+e);
          foutmelding = "Bestand opsturen is fout gegaan";
     return true;
     }

HOw to create a text file in the given path and delete it after the use?

Hi all,
I am trying to create a text file at the given path and delete the created file after the use.
I am using following code.:
import java.io.*;
// write binary data as characters
public class RanIO {
                                        public static void main(String f[])
                                                  // First illustrate append
                                                  String lineSep = "\n";
                                                  try {
                                                                 File temp= new File("C:/Ash","cute.txt");
                                                  boolean ch=temp.createNewFile();
                                                  if(ch)
                                                       System.out.println("file created");
                                                  else
                                                  System.out.println("file Not created");
                                                  //writing to file
                                             /*     PrintWriter p = new PrintWriter(new BufferedWriter(new FileWriter("cute.txt",true)));
                                                  p.print("Emp NO");
                                                  p.close();*/
                                                            // Open fileWriter in append mode
                                                                           FileWriter fos = new FileWriter(temp, true);
                                                                           BufferedWriter bw = new BufferedWriter(fos);
                                                                           PrintWriter pw = new PrintWriter(fos);
                                                                           double d=550;
                                                                      // lineSep = System.getProperty("line.separator");
                                                                      pw.print("Hello");
                                                                      //pw.print( lineSep );
                                                                      pw.print( d );
                                                                      pw.close();
                              boolean det=temp.delete();
                                             if(det)
                                                  System.out.println("File deleted");
                                             else
                                                  System.out.println("File not deleted");
                                             } catch (IOException ioe)
                                                            System.out.println( "Append IO error:" + ioe );
My problem:
1)
I am not able to write to the file. I want to know, where i am going wrong.
It is giving error message like
"Canot resolve Symbol: temp,"
But, FileWriter Constructor should accept a File type parameter.
here temp is a file parameter.
If i am not using file=new file();
i can't delete the file after the use. i.e if i use
PrintWriter p = new PrintWriter(new BufferedWriter(new FileWriter("cute.txt",true)));
how can i delete cute.txt after the use?
2)
I am not able to write to the text file. file is created but, a blank file.
"Hello" is not written into the text file.
can anyone help me in this regard
Thanks in advance
Ashvini

Thank you Ram,
But, i want to create a text file in Append mode.
for that i used
FileWriter fos = new FileWriter(temp,true); But, it is not accepting FileWriter constructor in
this format. if i use
FileWriter fos = new
FileWriter("c:/ash/cute.txt",true); it works fine. !!!!!Here's the javadoc
public FileWriter(File file,
boolean append)
throws IOExceptionConstructs a FileWriter object given a File object. If the second argument is true, then bytes will be written to the end of the file rather than the beginning.
Parameters:
file - a File object to write to
append - if true, then bytes will be written to the end of the file rather than the beginning
Throws:
IOException - if the file exists but is a directory rather than a regular file, does not exist but cannot be created, or cannot be opened for any other reason
Since:
1.4
Are you using jdk.13 or lower ?
>
ONe more doubt, Does flush method deletes a file?
if not, then i need to use
File temp=new File("c:/ash/cute.txt");
FileWriter fos = new FileWriter(temp,true); //which
is again a problem
if(temp.delete())
out.println("File is deleted");
}I don't know whether i am taking it wrong ! or
anything wrong with my coding ! but, after creating
and writing data into a text file. I must delete it
as it contains confidential informations.
Regards,
Thanks.'flush' writes to a file immediately. Else you should explicitly call 'flush' to write contents from buffer to underlying source.
javadoc again
PrintWriter
public PrintWriter(Writer out,
boolean autoFlush)Create a new PrintWriter.
Parameters:
out - A character-output stream
autoFlush - A boolean; if true, the println, printf, or format methods will flush the output buffer
cheers,
ram.
Question; What do you gain by opening a file, writing to it and deleting it in the same program ?

How do I add a jar file into the build path of the compiler?

Hey,
I'm trying to import a jar file into the build path of the compilation process, but it does not find the packages or the classes that are in it.
I think I don't add it right...
          ArrayList<String> options=new ArrayList<String>();
          options.add("-d");
          options.add(targetDirectory);
          options.add("-classpath");
          for(String str:includeDirectory)
               options.add(str);
          if (!compiler.getTask(writer, fileManager, diagnostics, options, classes, compilationUnits).call());
                ....and I've tried this way:
     public void setTargetDirectory(String targetDirectory) {
          this.targetDirectory = "-d " + targetDirectory;
     private void compile(Iterable<? extends JavaFileObject> compilationUnits) throws Exception {
          ArrayList<String> options = new ArrayList<String>();
          options.add(targetDirectory);
          String classPath="-cp ";// tried this also with "-classpath"
          for (String str : includeDirectory)
               classPath+=str+":";
          options.add(classPath);
     if (!compiler.getTask(writer, fileManager, diagnostics, options, classes, compilationUnits).call())
          // throw new Exception("Compilation Error");
     }Thanks in advance,
Adam.
Edited by: Adam-Z. on Feb 24, 2010 5:41 AM
Edited by: Adam-Z. on Feb 24, 2010 5:42 AM

Thank you for your reply,
Q: Are there .class files in that directory in that jar file? (the compiler doesn't ( can't )) look for directories, it can just look for specific files , and scan to get a list of all files matching certain criteria. So if there are no class files, it will say the package doesn't exist, even if there is a directory, possibly containing other files.yes there are class files in the jar, the tree structure:
j2MeDataChunkGenerator_Plugin\(lots of class files)
META-INF\manifest.mf
and thats it.
, your code will only work on windows because other platforms use a different path separator. You should use java.io.File.pathSeparator not explicit ';" when building your classpath. (this is unrelated to your problem, but you should correct it)will do, thanks.
Q: Is that error in your post formatted by your own diagnostics? (we could possibly help you better if we didn't have to guess!!)I would not post my own error code, this text is generated by the compiler diagnostic.
{code}
     System.err.println(" Error details: " + diagnostic.getMessage(null));
{code}
Q: Is line 3 of ImageCroper_Editor.java (sic) an import statement? (we could possibly help you better if we didn't have to guess!!)it is an import error... didn't the error message stated that it is an import problem? wired, I'm sure before it did. anyway it is an import error.
Also you don't show us what the variable includeDirectory is in terms of type, and contents, that might be helpful. (we could possibly help you better if we didn't have to guess!!)It has only one String object: "D:\%Important Documents\WorkSpaces\PacMan\ApplicationManager\Plug-in\Data Chunk Designer.jar"
the last file on the classpath list.
Q: Have you proven this? that i did post, in this long line of text.
Q: Is the compiler finding other classes (in other packages) in that same jar file?No. all the class files are in the jar, they all have entries that start with "j2MeDataChunkGenerator_Plugin\*.class", and since I get 47 errors I guess it does not load any other class.
thank you for you comments, the problem with having these errors, is that I can't even get a piece of information where this error is coming from, only that it is an import loading error package not found, what does that mean? that the jar was not loaded in compilation(no error about this), that the jar is corrupted(no error about this), that the path is incorrect(it is correct I made sure), that there is no such package in the jar(There is), that the compiler does not load the package(does it even do that?), really I can't even guess why this happens, I've been at this on and of all day today, really annoying.
Thanks,
Adam.

Logical file name or logical path name incorrectly defined

Dear All,
We are doing archival in our IDES for test purpose before we do it to our Production.
Steps Performed
Copied AM_ASSET archive object to ZAM_ASSET
Logical Path
Logical path    ZAM_ASSET
Name            Asset
Syntax group    UNIX       Unix compatible
Physical path   /archive/test/<FILENAME>
Logical File Name
   Logical file    ZAMASSET
   Name            Asset
   Physical file   FI_<MONTH>_<DAY>.txt
   Data format     ASC
   Applicat.area   AM
   Logical path    ZAM_ASSET
But when we run the WRITE though SARA , in the job log we get the following
Logical file name or logical path name incorrectly defined
When generating a file name for an archive file that is to be created, the system determined that the logical file name FIAA_ARCHIVE_DATA_FILE or the logical path name ARCHIVE_GLOBAL_PATH was defined incorrectly.
But we have maintained a Logical name ZAMASSET , so we are unable to change the location of archived file and as well as the format.
So is there any setting we need to maintain apart from the logical file name and logical file path.
Suggestions are highly appreciated.
Thanks in anticipation

hi,
follow this steps :
- transaction SARA
- enter authorization objects, eg SD_VBAK
- hit button CUSTOMIZING
- Archiving Object-Specific Customizing: execute Technical Setting
- field Logical File Name enter or select ARCHIVE_DATA_FILE
- leave CONTENT REPOSITORY as blank if you are not using 3rd party for storing (eg. IBM Tivoli)
- back to customizing
- from Basis Customizing, execute : Cross-Client File Names/Paths
- on Logical FIle Path Definition, highlight (select) ARCHIVE_GLOBAL_PATH on the right pane
- double click on the Assignment of Physical Paths to Logical Path on the left pane
- double click on OS used, eg. UNIX, define Physical Path where archive file (on WRITE process) will be stored
- save changes made
- double click Logical File Definition, Cross Client on the left pane
- double click ARCHIVE_DATA_FILE on the right pane
- make sure that logical path is already set to ARCHIVE_GLOBAL_PATH
- save changes made
this setting also can be done using transaction FILE
we have experienced on this case using SAP standard archiving (SARA, SARI) and everything is fine with this setting above.
hope it help you.
rgds,
Alfonsus Guritno

Out of memory error when Applet transferring a file of size 600 MB

Hi ,
I have an Applet which transfer files from client machine to server using streams.
I have a servlet which reads data sent from Applet.
This is working fine when i Transfer files of size 10 MB
when I chose a file of 600 MB the following error is thrown at Applet side after some time.
java.lang.OutOfMemoryError: Java heap space
     at java.util.Arrays.copyOf(Unknown Source)
     at java.io.ByteArrayOutputStream.write(Unknown Source)
     at sun.net.www.http.PosterOutputStream.write(Unknown Source)
     at java.io.DataOutputStream.write(Unknown Source)
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if(ae.getSource() == jbutton)
     JFileChooser jfc=new JFileChooser();
     jfc.showOpenDialog(null);
File f=jfc.getSelectedFile();
try
FileInputStream in = new FileInputStream(f);
byte[] buf=new byte[1024];
int bytesread = 0;
String toservlet = "http://9.122.18.115:8080/FileTransfer/FileUpload?"
                              + URLEncoder.encode("name") + "="
                              + URLEncoder.encode(f.getName());
URL servleturl = new URL(toservlet);
URLConnection servletconnection = servleturl.openConnection();
//servletconnection.setRequestMethod("GET");
servletconnection.setRequestProperty("Content-type",
            "application/octet-stream");
servletconnection.setDoInput(true);
servletconnection.setDoOutput(true);
servletconnection.setUseCaches(false);
servletconnection.setDefaultUseCaches(false);
DataOutputStream out = new DataOutputStream(servletconnection.getOutputStream ());
while( (bytesread = in.read( buf )) > -1 )
out.write( buf, 0, bytesread );
// out.flush(); // tried this to flush the data but didn't work
System.out.println("writing data");
out.flush();
out.close();
in.close();
DataInputStream inputFromClient = new DataInputStream(servletconnection.getInputStream() );
//get what you want from servlet
inputFromClient.close();
catch(Exception e)
e.printStackTrace();
}My Servlet Code
public void doPost(HttpServletRequest req,HttpServletResponse res)
ServletContext sc = this.getServletContext();
try
String path = "C:\\Downloads\\";
String fileName=req.getParameter("name");
System.out.println("name="+fileName);
File yourFile = new File(path+fileName);
System.out.println("name="+yourFile);
FileOutputStream toFile = new FileOutputStream( yourFile );
DataInputStream fromClient = new DataInputStream( req.getInputStream() );
byte[] buff = new byte[1024];
int cnt = 0;
int k=0;
long st,et;
st=System.currentTimeMillis();
while( (cnt = fromClient.read( buff )) > -1 ) {
toFile.write( buff, 0, cnt );
//System.out.println("writing data=="+ k++);
et=System.currentTimeMillis();
toFile.flush();
toFile.close();
fromClient.close();
int tt=(int) ((et-st)/1000);
System.out.println("total time for "+fileName+" Download ="+tt+" secs");
System.out.println("total time for "+fileName+" Download ="+(tt/60)+" mins");
catch(Exception e)
e.printStackTrace();
}Please Help me out...

yeah its working fine. Thank you very much..
But it would be better if we provide the chunklength rather than using a zero to upload the file fastly.

Solaris 10 - File System Size / Layout

Hello, I�m very new to Solaris OS � UNIX world and have a question around the file system.
I have SUN System with a single 36GB of Hard Disk, and eventually I want to install Oracle 10g on it as a development server for myself.
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Filesystem size used avail capacity Mounted on
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swap 1.5G 88K 1.5G 1% /var/run
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The /home/export folder is aprrox: 28GB which is now taking up most of the space, and I understand that this is for users, therefore my question is:
1.Should I have allocated a more space to the root file system for oracle during the OS install?
2. If so what should I do? (its not a production system and I know it will have to go somewhere on the single drive, I�m happy to rebuild as I�m still learning the OS).
Any suggestions?
Rgds
D

Thanks, I do have this but needed info on the file system layout... but I have the info now...
If you select the auto file system layout on Solaris 10 then it gives the remaining space to the home/export . I have since re-installed, customised the file system layout and have space for the database.
D

Download file of size 3GB failing with error

Hi all,
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Have you tried this// Retrieve storage account from connection string.
CloudStorageAccount storageAccount =
CloudStorageAccount.Parse(
CloudConfigurationManager.GetSetting("StorageConnectionString"));
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IronPort S170 WSA - Max file download size

Hello,
we're using an IronPort S170 WSA. Downloading big .iso (maybe other files, too) files fails. As far as I can find out, files of 2300MB or less can be downloaded, files of 3300MB or bigger fail to download (I haven't been able to try files with sizes between 2300MB - 3300MB). Using the same client without using the IronPort as a proxy, the download of the big files succeed.
The web page error message indicates:
Blocked by [companyname] Web Proxy
Category = Allowed%20URL%208080
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DVS Verdict = -
DVS Threat = -
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Description: Cisco IronPort S170
Product: Cisco IronPort S170 Web Security Appliance
Model: S170
Version: 7.7.0-500

Do you get an instant block for large files? Or does it try to download for a while, and then fails?
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How to change the default password file's name and path when the database created?

how to change the default password file's name and path when the database created?
null

Usage: orapwd file=<fname> password=<password> entries=<users>
where
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password - password for SYS and INTERNAL (mand),
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How to download a file from a server path?

Is it possible to attach files on the server to a web dynpro response with cl_wd_runtime_services=>attach_file_to_response? All the examples I have seen assume you have the file content as an XSTRING. But how can one just send a file from a known path?
Is it really necessary to first load the file (e.g. a 50Mb TIFF) as an XSTRING before downloading it? If so, how is this done?

Hi Marc,
Try using this FM to convert String to Xstring.
SCMS_STRING_TO_XSTRING
HR_KR_STRING_TO_XSTRING
and also check this...Error document.getElementById(...) in function 'SCMS_STRING_TO_XSTRING'
Might Helps you.
Cheers,
Kris.

After latest Photoshop CC update, I can no longer edit a raw file in Lightroom 4 as a smart object

I updated PS this morning, since then I am no longer able to edit a file in LIghtroom 4 as a smart object in Photoshop CC. Photoshop will open, but after some time passes, Lightroom 4 gives an error that Photoshop cannot be launched. If I choose "edit in photoshop" from lightroom, instead of "edit as smart object", the file opens fine.
Thanks for your time!
Windows 7 Pro
Photoshop CC
Lightroom 4.4

I updated PS this morning, since then I am no longer able to edit a file in LIghtroom 4 as a smart object in Photoshop CC. Photoshop will open, but after some time passes, Lightroom 4 gives an error that Photoshop cannot be launched. If I choose "edit in photoshop" from lightroom, instead of "edit as smart object", the file opens fine.
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Flat file connection: The file name "\server\share\path\file.txt" specified in the connection was not valid

I'm trying to execute a SSIS package via SQL agent with a flat file source - however it fails with Code: 0xC001401E The file name "\server\share\path\file.txt" specified in the connection was not valid.
It appears that the problem is with the rights of the user that's running the package (it's a proxy account). If I use a higher-privelege account (domain admin) to run the package it completes successfully. But this is not a long-term solution, and I can't
see a reason why the user doesn't have rights to the file. The effective permissions of the file and parent folder both give the user full control. The user has full control over the share as well. The user can access the file (copy, etc) outside the SSIS
package.
Running the package manually via DTExec gives me the same error - I've tried 32 and 64bit versions with the same result. But running as a domain admin works correctly every time.
I feel like I've been beating my head against a brick wall on this one... Is there some sort of magic permissions, file or otherwise, that are required to use a flat file target in an SSIS package?

Hi Rossco150,
I have tried to reproduce the issue in my test environment (Windows Server 2012 R2 + SQL Server 2008 R2), however, everything goes well with the permission settings as you mentioned. In my test, the permissions of the folders are set as follows:
\\ServerName\Temp --- Read
\\ServerName\Temp\Source --- No access
\\ServerName\Temp\Source\Flat Files --- Full control
I suspect that your permission settings on the folders are not absolutely as you said above. Could you double check the permission settings on each level of the folder hierarchy? In addition, check the “Execute as user” information from job history to make
sure the job was running in the proxy security context indeed. Which version of SSIS are you using? If possible, I suggest that you install the latest Service Pack for you SQL Server or even install the latest CU patch.
Regards,
Mike Yin
If you have any feedback on our support, please click
here
Mike Yin
TechNet Community Support

What are the reasons for using absolute paths to embedded objects in word?

Hello,
hopefully someone can answer me the question why word always stores the absolute paths to linked objects? We use Word for documentation and these documents are stored in a version control system and sometimes several persons are working on a document. Because
of the absolute paths stored in the document all users would need to have the same directory structure so that the links would work for every person.
For linked objects it's not even possible to change the absolute paths to relative paths.
Additionally I would like to know what the difference between a "Graphic" and "Picture" object is. A graphic can be inserted by selecting "INSERT" -> "Pictures" and a "Graphic" doesn't seem to have a field
function and always uses absolute paths. "Graphics" can be inserted by selecting "INSERT" -> "Quick Parts" -> "Field..." -> "IncludePicture" and it's possible to use relative paths for "IncludePicture"
objects and it's possible to show the field function of a "Picture" object.
Regards
Martin

This seems to be by design, not too much information about this. The thread below should be helpful for you to understand it:
https://social.technet.microsoft.com/Forums/office/en-US/6d4445e1-cdc5-4b3b-9355-9081c963fdd9/unable-to-use-relative-object-links-in-word-office-2010-sp1
Also, from the thread below, you will find some workarounds for this issue:
http://windowssecrets.com/forums/showthread.php/102080-Relative-Paths-in-Word-Fields-(All)?p=584769&viewfull=1#post584769
Aravindhan Battepati

File Complexity/Size Maintenance; Paths vs Compund Objects

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