I'm working on a computer with Windows XP. This computer restarts at least a couple of times each day. I downloaded the dump file generated to my personal computer (Windows 10) but I can't find how fix this issue. The Minidump file generated is always samething like:
`Loading Dump File [....Mini072422-02.dmp]
Mini Kernel Dump File: Only registers and stack trace are available
WARNING: Path element is empty
WARNING: Path element is empty
************* Path validation summary **************
Response Time (ms) Location
OK C:\WINDOWS\symbols
OK C:\WINDOWS\symbols\dll
WARNING: Path element is empty
WARNING: Path element is empty
Symbol search path is: C:\WINDOWS\symbols;C:\WINDOWS\symbols\dll;;
Executable search path is:
Unable to load image ntoskrnl.exe, Win32 error 0n2
*** WARNING: Unable to verify timestamp for ntoskrnl.exe
Windows XP Kernel Version 2600 (Service Pack 3) MP (2 procs) Free x86 compatible
Product: WinNt, suite: TerminalServer SingleUserTS
Machine Name:
Kernel base = 0x804d7000 PsLoadedModuleList = 0x8055d720
Debug session time: Sun Jul 24 19:30:39.421 2022 (UTC + 2:00)
System Uptime: 0 days 4:18:26.962
Unable to load image ntoskrnl.exe, Win32 error 0n2
*** WARNING: Unable to verify timestamp for ntoskrnl.exe
Loading Kernel Symbols
...............................................................
................................................................
...
Loading User Symbols
Loading unloaded module list
............
************* Symbol Loading Error Summary **************
Module name Error
ntoskrnl The system cannot find the file specified
You can troubleshoot most symbol related issues by turning on symbol loading diagnostics (!sym
noisy) and repeating the command that caused symbols to be loaded.
You should also verify that your symbol search path (.sympath) is correct.
For analysis of this file, run !analyze -v
Unable to load image iaStor.sys, Win32 error 0n2
*** WARNING: Unable to verify timestamp for iaStor.sys
eax=8a936990 ebx=8a936938 ecx=8a936000 edx=8a936990 esi=00000000 edi=8a9290e8
eip=ba67492a esp=8055127c ebp=80551298 iopl=0 nv up ei pl nz ac po nc
cs=0008 ss=0010 ds=0023 es=0023 fs=0030 gs=0000 efl=00010212
iaStor+0x992a:
ba67492a 8b4628 mov eax,dword ptr [esi+28h] ds:0023:00000028=????????`
Could you give any idea? Thanks in advance.
I'm trying to download a specific directory from Windows XP and Windows Server 2012 using a Get Directory method of SSHLibrary. This directory is on a different volume that the one in which the ssh connection it's established. For clarity when I open the connection this points to volume C:, the source directory is in volume D:. The issue that I'm seeing is that on my local machine the path to directory tries to include the volume letter resulting in something like: C:\path\to\robot\executable\D:\source\directory thus resulting in WindowsError: [Error 123] The filename, directory name, or volume label syntax is incorrect:
Is there a way to not have this issue but result in a path like C:\path\to\robot\executable\source\directory and successfully download the directory? I've tried doing Execute command d: before Get Directory but no luck there. Also is there a way to open a connection pointing to a specific volume?
Code I used:
*** Settings ***
Documentation Suite description
Library SSHLibrary timeout=120 seconds
*** Variables ***
${HOST_XP} remote.win.xp.machine
${USER_XP} user
${PASSWORD_XP} pass
${DIR} D:\\source\\
*** Test Cases ***
Test Win XP
SSHLibrary.open_connection ${HOST_XP}
SSHLibrary.login ${USER_XP} ${PASSWORD_XP}
SSHLibrary.get directory ${DIR}
Edit1: various typo
I actually found out the problem for Get Directory method. In path variables / should be used and not \. So my DIR variable should be ${DIR} D:/source/
When I compile my python program as a 'one file' with the VPython 7 library, I get the following error in the command line box at run time. Not sure what to add to my spec file or code to collect file when program is frozen.
FileNotFoundError: [Errno 2] No such file or directory: 'C\Users\Name\AppData\Local\Temp\_MEI106402\vpython\vpython_libraries\glow.min.js'
I tried adding file to spec file but no luck.
a.datas += [ ('glow.min.js', 'C:\Users\Name\AppData\Local\Programs\Python\Python37\Lib\site-packages\vpython\vpython_libraries\glow.min.js', 'Data')]
I try to unzip file.zip with files (a, b, c) in pentaho kettle (file management -> unzip file). it working fine.
But if i try to unzip file.zip with files (a, b, ж), for example, i have errors:
2016/01/18 17:46:17 - cfgbuilder - Warning: The configuration parameter [org] is not supported by the default configuration builder for scheme: sftp
2016/01/18 17:46:17 - Unzip file - ERROR (version 6.0.1.0-386, build 1 from 2015-12-03 11.37.25 by buildguy) : Could not unzip file [file:///D:/projects/loaders/loader_little_files/src.zip]. Exception : [MALFORMED]
2016/01/18 17:46:17 - Unzip file - ERROR (version 6.0.1.0-386, build 1 from 2015-12-03 11.37.25 by buildguy) : java.lang.IllegalArgumentException: MALFORMED
2016/01/18 17:46:17 - Unzip file - at java.util.zip.ZipCoder.toString(ZipCoder.java:58)
2016/01/18 17:46:17 - Unzip file - at java.util.zip.ZipFile.getZipEntry(ZipFile.java:566)
2016/01/18 17:46:17 - Unzip file - at java.util.zip.ZipFile.access$900(ZipFile.java:60)
2016/01/18 17:46:17 - Unzip file - at java.util.zip.ZipFile$ZipEntryIterator.next(ZipFile.java:524)
2016/01/18 17:46:17 - Unzip file - at java.util.zip.ZipFile$ZipEntryIterator.nextElement(ZipFile.java:499)
2016/01/18 17:46:17 - Unzip file - at java.util.zip.ZipFile$ZipEntryIterator.nextElement(ZipFile.java:480)
2016/01/18 17:46:17 - Unzip file - at org.apache.commons.vfs2.provider.zip.ZipFileSystem.init(ZipFileSystem.java:91)
2016/01/18 17:46:17 - Unzip file - at org.apache.commons.vfs2.provider.AbstractVfsContainer.addComponent(AbstractVfsContainer.java:53)
2016/01/18 17:46:17 - Unzip file - at org.apache.commons.vfs2.provider.AbstractFileProvider.addFileSystem(AbstractFileProvider.java:103)
2016/01/18 17:46:17 - Unzip file - at org.apache.commons.vfs2.provider.AbstractLayeredFileProvider.createFileSystem(AbstractLayeredFileProvider.java:88)
2016/01/18 17:46:17 - Unzip file - at org.apache.commons.vfs2.provider.AbstractLayeredFileProvider.findFile(AbstractLayeredFileProvider.java:61)
2016/01/18 17:46:17 - Unzip file - at org.apache.commons.vfs2.impl.DefaultFileSystemManager.resolveFile(DefaultFileSystemManager.java:790)
2016/01/18 17:46:17 - Unzip file - at org.apache.commons.vfs2.impl.DefaultFileSystemManager.resolveFile(DefaultFileSystemManager.java:712)
2016/01/18 17:46:17 - Unzip file - at org.pentaho.di.core.vfs.KettleVFS.getFileObject(KettleVFS.java:151)
2016/01/18 17:46:17 - Unzip file - at org.pentaho.di.core.vfs.KettleVFS.getFileObject(KettleVFS.java:106)
2016/01/18 17:46:17 - Unzip file - at org.pentaho.di.job.entries.unzip.JobEntryUnZip.unzipFile(JobEntryUnZip.java:618)
2016/01/18 17:46:17 - Unzip file - at org.pentaho.di.job.entries.unzip.JobEntryUnZip.processOneFile(JobEntryUnZip.java:516)
2016/01/18 17:46:17 - Unzip file - at org.pentaho.di.job.entries.unzip.JobEntryUnZip.execute(JobEntryUnZip.java:461)
2016/01/18 17:46:17 - Unzip file - at org.pentaho.di.job.Job.execute(Job.java:730)
2016/01/18 17:46:17 - Unzip file - at org.pentaho.di.job.Job.execute(Job.java:873)
2016/01/18 17:46:17 - Unzip file - at org.pentaho.di.job.Job.execute(Job.java:873)
2016/01/18 17:46:17 - Unzip file - at org.pentaho.di.job.Job.execute(Job.java:873)
2016/01/18 17:46:17 - Unzip file - at org.pentaho.di.job.Job.execute(Job.java:546)
2016/01/18 17:46:17 - Unzip file - at org.pentaho.di.job.Job.run(Job.java:435)
I'am using windows 7, when i create "ж" file.
I'am trying to rename file in linux to "ж" - the result has not changed.
How can i do this? Any hidden setting?
Thanks!
Non utf-8 encoding in zip files.
Taken from here. https://blogs.oracle.com/xuemingshen/entry/non_utf_8_encoding_in
Important parts
The Zip specification (historically) does not specify what character encoding to be used for the embedded file names
Jar specification meanwhile explicitly specifies to use UTF-8 as the encoding to encode and decode all file names and comments in Jar files. Our java.util.jar and java.util.zip implementation therefor strictly followed Jar specification to use UTF-8 as the sole encoding when dealing with the file names and comments stored in Jar/Zip files.
Windows NFTS filesystem encoding UTF-16. Cyrillic symbols in file names cause problems in java application. Troubles will arise in use some third party tools to create zip archive (unless u use java based tools - which rarely) and then unzip them using java tools like PDI.
Good staff for Linux users, ext4 use by default UTF-8 (actually it doesn't rely on encoding just byte sequence, but GUI like gnome (environment where u create files whatever shell, or gnome nautilus file manager) assume UTF-8 to decode symbols to write file name on disk. QT relies on locale. Of cause there are ways to override but by default as I know UTF-8 become wide used as default locale.
Conclusion:
zip file created in linux(tested in ubuntu) can be unzipped using PDI.
zip file created using JavaAPI can be unzipped anywhere using PDI
zip file created on Windows can cause trouble unzipped using PDI
How to decompress zip file created on Windows 8.1, using 7zip. Files have names contain cyrilic symbols. Zip archive contains 3 files inside named:
а.txt
ж.txt
ё.txt
Fortunately all needed libraries (Apache commons-compress and commons-io) are in directory PENTAHO_HOME/lib, so u don't have to add extra libraries to kettle.
Here is code underneath, for "User Defined Java Class" step
import java.io.File;
import java.io.FileOutputStream;
import java.io.InputStream;
import java.io.OutputStream;
import java.util.Enumeration;
import org.apache.commons.compress.archivers.zip.ZipArchiveEntry;
import org.apache.commons.compress.archivers.zip.ZipFile;
import org.apache.commons.io.IOUtils;
public boolean processRow(StepMetaInterface smi, StepDataInterface sdi) throws KettleException{
Object[] r = getRow();
r = createOutputRow(r, data.outputRowMeta.size());
String fname = getVariable("FNAME", null);
String outDir = getVariable("OUT", null);
System.out.println(fname + " " + outDir);
try {
java.io.File inputFile = new java.io.File(fname);
ZipFile zipFile = new ZipFile(inputFile, "cp866", false);
Enumeration enumEntry = zipFile.getEntries();
int i = 0;
while(enumEntry.hasMoreElements()){
ZipArchiveEntry entry = (ZipArchiveEntry) enumEntry.nextElement();
String entryName = entry.getName();
System.out.println(entryName);
OutputStream os = new FileOutputStream(new File(outDir, Integer.valueOf(++i) + entryName));
InputStream is = zipFile.getInputStream(entry);
IOUtils.copy(is, os);
is.close();
os.close();
}
} catch (Exception exc) {
System.out.println("Faild to unzip");
exc.printStackTrace();
}
putRow(data.outputRowMeta, r);
return true;
}
Important parts of code are:
String fname = getVariable("FNAME", null);
String outDir = getVariable("OUT", null);
Those mean that 2 variables should be available in transformation
FNAME - absolute path to ZipFile,
OUT - directory where need to extract files
In this line:
ZipFile zipFile = new ZipFile(inputFile, "cp866", false);
"cp866" means encoding used by 7zip for zipfile entries(cp866 on windows). If u use another zipper then u might need to change encoding. Here is some notice https://commons.apache.org/proper/commons-compress/zip.html. Part Recommendations for Interoperability. U can write own algorith to identify encoding, rely on for example on known part of name of files in zip archive.
Anyway I think most probably this kettle job/tranformation will use zip file from single certain source, and just need to identify and set proper encoding of zip file in code.
This line:
Integer.valueOf(++i) + entryName)
Why file name generated using integer? If wrong encoding is used then ZipFile will decode filename of zip entries to [].txt (ZipFile can't decode а.txt, ж.txt so it will replace symbols 'а', 'ж' with '[]'). Which lead to (if u have wrong encoding and filenames have same length and written in cyrilic) each enty in loop will overwrite same file and u will get in the end, single file named [].txt.
With counter in file name u will guaranty all files will have different name even if u not able to decode correct file name.
1[].txt
2[].txt
3[].txt
Anyway if u know exactly encoding then just remove this part of code to eliminate numbers in file name.
only one worked for me in Debian Jessie - install WinRAR into wine and choose there file names encoding
I have a problem this morning with my symfony project.
I develop an application online sales that works without problem on my pc running on a ubuntu system.
I 'bought a dedicated server (centos) in which I wanted to put my application.
so I transferred all my project in the FTP serveur.mais qd I want to accede to the site with mozilla I get this error
Warning: require_once (/opt/lampp/htdocs Symfony/web/.. /app/bootstrap.php.cache..): Failed to open stream: No such file or folder of this type in /opt/lampp/htdocs/Symfony/web/app.php on line 6
Fatal error: require_once (): Failed opening required '/opt/lampp/htdocs/Symfony/web/../app/bootstrap.php.cache..' (Include_path = ':/opt/lampp/lib/php.') In / opt / lampp / htdocs / Symfony / web / app.php on line 6
and when I do php app / console in the terminal told me Could not open input file: app / console
[root # server Symfony] # sudo php app / console cache: clear - env = prod
Could not open input file: app / console
[root # server Symfony] # php app / console
Could not open input file: app / console
please i need help
do a :
composer update
and eventually an :
app/console cache:clear -env=prod
not
app / console cache: clear - env = prod
I know this is an old topic but i want to help.
You can save your document with utf-8 encoding option.
If you are on windows use Notepad++.
Encoding > Encode in UTF-8
If you are mac or linux use SublimeText2.
File > Save With Encoding > UTF-8