I found several code snippets for running cmd commands through a Java class, but I wasn't able to understand it.
This is code for opening the cmd
public void excCommand(String new_dir){
Runtime rt = Runtime.getRuntime();
try {
rt.exec(new String[]{"cmd.exe","/c","start"});
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
And I found some other links for adding other commands such as cd
http://www.coderanch.com/t/109753/Linux-UNIX/exec-command-cd-command-java
How to open the command prompt and insert commands using Java?
Can anyone help me to understand how to cd a directory such as:
cd C:\Program Files\Flowella
then run other commands on that directory?
One way to run a process from a different directory to the working directory of your Java program is to change directory and then run the process in the same command line. You can do this by getting cmd.exe to run a command line such as cd some_directory && some_program.
The following example changes to a different directory and runs dir from there. Admittedly, I could just dir that directory without needing to cd to it, but this is only an example:
import java.io.*;
public class CmdTest {
public static void main(String[] args) throws Exception {
ProcessBuilder builder = new ProcessBuilder(
"cmd.exe", "/c", "cd \"C:\\Program Files\\Microsoft SQL Server\" && dir");
builder.redirectErrorStream(true);
Process p = builder.start();
BufferedReader r = new BufferedReader(new InputStreamReader(p.getInputStream()));
String line;
while (true) {
line = r.readLine();
if (line == null) { break; }
System.out.println(line);
}
}
}
Note also that I'm using a ProcessBuilder to run the command. Amongst other things, this allows me to redirect the process's standard error into its standard output, by calling redirectErrorStream(true). Doing so gives me only one stream to read from.
This gives me the following output on my machine:
C:\Users\Luke\StackOverflow>java CmdTest
Volume in drive C is Windows7
Volume Serial Number is D8F0-C934
Directory of C:\Program Files\Microsoft SQL Server
29/07/2011 11:03 <DIR> .
29/07/2011 11:03 <DIR> ..
21/01/2011 20:37 <DIR> 100
21/01/2011 20:35 <DIR> 80
21/01/2011 20:35 <DIR> 90
21/01/2011 20:39 <DIR> MSSQL10_50.SQLEXPRESS
0 File(s) 0 bytes
6 Dir(s) 209,496,424,448 bytes free
You can try this:-
Process p = Runtime.getRuntime().exec(command);
If you want to perform actions like cd, then use:
String[] command = {command_to_be_executed, arg1, arg2};
ProcessBuilder builder = new ProcessBuilder(command);
builder = builder.directory(new File("directory_location"));
Example:
String[] command = {"ls", "-al"};
ProcessBuilder builder = new ProcessBuilder(command);
builder = builder.directory(new File("/ngs/app/abc"));
Process p = builder.start();
It is important that you split the command and all arguments in separate strings of the string array (otherwise they will not be provided correctly by the ProcessBuilder API).
Here is a more complete implementation of command line execution.
Usage
executeCommand("ls");
Output:
12/27/2017 11:18:11:732: ls
12/27/2017 11:18:11:820: build.gradle
12/27/2017 11:18:11:820: gradle
12/27/2017 11:18:11:820: gradlew
12/27/2017 11:18:11:820: gradlew.bat
12/27/2017 11:18:11:820: out
12/27/2017 11:18:11:820: settings.gradle
12/27/2017 11:18:11:820: src
Code
private void executeCommand(String command) {
try {
log(command);
Process process = Runtime.getRuntime().exec(command);
logOutput(process.getInputStream(), "");
logOutput(process.getErrorStream(), "Error: ");
process.waitFor();
} catch (IOException | InterruptedException e) {
e.printStackTrace();
}
}
private void logOutput(InputStream inputStream, String prefix) {
new Thread(() -> {
Scanner scanner = new Scanner(inputStream, "UTF-8");
while (scanner.hasNextLine()) {
synchronized (this) {
log(prefix + scanner.nextLine());
}
}
scanner.close();
}).start();
}
private static SimpleDateFormat format = new SimpleDateFormat("MM/dd/yyyy hh:mm:ss:SSS");
private synchronized void log(String message) {
System.out.println(format.format(new Date()) + ": " + message);
}
My example (from real project)
folder — File.
zipFile, filesString — String;
final String command = "/bin/tar -xvf " + zipFile + " " + filesString;
logger.info("Start unzipping: {} into the folder {}", command, folder.getPath());
final Runtime r = Runtime.getRuntime();
final Process p = r.exec(command, null, folder);
final int returnCode = p.waitFor();
if (logger.isWarnEnabled()) {
final BufferedReader is = new BufferedReader(new InputStreamReader(p.getInputStream()));
String line;
while ((line = is.readLine()) != null) {
logger.warn(line);
}
final BufferedReader is2 = new BufferedReader(new InputStreamReader(p.getErrorStream()));
while ((line = is2.readLine()) != null) {
logger.warn(line);
}
}
The easiest way would be to use Runtime.getRuntime.exec().
For example, to get a registry value for the default browser on Windows:
String command = "REG QUERY HKEY_CLASSES_ROOT\\http\\shell\\open\\command";
try
{
Process process = Runtime.getRuntime().exec(command);
} catch (IOException e)
{
e.printStackTrace();
}
Then use a Scanner to get the output of the command, if necessary.
Scanner kb = new Scanner(process.getInputStream());
Note: the \ is an escape character in a String, and must be escaped to work properly (hence the \\).
However, there is no executable called cd, because it can't be implemented in a separate process.
The one case where the current working directory matters is executing an external process (using ProcessBuilder or Runtime.exec()). In those cases you can specify the working directory to use for the newly started process explicitly.
Easiest way for your command:
System.setProperty("user.dir", "C:\\Program Files\\Flowella");
Try this:
Process runtime = Runtime.getRuntime().exec("cmd /c start notepad++.exe");
Once you get the reference to Process, you can call getOutpuStream on it to get the standard input of the cmd prompt. Then you can send any command over the stream using write method as with any other stream.
Note that it is process.getOutputStream() which is connected to the stdin on the spawned process. Similarly, to get the output of any command, you will need to call getInputStream and then read over this as any other input stream.
Stopping and Disabling a service can be done via below code:
static void sdService() {
String[] cmd = {"cmd.exe", "/c", "net", "stop", "MSSQLSERVER"};
try {
Process process = new ProcessBuilder(cmd).start();
process.waitFor();
String line = null;
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(process.getInputStream()));
while((line = bufferedReader.readLine()) != null) {
System.out.println(line);
}
line = null;
bufferedReader = null;
Process p = Runtime.getRuntime().exec("sc config MSSQLSERVER start= disabled");
p.waitFor();
bufferedReader = new BufferedReader(new InputStreamReader(p.getInputStream()));
while((line = bufferedReader.readLine()) != null) {
System.out.println(line);
}
} catch (Exception e) {
e.printStackTrace();
}
}
Enabling and Starting a service can be done via below code
static void esService() {
String[] cmd = {"cmd.exe", "/c", "net", "start", "MSSQLSERVER"};
try {
Process p = Runtime.getRuntime().exec("sc config MSSQLSERVER start= auto");
//Process p = Runtime.getRuntime().exec("sc config MSSQLSERVER start= demand");
p.waitFor();
String line = null;
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(p.getInputStream()));
while((line = bufferedReader.readLine()) != null) {
System.out.println(line);
}
line = null;
bufferedReader = null;
Process process = new ProcessBuilder(cmd).start();
process.waitFor();
bufferedReader = new BufferedReader(new InputStreamReader(process.getInputStream()));
while((line = bufferedReader.readLine()) != null) {
System.out.println(line);
}
} catch (Exception e) {
e.printStackTrace();
}
}
Executing command from any folder can be done via below code.
static void runFromSpecificFolder() {
try {
ProcessBuilder processBuilder = new ProcessBuilder("cmd.exe", "/c", "cd \"C:\\Users\\himan\\Desktop\\Java_Test_Deployment\\jarfiles\" && dir");
//processBuilder.directory(new File("C://Users//himan//Desktop//Java_Test_Deployment//jarfiles"));
processBuilder.redirectErrorStream(true);
Process p = processBuilder.start();
p.waitFor();
String line = null;
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(p.getInputStream()));
while((line = bufferedReader.readLine()) != null) {
System.out.println(line);
}
} catch (Exception e) {
e.printStackTrace();
}
}
public static void main(String args[]) {
sdService();
runFromSpecificFolder();
esService();
}
You can't run cd this way, because cd isn't a real program; it's a built-in part of the command-line, and all it does is change the command-line's environment. It doesn't make sense to run it in a subprocess, because then you're changing that subprocess's environment — but that subprocess closes immediately, discarding its environment.
To set the current working directory in your actual Java program, you should write:
System.setProperty("user.dir", "C:\\Program Files\\Flowella");
public class Demo {
public static void main(String args[]) throws IOException {
Process process = Runtime.getRuntime().exec("/Users/******/Library/Android/sdk/platform-tools/adb" + " shell dumpsys battery ");
BufferedReader in = new BufferedReader(new InputStreamReader(process.getInputStream()));
String line = null;
while (true) {
line = in.readLine();
if (line == null) { break; }
System.out.println(line);
}
}
}
The simplest and shortest way is to use CmdTool library.
new Cmd()
.configuring(new WorkDir("C:/Program Files/Flowella"))
.command("cmd.exe", "/c", "start")
.execute();
You can find more examples here.
one of the way to execute cmd from java !
public void executeCmd() {
String anyCommand="your command";
try {
Process process = Runtime.getRuntime().exec("cmd /c start cmd.exe /K " + anyCommand);
} catch (IOException e) {
e.printStackTrace();
}
}
Here the value adder is use of ampersands to batch commands and correct format for change drive with cd.
public class CmdCommander {
public static void main(String[] args) throws Exception {
//easyway to start native windows command prompt from Intellij
/*
Rules are:
1.baseStart must be dual start
2.first command must not have &.
3.subsequent commands must be prepended with &
4.drive change needs extra &
5.use quotes at start and end of command batch
*/
String startQuote = "\"";
String endQuote = "\"";
//String baseStart_not_taking_commands = " cmd /K start ";
String baseStart = " cmd /K start cmd /K ";//dual start is must
String first_command_chcp = " chcp 1251 ";
String dirList = " &dir ";//& in front of commands after first command means enter
//change drive....to yours
String changeDir = " &cd &I: ";//extra & makes changing drive happen
String javaLaunch = " &java ";//just another command
String javaClass = " Encodes ";//parameter for java needs no &
String javaCommand = javaLaunch + javaClass;
//build batch command
String totalCommand =
baseStart +
startQuote +
first_command_chcp +
//javaCommand +
changeDir +
dirList +
endQuote;
System.out.println(totalCommand);//prints into Intellij terminal
runCmd(totalCommand);
//Thread t = Thread.currentThread();
//t.sleep(3000);
System.out.println("loppu hep");//prints into Intellij terminal
}
public static void runCmd(String command) throws Exception {
Runtime rt = Runtime.getRuntime();
Process proc = rt.exec(command);
}
}
I was trying to create a feature to create hyperlinks in the console while the information is added.
While debugging, I noticed that the first hyperlink is succesfully created, but when the next line is printed to console, the hyperlink dissapears.
The code which I use to create the hyperlinks is:
String hyperLinkText = "test";
myConsole.printToConsole(test);
String myFile = new File("D:/Test/testFile.txt");
URI location = myFile.toURI();
IFile[] files = project.getWorkspace().getRoot().findFilesForLocationURI(location);
HyperlinkInformation hyperlinkInformation = new HyperlinkInformation(myConsole, myConsole.getDocument().get().length(), hyperLinkText.length());
FileHyperlink fileHyperlink = new FileHyperlink(hyperlinkInformation,
files[0], 1);
try {
hyperlinkInformation.getConsole().addHyperlink(fileHyperlink,
fileHyperlink.getOffset(), fileHyperlink.getLength());
} catch (BadLocationException e) {
//
}
The function addHyperlink:
public void addHyperlink(IHyperlink hyperlink, int offset, int length) throws BadLocationException {
IDocument document = getDocument();
ConsoleHyperlinkPosition hyperlinkPosition = new ConsoleHyperlinkPosition(hyperlink, offset, length);
try {
document.addPosition(ConsoleHyperlinkPosition.HYPER_LINK_CATEGORY, hyperlinkPosition);
fConsoleManager.refresh(this);
} catch (BadPositionCategoryException e) {
ConsolePlugin.log(e);
}
}
Is this the normal behaviour or am I missing something?
I'm using gwtupload.client.MultiUploader to upload zip files on the server in GWT. Then on the server I transform zip file to array of bytes to insert into database. As the result 50% of files in database are corrupted. Here a little bit of my code.
#UiField(provided = true)
MultiUploader muplDef;
public MyClass(){
muplDef = new MultiUploader();
muplDef.setValidExtensions("zip");
muplDef.addOnFinishUploadHandler(onFinishUploaderHandler);
muplDef.addOnCancelUploadHandler(onCancelUploaderHander);
}
private final IUploader.OnFinishUploaderHandler onFinishUploaderHandler = new IUploader.OnFinishUploaderHandler() {
#SuppressWarnings("incomplete-switch")
#Override
public void onFinish(IUploader uploader) {
switch (uploader.getStatus()) {
case SUCCESS:
attachZip = true;
}
}
};
private final IUploader.OnCancelUploaderHandler onCancelUploaderHander = new IUploader.OnCancelUploaderHandler() {
#Override
public void onCancel(IUploader uploader) {
attachZip = false;
}
};
Byte Array
String fileName = "D:\1.zip";
File f = new File(fileName);
byte[] edocBinary = new byte[(int) f.length()];
RandomAccessFile ff;
try {
ff = new RandomAccessFile(f, "r");
ff.readFully(edocBinary);
} catch (FileNotFoundException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
My questions are: Files can be correpted even if I have OnFinishUploaderHandler and case:SUCCESS? There are other cases like ERROR, maybe this case will check the file? Or the problem is with the transformation to byte array? Can you provide me some advices, thanks.
As you said, you had two steps :
1- Uploading the zip file
2- Inserting the zip file in the database
If the step 1 is done correctly, you're gonna get a success showing that the file is transfered correcly from client to the server, what you do after that is not managed by GwtUpload.
So I guess the corruption happened when you try to insert the file in the database. If you are using MySQL Try this http://www.codejava.net/java-se/jdbc/insert-file-data-into-mysql-database-using-jdbc
I have following code. On making changes in the text file when I open the file I can't see any change in fact the text of text file is removed.
private void btnGo_Click(object sender, EventArgs e)
{
string content;
string newword = "Tanu";
string oldword = "Sunrise";
System.IO.StreamReader file =
new System.IO.StreamReader(txtFilePath.Text);
while ((content = file.ReadLine()) != null)
{
if (content == "Project name = Sunrise ")
{
string newtxt = Regex.Replace(content, oldword, newword);
content = content.Replace(content, newtxt);
}
// counter++;
}
file.Close();
using (StreamWriter writer = new StreamWriter(txtFilePath.Text))
{
writer.Write(content);
writer.Close();
Error is in your while loop.
while ((content = file.ReadLine()) != null)
after replacing content with newtxt while is executing again and is trying to read next line in file which is setting content to null and so when you come out of while loop your content is null.
try reading file content into a different variable or use say newtxt itself to write to file.
I use a servlet to access a folder outside the web container to load some graphics to web application by using GWT. I use the following snippet in servlet to test the idea:
String s = null;
File inputFile = new File("C:\\Documents and Settings\\User\\My Documents\\My Pictures\\megan-fox.jpg");
FileInputStream fin = null;
try {
fin = new FileInputStream(inputFile);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
byte c[] = new byte[(int) inputFile.length()];
try {
fin.read(c);
} catch (IOException e) {
e.printStackTrace();
}
try {
fin.close();
} catch (IOException e1) {
e1.printStackTrace();
}
String imgFolderPath = getServletContext().getRealPath("/")+"img";
File imgFolder = new File(imgFolderPath);
imgFolder.mkdir();
File newImage = new File("megan-fox.jpg");
FileOutputStream fout = null;
try {
fout = new FileOutputStream(newImage);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
try {
fout.write(c);
} catch (IOException e) {
e.printStackTrace();
}
try {
fout.close();
} catch (IOException e) {
e.printStackTrace();
}
boolean success = newImage.renameTo(new File(imgFolderPath, newImage.getName()));
The code in servlet reads the image file from the specified folder in hard disk, creates a new folder called 'img' in war folder and copies to it the jpg file. Then it returns to the client the path to the image (for now hardcoded as) '/img/megan-fox.jpg'.
The client then uses the Image class in GWT with the returned path-string to display the image, like in the following snippet:
public void onSuccess(String result) {
String myImage = result;
image = new Image(myImage);
RootPanel.get().add(image);
closeButton.setFocus(true);
}
I need to know if there is a way to achieve the same result without using the 'intermediate' step of creating a folder in the web container root (optional) and copying the file there in order to access it with Image GWT class and display it?
updated: The original servlet class.
public class GreetingServiceImpl extends RemoteServiceServlet implements
GreetingService {
// This method is called by the servlet container to process a GET request.
public void doGet(HttpServletRequest req, HttpServletResponse resp) throws IOException {
// Get the absolute path of the image
ServletContext sc = getServletContext();
// i want to load the image in the specified folder (outside the web container)
String filename = sc.getRealPath("C:\\Documents and Settings\\User\\My Documents\\My Pictures\\megan-fox.jpg");
// Get the MIME type of the image
String mimeType = sc.getMimeType(filename);
if (mimeType == null) {
sc.log("Could not get MIME type of "+filename);
resp.setStatus(HttpServletResponse.SC_INTERNAL_SERVER_ERROR);
return;
}
// Set content type
resp.setContentType(mimeType);
// Set content size
File file = new File(filename);
resp.setContentLength((int)file.length());
// Open the file and output streams
FileInputStream in = new FileInputStream(file);
OutputStream out = resp.getOutputStream();
// Copy the contents of the file to the output stream
byte[] buf = new byte[1024];
int count = 0;
while ((count = in.read(buf)) >= 0) {
out.write(buf, 0, count);
}
in.close();
out.close();
}
// This is the method that is called from the client using GWT-RPC
public String greetServer(String input) throws IllegalArgumentException {
HttpServletRequest req = this.getThreadLocalRequest();
HttpServletResponse res = this.getThreadLocalResponse();
try {
doGet(req, res);
} catch (IOException e) {
e.printStackTrace();
}
// actually i dont know what that means but i thought i would have to returned something like the image's url?
return res.encodeURL("/img/image0.png");
}
}
I logically misused the method that was proposed to solve my problem. What is the correct way?
Sure, just have your servlet serve the image directly:
Set the Content-Type header to image/jpeg.
Write out image file contents to servlet response writer.
Here is an example.