我想在进程生成时读取它的标准输出。 该过程将发送进度指示器的信息,因此我一次获取所有信息是没有意义的,我这样做了,这就是问题所在。我尝试按照帖子中的建议使用 Scanner 类,但只有在该过程完成后我仍然得到输出。 我意识到这个问题以前曾被问过,但尚未得到解答。
您可能需要先查看 StreamGobblerOutput 类。
public List<String> executeCall(String fileName)
{
StringBuilder sbOutput = new StringBuilder();
StringBuilder sbError = new StringBuilder();
File file = new File(fileName);
try ( BufferedReader br = new BufferedReader(new FileReader(file)) ) {
String line;
while ((line = br.readLine()) != null) {
String [] parts = line.split("\\s");
if(parts.length<2) {
sbError.append("Command too short for call: " + parts[0]);
continue;
}
List<String> args = new ArrayList<String>();
args.add ("sfb.exe");
for(int i = 1; i <parts.length; ++i) {
args.add (parts[i]);
}
args.add (sfbPassword);
ProcessBuilder pb = new ProcessBuilder (args);
pb.directory(new File(Support.getJustThePathFromFile(file)));
Map<String, String> envs = pb.environment();
String path = envs.get("Path");
envs.put("Path", Paths.get(".").toAbsolutePath().normalize().toString() + ";" +path);
//pb.redirectOutput(new Redirect() {});
Process p = pb.start();
String outputPathPrefix = pb.directory().getCanonicalPath();
// any output?
StreamGobblerOutput outputGobbler = new StreamGobblerOutput(p.getInputStream(), outputPathPrefix);
outputGobbler.start();
// any errors?
StreamGobblerError errorGobbler = new StreamGobblerError(p.getErrorStream());
errorGobbler.start();
try
{
p.waitFor();
}
catch (InterruptedException e)
{
// TODO Auto-generated catch block
e.printStackTrace();
}
sbOutput = outputGobbler.getOutput();
sbError = errorGobbler.getErrors();
String rootPath= Support.getJustThePathFromFile(new File(fileName));
File rootFile = new File(rootPath + "/..");
String rootFolder = rootFile.getCanonicalFile().getName();
System.err.println("rootFolder: " + rootFolder);
mainApp.addModifiedFiles(outputGobbler.getModifiedFileNames(), rootFolder);
}
} catch ( IOException ex) {
sbError.append(ex.getMessage());
}
mainApp.addOutput(sbOutput.toString());
mainApp.addError(sbError.toString());
return;
}
private class StreamGobblerOutput extends Thread {
private InputStream is;
private String outputPathPrefix;
private StringBuilder sbOutput;
private List<String> modifiedFileNames;
private Scanner scanner;
private StreamGobblerOutput(InputStream is, String outputPathPrefix) {
this.is = is;
this.outputPathPrefix = outputPathPrefix;
sbOutput = new StringBuilder();
modifiedFileNames = new ArrayList<String>();
scanner = new Scanner(is);
}
public StringBuilder getOutput() {
return sbOutput;
}
public List<String> getModifiedFileNames() {
return modifiedFileNames;
}
@Override
public void run() {
//create pattern
Pattern patternProgress = Pattern.compile("\\((\\d+)%\\)");
//InputStreamReader isr = new InputStreamReader(is);
//BufferedReader br = new BufferedReader(isr);
String ligne = null;
while (scanner.hasNextLine()) {
ligne = scanner.nextLine();
sbOutput.append(ligne);
sbOutput.append("\r\n");
//bw.write("\r\n");
Matcher mProgress = patternProgress.matcher(ligne);
if (mProgress.find()) {
int percentage = Integer.parseInt(mProgress.group(1));
System.err.println("percentage=" + percentage);
mainApp.mainWindowController.setProgressExecute(percentage/100.0);
}
}
mainApp.mainWindowController.setProgressExecute(1.0);
if (scanner != null) {
scanner.close();
}
}
}
private class StreamGobblerError extends Thread {
private InputStream is;
private StringBuilder sbError;
private Scanner scanner;
private StreamGobblerError(InputStream is) {
this.is = is;
sbError = new StringBuilder();
scanner = new Scanner(is);
}
public StringBuilder getErrors() {
return sbError;
}
@Override
public void run() {
//InputStreamReader isr = new InputStreamReader(is);
//BufferedReader br = new BufferedReader(isr);
String ligne = null;
while (scanner.hasNextLine()) {
ligne = scanner.nextLine();
sbError.append(ligne);
sbError.append("\r\n");
}
if (scanner != null) {
scanner.close();
}
}
}
更新:我尝试将输出重定向到文件并从中读取,但这似乎遇到了与之前的实现相同的缓冲问题:我只得到两个数据点。
作为解决方法,我必须要求 .exe 的创建者在显示进度的每行中包含 4100 个额外字符。
最佳答案
如果您的外部进程是基于 C/C++ (stdio) 的,则这很可能是 block 缓冲问题:
stdio-based programs as a rule are line buffered if they are running interactively in a terminal and block buffered when their stdout is redirected to a pipe. In the latter case, you won't see new lines until the buffer overflows or flushed.
参见this answer了解更多详细信息以及一些可能的解决方法。
另请注意,根据this ,行缓冲不是 Win32 上的一个选项:
_IOLBF For some systems, this provides line buffering. However, for Win32, the behavior is the same as _IOFBF - Full Buffering.
因此,如果您选择修改“exe”程序以使用 setvbuf
设置正确的输出模式,则必须使用:
_IONBF No buffer
相反。
关于Java 读取进程输出,无缓冲/实时,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/45075899/