You should use BufferedReader with FileInputStreamReader if your read from a file BufferedReader reader = new BufferedReader(new FileInputStreamReader(pathToFile)); or with InputStreamReader if you read from any other InputStream BufferedReader reader = new BufferedReader(new InputStreamReader(inputStream)); Then use its readLine() method in a loop while(reader.ready()) { String line = reader.readLine(); } But if you really love InputStream …
I need to store data into files inside .jar file and read it again No you don’t. Instead store the ‘default’ file inside the Jar. If it is changed, store the altered file in another place. One common place is a sub-directory of user.home. When checking for the file, first check the existence of an …
It’s looking for more data because nothing’s told it that there won’t be more data. The other end of the network could send more data at any time. It’s not clear whether you’re designing the client/server protocol or just trying to implement it, but typically there are three common ways of detecting the end of …
Since Java 11, you could use a static method InputStream.nullInputStream(): Returns a new InputStream that reads no bytes. The returned stream is initially open. The stream is closed by calling the close() method. Subsequent calls to close() have no effect.
You should use BufferedReader with FileInputStreamReader if your read from a file BufferedReader reader = new BufferedReader(new FileInputStreamReader(pathToFile)); or with InputStreamReader if you read from any other InputStream BufferedReader reader = new BufferedReader(new InputStreamReader(inputStream)); Then use its readLine() method in a loop while(reader.ready()) { String line = reader.readLine(); } But if you really love InputStream …
From reading UNIXProcess.java, this is what happens: We need to distinguish between two states: either process is still alive, or it is dead. If the process is alive, by closing OutputStream (goes to stdin of the process), you are telling the process that there is no more input for it. By closing InputStreams (stdout, stderr …
Use ByteArrayOutputStream with ZipOutputStream to accomplish the task. you can use ZipEntry to specify the files to be included into the zip file. Here is an example of using the above classes, String s = “hello world”; ByteArrayOutputStream baos = new ByteArrayOutputStream(); try(ZipOutputStream zos = new ZipOutputStream(baos)) { /* File is not on the disk, …
For files generated in memory, you may use ByteArrayResource. Just convert your InputStream object using IOUtils from the Apache Commons IO library. It is quite simple: helper.addAttachment(“attachement”, new ByteArrayResource(IOUtils.toByteArray(inputStream)));
The previous answers are not wrong, but they don’t go into the depth that I like to see. There are cases when dealing with low level InputStream is not only desirable, but necessary, the most common example being streaming a large file from source (some web server) to destination (a database). If you try to …
How about: Stream.continually(is.read).takeWhile(_ != -1).map(_.toByte).toArray Update: use LazyList instead of Stream (since Stream is deprecated in Scala 3) LazyList.continually(is.read).takeWhile(_ != -1).map(_.toByte).toArray