问题描述
Path pp = FileSystems.getDefault().getPath("logs","access.log");
final int BUFFER_SIZE = 1024*1024; //this is actually bytes
FileInputStream fis = new FileInputStream(pp.toFile());
byte[] buffer = new byte[BUFFER_SIZE];
int read = 0;
while( ( read = fis.read( buffer ) ) > 0 ){
// call your other methodes here...
}
fis.close();
解决方法
多年来,我都面临着同样的情况。我的最终解决方案是使用接口列表中的方法.sublist()
,您可以使用该方法:
第一步:读取给定文件中的所有行
String textfileRow = null;
List<String> fileLines = new ArrayList<String>();
BufferedReader fileContentBuffer = null;
fileContentBuffer = new BufferedReader(new FileReader(<your file>));
while ((textfileRow = fileContentBuffer.readLine()) != null)
{
fileLines.add(textfileRow);
}
第二步:从给定大小的先前创建的列表中创建块
int final CHUNKSIZE = <your needed chunk size>;
int lineIndex = 0;
while (lineIndex < fileLines.size())
{
int chunkEnd = lineIndex + CHUNKSIZE;
if (chunkEnd >= fileLines.size())
{
chunkEnd = fileLines.size();
}
List<Type you need> mySubList = fileLines.subList(lineIndex,chunkEnd);
//What ever you want do to...
lineIndex = chunkEnd;
}
在我的项目中,我将其与长达2万行的csv文件一起使用,并且效果很好。
编辑:我在标题中看到了对文本文件的请求,因此我更改了读取文本文件的方式。
,旧方法:使用BufferedReader的readLine方法代替原始的FileInputStream
。
Path path = // access your path...;
List<String> buffer = new ArrayList<>();
try (BufferedReader in = new BufferedReader(new FileReader(path.toFile))) {
String nextLine;
do {
buffer.clear();
for (int i=0; i < chunkSize; i++) {
// note that in.readLine() returns null at end-of-file
if ((nextLine = in.readLine()) == null) break;
buffer.add(next);
}
processChunk(buffer); // note size may be less than desiredChunkSize
} while (nextLine != null);
} catch (IOException ioe) {
// handle exceptions here
}
// all resources will be automatically closed before this line is reached
较新的方法:使用Files.lines访问懒散的行流:
Path path = // access your path...;
final AtomicInteger c = new AtomicInteger();
Files.lines(path)
.collect(Collectors.groupingBy(e -> c.getAndIncrement()/chunkSize))
.forEach(chunk -> processChunk(chunk));
// all resources will be automatically closed before this line is reached
免责声明:我也没有测试;但是两种方法应该都可以。