尝试在大文件中查找单词。文件逐行读取。读取redLine异常时的方式。这有什么办法吗?你可以把它作为字符串在地板上阅读吗?
for(String line; (line = fileOut.readLine()) != null; ){
if(line.contains(commandString))
System.out.println(count + ": " + line);
count++;
}
java.lang.OutOfMemoryError:
UDP:
这是我糟糕的代码:
static String file = "files.txt";
static String commandString = "first";
static int count = 1;
public static void main(String[] args) throws IOException
{
try(BufferedReader fileOut = new BufferedReader(new InputStreamReader(new FileInputStream(file), "Cp1251")) ){
for(String line; (line = fileOut.readLine()) != null; ){
if(line.contains(commandString))
System.out.println(count + ": " + line);
count++;
}
System.out.println("before wr close :" + Runtime.getRuntime().freeMemory());
fileOut.close();
}catch(Exception e) {
System.out.println(e);
}
}
答案 0 :(得分:1)
搜索单词时,您可以按字节顺序读取文件,而无需在内存中保存多个文件的单个字节。 逐字节读取,每次,一个字节等于搜索字的第一个字节,开始第二个循环并读取后面的字节,并检查下一个字节是否等于字中的下一个字节,依此类推。 举个例子,我根据您的需要修改了样本 我已经省略了文件的输出,因为我不知道,如果你想输出所有的行或只是那些包含你的关键字的行,后者可能会像逐行读取代码一样有问题
static String fileName = "files.txt";
static byte[] searchString = { 'f', 'i', 'r', 's', 't' };
static int count = 0;
static long position = 1;
public static void main(String[] args) throws IOException {
try (FileInputStream file = new FileInputStream(fileName)) {
byte read[] = new byte[1];
outerLoop: while (-1 < file.read(read, 0, 1)) {
position++;
if (read[0] == searchString[0]) {
int matches = 1;
for (int i = 1; i < searchString.length; i++) {
if (-1 > file.read(read, 0, 1)) {
break outerLoop;
}
position++;
if (read[0] == searchString[i]) {
matches++;
} else {
break;
}
}
if (matches == searchString.length) {
System.out.println((++count)+". found at position "+ (position-matches));
}
}
}
file.close();
} catch (Exception e) {
e.printStackTrace();
}
}