我写了一个程序，根据两个常量文件，将随机数生成两个文本文件，随机写入三分之一。现在我需要逐行读取每个文本文件，并将它们放在一起。该计划是here找到的建议对我的情况没有任何帮助。当我尝试这种方法时，它只读取所有行，直到它完成而不允许我暂停它，转到不同的文件等选项。
理想情况下，我想找到一些方法来阅读下一行，然后再去那行。就像某种变量可以让我在阅读中占有一席之地。
public static void mergeProductCodesToFile(String prefixFile,
String inlineFile,
String suffixFile,
String productFile) throws IOException
{
try (BufferedReader br = new BufferedReader(new FileReader(prefixFile)))
{
String line;
while ((line = br.readLine()) != null)
{
try (PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter(productFile, true))))
{
out.print(line); //This will print the next digit to the right
}
catch (FileNotFoundException e)
{
System.err.println("File error: " + e.getMessage());
}
}
}
}
编辑：根据以下内容创建的数字。基本上，常量告诉它要在每行中创建多少位数以及要创建多少行。现在我需要将这些组合在一起而不删除任何文本文件中的任何内容。
public static void writeRandomCodesToFile(String codeFile,
char fromChar, char toChar,
int numberOfCharactersPerCode,
int numberOfCodesToGenerate) throws IOException
{
for (int i = 1; i <= PRODUCT_COUNT; i++)
{
int I = 0;
if (codeFile == "inline.txt")
{
for (I = 1; I <= CHARACTERS_PER_CODE; I++)
{
int digit = (int)(Math.random() * 10);
try (PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter(codeFile, true))))
{
out.print(digit); //This will print the next digit to the right
}
catch (FileNotFoundException e)
{
System.err.println("File error: " + e.getMessage());
System.exit(1);
}
}
}
if ((codeFile == "prefix.txt") || (codeFile == "suffix.txt"))
{
for (I = 1; I <= CHARACTERS_PER_CODE; I++)
{
Random r = new Random();
char digit = (char)(r.nextInt(26) + 'a');
digit = Character.toUpperCase(digit);
try (PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter(codeFile, true))))
{
out.print(digit);
}
catch (FileNotFoundException e)
{
System.err.println("File error: " + e.getMessage());
System.exit(1);
}
}
}
//This will take the text file to the next line
if (I >= CHARACTERS_PER_CODE)
{
{
Random r = new Random();
char digit = (char)(r.nextInt(26) + 'a');
try (PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter(codeFile, true))))
{
out.println(""); //This will return a new line for the next loop
}
catch (FileNotFoundException e)
{
System.err.println("File error: " + e.getMessage());
System.exit(1);
}
}
}
}
System.out.println(codeFile + " was successfully created.");
}// end writeRandomCodesToFile()
答案 0 :(得分：1)
尊重您的代码，它将是这样的：
public static void mergeProductCodesToFile(String prefixFile, String inlineFile, String suffixFile, String productFile) throws IOException {
try (BufferedReader prefixReader = new BufferedReader(new FileReader(prefixFile));
BufferedReader inlineReader = new BufferedReader(new FileReader(inlineFile));
BufferedReader suffixReader = new BufferedReader(new FileReader(suffixFile))) {
StringBuilder line = new StringBuilder();
String prefix, inline, suffix;
while ((prefix = prefixReader.readLine()) != null) {
//assuming that nothing fails and the files are equals in # of lines.
inline = inlineReader.readLine();
suffix = suffixReader.readLine();
line.append(prefix).append(inline).append(suffix).append("\r\n");
// write it
...
}
} finally {/*close writers*/}
}
可能会抛出一些例外。
我希望你不要在一个方法中实现它。 您也可以使用迭代器，或者非常简单的读者类（方法）。
我不会使用List来加载数据至少我保证文件的大小很小，我可以节省内存使用量。
答案 1 :(得分：1)
我们通过存储数据和交错来讨论我的方法。就像塞尔吉奥在他的回答中所说，确保内存在文件大小和数据结构将使用多少内存方面不是问题。
//the main method we're working on
public static void mergeProductCodesToFile(String prefixFile,
String inlineFile,
String suffixFile,
String productFile) throws IOException
{
try {
List<String> prefix = read(prefixFile);
List<String> inline = read(inlineFile);
List<String> suffix = read(productFile);
String fileText = interleave(prefix, inline, suffix);
//write the single string to file however you want
} catch (...) {...}//do your error handling...
}
//helper methods and some static variables
private static Scanner reader;//I just prefer scanner. Use whatever you want.
private static StringBuilder sb;
private static List<String> read(String filename) throws IOException
{
List<String> list = new ArrayList<String>;
try (reader = new Scanner(new File(filename)))
{
while(reader.hasNext())
{ list.add(reader.nextLine()); }
} catch (...) {...}//catch errors...
}
//I'm going to build the whole file in one string, but you could also have this method return one line at a time (something like an iterator) and output it to the file to avoid creating the massive string
private static String interleave(List<String> one, List<String> two, List<String> three)
{
sb = new StringBuilder();
for (int i = 0; i < one.size(); i++)//notice no checking on size equality of words or the lists. you might want this
{
sb.append(one.get(i)).append(two.get(i)).append(three.get(i)).append("\n");
}
return sb.toString()
}
显然，在内存和性能方面仍有一些需要;此外，还有一些方法可以使其在其他情况下更具可扩展性，但这是一个很好的起点。使用c＃，我可以更轻松地使用迭代器使交错一次为您提供一行，从而可能节省内存。只是一个不同的想法！