名称要求为AS011_20210801-99.txt
20210801为递增值,其余为不变值。
做连续的一百个文件。
代码越简单越好,用batch文件写出来也可。
import java.io.BufferedWriter;
import java.io.FileWriter;
import java.text.SimpleDateFormat;
import java.util.Calendar;
import java.util.Date;
public class Main {
public static void main(String[] args) throws Exception {
SimpleDateFormat sdf = new SimpleDateFormat("yyyyMMdd");
Date date = sdf.parse("20210801");
Calendar c = Calendar.getInstance();
c.setTime(date);
for (int i = 0; i < 100; i++) {
String time = sdf.format(c.getTime());
try (BufferedWriter bw = new BufferedWriter(new FileWriter("AS011_"+time+"-99.txt"))) {
}
c.add(Calendar.DATE, 1);
}
}
}
创建文件夹,将文件后缀格式改为bat,将以下代码复制到文件内,保存之后,双击运行即可。
@echo off
for /l %%i in (20210801,1,20210901) do echo=>AS011_%%i-99.txt
pause
运行之后的截图:
如果有帮助到你,希望能够采纳支持一下,谢谢!
public class Test {
public static void main(String[] args) {
for (int i = 1; i<= 100; i++)
{
String fileName = "AS011_20210";
fileName = fileName + String.valueOf(800 + i) + "-99.txt";
File testFile = new File("D:" + File.separator + fileName);
try {
testFile.createNewFile();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
}
如果我的回答对你有帮你,麻烦点一下我的回答右上角的采纳,谢谢。
代码如上,运行效果如下:
题主应该需要这个:
望采纳,谢谢
@echo off
set "year=%date:~0,4%"
set "month=%date:~5,2%"
set "day=%date:~8,2%"
set "hour_ten=%time:~0,1%"
set "hour_one=%time:~1,1%"
set "minute=%time:~3,2%"
set "second=%time:~6,2%"
if "%hour_ten%" == " " (
type nul > %year%%month%%day%0%hour_one%%minute%%second%.txt
) else (
type nul > %year%%month%%day%%hour_ten%%hour_one%%minute%%second%.txt
)
pause
public class Test {
public static void main(String[] args) throws Exception {
SimpleDateFormat sdf = new SimpleDateFormat("yyyyMMdd");
Date date = sdf.parse("20210801");
Calendar c = Calendar.getInstance();
c.setTime(date);
for (int i = 0; i < 100; i++) {
String time = sdf.format(c.getTime());
try (BufferedWriter bw = new BufferedWriter(new FileWriter("AS011_"+time+"-99.txt"))) {
// 此处给文本赋值,默认空白
}
c.add(Calendar.DATE, 1);
}
}
}