dddddd-1.0-SNAPSHOT.jar中没有主清单属性

当一个JAR文件中没有主清单属性时，意味着该JAR文件无法直接作为可执行的Java应用程序运行。主清单属性是指在JAR文件的META-INF/MANIFEST.MF文件中定义的Main-Class属性，它指定了JAR文件中包含的可执行类。

如果你想要运行这个JAR文件，你可以通过命令行来执行它，指定要运行的类名。例如，假设你的JAR文件名为dddddd-1.0-SNAPSHOT.jar，其中包含了一个名为com.example.Main的可执行类，你可以使用以下命令来运行它：

java -cp dddddd-1.0-SNAPSHOT.jar com.example.Main

这将使用java命令来执行JAR文件中的com.example.Main类。

相关问题

Description Consider the following 5 picture frames placed on an 9 x 8 array. ........ ........ ........ ........ .CCC.... EEEEEE.. ........ ........ ..BBBB.. .C.C.... E....E.. DDDDDD.. ........ ..B..B.. .C.C.... E....E.. D....D.. ........ ..B..B.. .CCC.... E....E.. D....D.. ....AAAA ..B..B.. ........ E....E.. D....D.. ....A..A ..BBBB.. ........ E....E.. DDDDDD.. ....A..A ........ ........ E....E.. ........ ....AAAA ........ ........ EEEEEE.. ........ ........ ........ ........ 1 2 3 4 5 Now place them on top of one another starting with 1 at the bottom and ending up with 5 on top. If any part of a frame covers another it hides that part of the frame below. Viewing the stack of 5 frames we see the following. .CCC.... ECBCBB.. DCBCDB.. DCCC.B.. D.B.ABAA D.BBBB.A DDDDAD.A E...AAAA EEEEEE.. In what order are the frames stacked from bottom to top? The answer is EDABC. Your problem is to determine the order in which the frames are stacked from bottom to top given a picture of the stacked frames. Here are the rules: 1. The width of the frame is always exactly 1 character and the sides are never shorter than 3 characters. 2. It is possible to see at least one part of each of the four sides of a frame. A corner shows two sides. 3. The frames will be lettered with capital letters, and no two frames will be assigned the same letter. Input Each input block contains the height, h (h<=30) on the first line and the width w (w<=30) on the second. A picture of the stacked frames is then given as h strings with w characters each. Your input may contain multiple blocks of the format described above, without any blank lines in between. All blocks in the input must be processed sequentially. Output Write the solution to the standard output. Give the letters of the frames in the order they were stacked from bottom to top. If there are multiple possibilities for an ordering, list all such possibilities in alphabetical order, each one on a separate line. There will always be at least one legal ordering for each input block. List the output for all blocks in the input sequentially, without any blank lines (not even between blocks). Sample Input 9 8 .CCC.... ECBCBB.. DCBCDB.. DCCC.B.. D.B.ABAA D.BBBB.A DDDDAD.A E...AAAA EEEEEE.. Sample Output EDABC

这是一道经典的拓扑排序问题。我们可以将每个图形看作一个节点，如果一个图形可以覆盖另一个图形，那就从覆盖的图形指向被覆盖的图形。最后得到的图是一个 DAG，我们可以对其进行拓扑排序。拓扑排序的结果就是图形的叠放顺序。

代码实现可以使用邻接表存储 DAG，然后使用 Kahn 算法进行拓扑排序。对于存在多个拓扑序列的情况，我们可以使用字典序最小的方式输出所有的序列。

以下是 Python 代码实现：

文件格式为 aaaa_bbbbb_cccccc_dddddd.csv,如何批量修改文件名格式为abaa_bbbbb_dddddd.csv

你可以使用批处理脚本来批量修改文件名的格式。假设你的文件名格式都是"aaaa_bbbbb_cccccc_dddddd.csv"，你想要将其修改为"abaa_bbbbb_dddddd.csv"的格式。以下是一个示例脚本：

@echo off
setlocal enabledelayedexpansion

set "source_folder=C:\path\to\your\folder"

for %%F in ("%source_folder%\*.csv") do (
    set "filename=%%~nF"
    set "extension=%%~xF"
    
    set "part1=!filename:~0,1!"
    set "part2=!filename:~1,5!"
    set "part3=!filename:~7,6!"
    set "part4=!filename:~14!"

    set "new_filename=!part1!!part2!!part3!!part4!"

    ren "%%F" "!new_filename!.csv"
)

echo File name format conversion completed.

将上述脚本保存为一个批处理脚本文件（例如，rename_format.bat），然后将`source_folder`变量的值替换为你要修改的目录的路径。

该脚本会遍历指定目录下所有的csv文件，按照"aaaa_bbbbb_cccccc_dddddd.csv"的格式进行重命名，将其修改为"abaa_bbbbb_dddddd.csv"的格式。脚本会将文件名拆分为四个部分，然后重新组合成新的文件名。

运行脚本后，输出窗口会显示"File name format conversion completed."，表示文件名格式转换完成。请确保在运行脚本之前备份好你的文件，以防止意外情况。