在Java中将文件路径作为参数传递-PassingfilepathasanargumentinJava

4  

You need to do this:

你需要这样做:

public static void main(String[] args) {
    String path = args[0];
    // ... 
    File CP_file = new File(path);
    // ... 
}        

1  

If you want to replace the hard coded path with one that you are passing via the command line, you should just be able to pass it in as a String. Your code will not read:

如果要将硬编码路径替换为通过命令行传递的路径,则应该能够将其作为String传递。您的代码不会读取:

...
File CP_file = new File(arg[0]);   //Assuming that the path is the first argument
...

Be sure to quote the path on the CLI, especially if it contains white space or other special characters.

请务必在CLI上引用路径,特别是如果它包含空格或其他特殊字符。

1  

The complete answer would be something like this :

完整的答案是这样的:

The below code by path means /etc/dir/fileName.txt

以下代码的路径表示/etc/dir/fileName.txt

public static void main(String[] args) {
String path = args[0];
// ... 
File CP_file = new File(path);
// ... 

}

But if you want to give just a path and from that path your code to read all your files contained by that directory you would require the above to be exported as a jar file code plus a bat file/ script : bat file example :

但是如果你想只提供一个路径,并且从那个路径你的代码读取该目录包含的所有文件,你需要将上面的内容导出为jar文件代码加上bat文件/脚本:bat文件示例:

FOR %%i IN  (C:/user/dir/*.zip) do (java -cp application.jar;dependencies.jar;...;%%i)

Run the script and your code will run the file / files that are on the path C:/user/dir/*.zip

运行脚本,您的代码将运行路径C:/ user / dir / * .zip上的文件/文件

1  

Your problem has two sides: how to pass the parameter from the command line, and how to read it inside your code.

您的问题有两个方面:如何从命令行传递参数,以及如何在代码中读取它。

Passing arguments to a Java program

All arguments meant for the actual Java class and not for the JVM, should be put after the class name, like this:

所有参数都是针对实际的Java类而不是针对JVM的,应该放在类名后面,如下所示:

C:\YOUR\WORKSPACE> java your.package.YouMainClass "C:\Users\XYZ\workspace\Customer_Product_info.txt"`

Things to watch out for:

需要注意的事项:

• Slashes / vs backslashes \: since you're on a Windows system, I'd rather use backslashes for your path, specially if you are including the drive letter. Java can work with both variants, but it's better to follow your SO conventions.

斜杠/ vs反斜杠\:因为你在Windows系统上,我宁愿使用反斜杠作为你的路径,特别是如果你包括驱动器号。 Java可以使用这两种变体,但最好遵循您的SO约定。

• Double quotes " to allow for spaces: you'll need to enclose your path in double quotes if any directories contain spaces in their names, so just double quote it every time.

双引号“允许空格:如果任何目录的名称中包含空格,则需要将路径括在双引号中,因此每次都要双引号。

• Remove final backslash: this only applies to directory paths (file paths can't end in a backslash in Windows). If you write a directory path like this: "C:\My path\XYZ\", the last double quote will be included as part of the path, because of the previous backslash escaping it \". Instead, "C:\My path\XYZ" will do fine.

删除最后的反斜杠:这仅适用于目录路径(文件路径不能以Windows中的反斜杠结尾)。如果你编写一个这样的目录路径:“C:\ My path \ XYZ \”,最后的双引号将作为路径的一部分包含在内,因为前面的反斜杠转义它“。而是,”C:\ My path \ XYZ“会很好。

Reading arguments from your main(String[]) method

Now this one is simple: as others have pointed out, the String with your path should now be in args[0]:

现在这个很简单:正如其他人指出的那样,带有路径的String现在应该在args [0]中:

public static void main(String[] args) {

    fileReader fr = new fileReader();
    getList lists = new getList();

    if (args[0] == null || args[0].trim().isEmpty()) {
        System.out.println("You need to specify a path!");
        return;
    } else {
        File CP_file = new File(args[0]);
        int count = fr.fileSizeInLines(CP_file);
        System.out.println("Total number of lines in the file are: "+count);

        List lines = fr.strReader(CP_file);

        ....
    }
}

I added some null-checking to avoid problems like the ArrayIndexOutOfBounds you run into.

我添加了一些空值检查,以避免遇到像您遇到的ArrayIndexOutOfBounds之类的问题。