Java 从完整路径获取没有扩展名的文件名
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Get filename without extension from full path
提问by dedmar
I am making a program to store data from excel files in database. I would like the user to give in console the full path of the file and after the program to take only the file name to continue.
我正在制作一个程序来将excel文件中的数据存储在数据库中。我希望用户在控制台中给出文件的完整路径，并在程序之后只取文件名继续。
The code for loading the full path is:
加载完整路径的代码是：
String strfullPath = "";
Scanner scanner = new Scanner(System.in);
System.out.println("Please enter the fullpath of the file");
strfullPath = scanner.nextLine();
String file = strfullPath.substring(strfullPath.lastIndexOf('/') + 1);
System.out.println(file.substring(0, file.indexOf('.')));
After that I would like to have: String filename = .......
在那之后，我想要： String filename = .......
The full path that the user would type would be like this: C:\\Users\\myfiles\\Documents\\test9.xls
用户输入的完整路径如下： C:\\Users\\myfiles\\Documents\\test9.xls
The filename that I would create would take only the name without the .xls!
Could anyone help me how I would do this?
我将创建的文件名将只采用没有.xls! 谁能帮助我如何做到这一点？
How i would do it if i would like to take as filename "test9.xls" ? –
如果我想将文件名“test9.xls”作为文件名，我会怎么做？——
回答by Lukas Eichler
You can call the file.getName()method that returns the name of the file as String. Then you cut the extension.
您可以调用将file.getName()文件名作为字符串返回的方法。然后你切断了扩展。
String fileName = file.getName();
fileName = fileName.substring(0, fileName.lastIndexOf(".")+1);
回答by CloudyMarble
You can do it like this:
你可以这样做：
String fname = file.getName();
int pos = fname.lastIndexOf(".");
if (pos > 0) {
fname = fname.substring(0, pos);
}
or you can use the apache.commons.io.FilenameUtils:
或者您可以使用apache.commons.io.FilenameUtils：
String fileNameWithOutExt = FilenameUtils.removeExtension(fileNameWithExt);
回答by Duncan Jones
You could use the Fileclass to get the file name:
您可以使用File该类来获取文件名：
File userFile = new File(strfullPath);
String filename = userFile.getName();
Using a Fileobject has numerous benefits, including the ability to test the file exists:
使用File对象有很多好处，包括能够测试文件是否存在：
if (userFile.isFile()) {
// Yay, it's a valid file (not a directory and not an invalid path)
}
You also need to check the file has an extension before you try and strip it:
在尝试删除文件之前，您还需要检查文件是否具有扩展名：
if (filename.indexOf(".") > 0) {
filename = filename.substring(0, filename.lastIndexOf("."));
}
回答by Oibaf it
I usually use this solution described in other post:
我通常使用其他帖子中描述的这个解决方案：
import org.apache.commons.io.FilenameUtils;
String basename = FilenameUtils.getBaseName(fileName);
回答by Maheshbabu Jammula
if (!filename.equals(""))
{
String [] fileparts = filename.split("\.");
String filename = fileparts[0]; //Get first part
}
