如何打开文件弹出 [英] How to get open file popup

问题描述

现在，我有一个设置类路径，但我想打开一个打开的文件，用户选择打开哪个文件。我试过JFileChooser，但迄今为止还没有成功。这是我的代码：

pre $ public static void main（String [] args）throws IOException {


JFileChooser chooser = new JFileChooser（）;

int returnValue = chooser.showOpenDialog（null）;
if（returnValue == JFileChooser.APPROVE_OPTION）{
File file = chooser.getSelectedFile（）;
}

//我不希望这是硬编码的：
String filePath =/Users/Bill/Desktop/hello.txt;


我应该如何去做这件事？

文件文件的范围。

尝试在if块之外声明文件。

 文件file =空值; 
 if（returnValue == JFileChooser.APPROVE_OPTION）{
 file = chooser.getSelectedFile（）; 
 
 if（file！= null）
 {
 String filePath = file.getPath（）; 
} 
  

Right now, I have a set class path, but I want to have an open file pop up and the user chooses which file to open. I've tried JFileChooser, but haven't been successful so far. Here's my code:

public static void main(String[] args) throws IOException {


    JFileChooser chooser = new JFileChooser();

            int returnValue = chooser.showOpenDialog( null ) ;
    if( returnValue == JFileChooser.APPROVE_OPTION ) {
        File file = chooser.getSelectedFile() ;
    }

    // I don't want this to be hard-coded:
    String filePath = "/Users/Bill/Desktop/hello.txt";

How should I go about doing this?

解决方案

I think the problem is the scope of File file.

Try Declaring file outside the if-block.

 File file = null;
 if( returnValue == JFileChooser.APPROVE_OPTION ) {
        file = chooser.getSelectedFile() ;
 }
 if(file != null)
 {
      String filePath = file.getPath();
 } 

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