[英]Java unreported exception confusing me
大家好,我一直在為GUI設計一些按鈕,我決定實施一些以前的代碼。
但是,嘗試編譯時出現錯誤。 在我的代碼的第141行(特別是最后一個按鈕)中,我被告知我有一個未報告的IOException
,必須將其捕獲或聲明為拋出。
我的代碼如下:
public void actionPerformed(ActionEvent ae) {
if ((ae.getSource() == button5) && (!connected)) {
try {
s = new Socket("127.0.0.1", 2020);
pw = new PrintWriter(s.getOutputStream(), true);
} catch (UnknownHostException uhe) {
System.out.println(uhe.getMessage());
} catch (IOException ioe) {
System.out.println(ioe.getMessage());
}
connected = true;
t = new Thread(this);
//b.setEnabled(false);
button5.setLabel("Disconnect");
t.start();
} else if ((ae.getSource() == button5) && (connected)) {
connected = false;
try {
s.close(); //no buffering so, ok
} catch (IOException ioe) {
System.out.println(ioe.getMessage());
}
//System.exit(0);
button5.setLabel("Connect");
} else {
temp = tf.getText();
pw.println(temp);
tf.setText("");
}
if (ae.getActionCommand().equals("Save it")) {
Scanner scan = new Scanner(System.in);
try {
PrintWriter pw = new PrintWriter(new FileWriter(new File("test.txt")));
for (;;) {
String temp = scan.nextLine();
if (temp.equals("")) {
break;
}
pw.println(temp);
}
pw.close();
} catch (IOException ioe) {
System.out.println("IO Exception! " + ioe.getMessage());
}
} else if (ae.getActionCommand().equals("Load it")) {
Scanner scan = new Scanner(System.in);
try {
BufferedReader br = new BufferedReader(new FileReader(new File("test.txt")));
String temp = "";
while ((temp = br.readLine()) != null) {
System.out.println(temp);
}
br.close();
} catch (FileNotFoundException fnfe) {
System.out.println("Input file not found.");
} catch (IOException ioe) {
System.out.println("IO Exception! " + ioe.getMessage());
}
} else if (ae.getActionCommand().equals("Clear it")) {
ta.setText("");
} else {
PrintWriter pw = new PrintWriter(new FileWriter(new File("test.txt")));
}
}
只需在以下代碼中添加try / catch塊(發布內容的結尾):
else{
PrintWriter pw = new PrintWriter (
new FileWriter(
new File("test.txt")));
}}
像這樣:
else{
try{
PrintWriter pw = new PrintWriter (new FileWriter(new File("test.txt")));
} catch (Exception e) {
e.printStackTrace();
}
}}
通常,任何IO操作都可能導致異常。 根據您的需要,最簡單的解決方案是throws IOException
到您看到問題的方法的頂部,但這不是很好的做法,在這種情況下不起作用。 將try / catch塊放在問題線上,並包含有意義的錯誤消息,可能是最好的方法。
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