java parseint - 线程异常 "AWT-EventQueue-0" java.lang.NumberFormatException：对于输入字符串：“”

Question

我正在用 netbeans 编写数独程序，代码如下：

private void jButton1ActionPerformed(java.awt.event.ActionEvent evt) {                                         

    int[][] sudoku = new int [4][4];


    sudoku[0][0] = Integer.parseInt(t1.getText());
    sudoku[1][0] = Integer.parseInt(t2.getText());
    sudoku[2][0] = Integer.parseInt(t5.getText());
    sudoku[3][0] = Integer.parseInt(t6.getText());
    sudoku[0][1] = Integer.parseInt(t3.getText());
    sudoku[1][1] = Integer.parseInt(t4.getText());
    sudoku[2][1] = Integer.parseInt(t7.getText());
    sudoku[3][1] = Integer.parseInt(t8.getText());
    sudoku[0][2] = Integer.parseInt(t9.getText());
    sudoku[1][2] = Integer.parseInt(t10.getText());
    sudoku[2][2] = Integer.parseInt(t13.getText());
    sudoku[3][2] = Integer.parseInt(t14.getText());
    sudoku[0][3] = Integer.parseInt(t11.getText());
    sudoku[1][3] = Integer.parseInt(t12.getText());
    sudoku[2][3] = Integer.parseInt(t15.getText());
    sudoku[3][3] = Integer.parseInt(t16.getText());

    if (sudoku[1][1] == sudoku[2][1])
    { 
        jLabel1.setText("true");
    }

}

而且每次都会报错：
线程 "AWT-EventQueue-0" java.lang.NumberFormatException 中的异常：对于输入字符串：“”

如果你们能弄清楚哪里出了问题，也许会写try catch exception（因为我不知道怎么写）就太好了谢谢

Answer 1

如果您认为在您的代码中“”应该与“0”同义，那么创建一个小方法

int getNumber (String str) {

   if (str.equals ("")) {
        return 0;
   } 

   return Integer.parseInt (str);
}

然后可以通过您的代码调用此方法

sudoku[0][0] = getNumber(t1.getText());

java parseint - 线程异常 "AWT-EventQueue-0" java.lang.NumberFormatException：对于输入字符串：“”

java parseint - Exception in thread "AWT-EventQueue-0" java.lang.NumberFormatException: For input string: ""

java

compiler-errors