我用swing创建了一个简单的应用程序,当涉及到arraylists和listiterator时,我是新手。我正在尝试做的是一个按钮,该按钮可让您更新有关数组列表中人员的信息。我应该这样做,而无需使用for或for-each循环。这是我的代码:
String searchedName = JOptionPane.showInputDialog(null, "Type in the first name of the person you want to update:");
ListIterator<Student> listIt = peopleArray.listIterator();
while(listIt.hasNext()){
String firstname = listIt.next().getFirstname();
if(searchedName.equalsIgnoreCase(firstname){
String newFirstname = JOptionPane.showInputDialog(null, "Input new first name:");
String newLastname = JOptionPane.showInputDialog(null, "Type in new last name:");
String newEmail = JOptionPane.showInputDialog(null, "Type in new email:");
String newOccupation = JOptionPane.showInputDialog(null, "Type in new occupation:");
listIt.previous().setFirstname(newFirstname);
listIt.next().setLastname(newLastname);
listIt.previous().setEmail(newEmail);
listIt.next().setOccupation(newOccupation);
}
}
该代码有效,但看起来很奇怪。我是否必须像这样跳回第四个位置(listIt.next()然后listIt.previous())还是有更好的方法?
// N
最佳答案
您不必来回跳动,确保只调用一次next()。那这个呢:
String searchedName = JOptionPane.showInputDialog(null, "Type in the first name of the person you want to update:");
ListIterator<Student> listIt = peopleArray.listIterator();
while(listIt.hasNext()){
Student studentToUpdate = listIt.next();
if(searchedName.equalsIgnoreCase(studentToUpdate.getFirstname()){
String newFirstname = JOptionPane.showInputDialog(null, "Input new first name:");
String newLastname = JOptionPane.showInputDialog(null, "Type in new last name:");
String newEmail = JOptionPane.showInputDialog(null, "Type in new email:");
String newOccupation = JOptionPane.showInputDialog(null, "Type in new occupation:");
studentToUpdate.setFirstname(newFirstname);
studentToUpdate.next().setLastname(newLastname);
studentToUpdate.setEmail(newEmail);
studentToUpdate.setOccupation(newOccupation);
}
}