我刚刚开始学习Android,我制作了一个代码,给你一个方程式,你需要解决它,这很简单但是如果有人按下检查按钮并且编辑文本为空,那么应用程序崩溃了,我希望它能够计算这是错误..这是我的代码:
public void Generate(View v) {
x1=10+(int)((99-10+1)*Math.random());
x2=10+(int)((99-10+1)*Math.random());
tv1.setText("" + x1 + "+" + x2 + "=" + "?");
}
public void Check(View view) {
answer1 = et1.getText().toString();
answer = Integer.parseInt(answer1);
if (answer == (x1 + x2)) {
count1++;
count2++;
Toast.makeText(getApplicationContext(), "Correct !",
Toast.LENGTH_SHORT).show();
}
if(answer!=(x1+x2))
{
count1++;
count3++;
Toast.makeText(getApplicationContext(), "Wrong !",
Toast.LENGTH_SHORT).show();
}
if(answer1=="")
{
count1++;
count3++;
Toast.makeText(getApplicationContext(), "Don't try to cheat !",
Toast.LENGTH_SHORT).show();
}
et1.setText("");
tv2.setText("Number of questions : "+count1);
tv3.setText("Right answers : "+count2);
tv4.setText("Wrong answers : "+count3);
Generate(view);
}
答案 0 :(得分:0)
问题是您正在将et1.getText()转换为String,如果返回null值,则引发空指针异常。所以需要实现检查值是否为null:
if(et1.getText()!=null){
answer1 = et1.getText().toString();
if(et1.getText().toString().length()>0){
answer = Integer.parseInt(answer1);
}
if (answer == (x1 + x2)) {
count1++;
count2++;
Toast.makeText(getApplicationContext(), "Correct !",
Toast.LENGTH_SHORT).show();
}
if(answer!=(x1+x2))
{
count1++;
count3++;
Toast.makeText(getApplicationContext(), "Wrong !",
Toast.LENGTH_SHORT).show();
}
if(answer1=="")
{
count1++;
count3++;
Toast.makeText(getApplicationContext(), "Don't try to cheat !",
Toast.LENGTH_SHORT).show();
}
et1.setText("");
tv2.setText("Number of questions : "+count1);
tv3.setText("Right answers : "+count2);
tv4.setText("Wrong answers : "+count3);
Generate(view);
}
另一件事是,如果String为空,那么它将无法解析为int,因此您还需要检查
或者
if(et1.getText().toString().length()>0){
answer = Integer.parseInt(answer1);
}
或
if(!TextUtils.isEmpty(answer1)){
answer = Integer.parseInt(answer1);
}
答案 1 :(得分:0)
您可以检查EditText是否为空。在您的代码
中执行以下操作if (!et1.getText().toString().matches(""))
{
answer1 = et1.getText().toString();
}
else
{
//No value entered
}
或者只是
if (!et1.getText().toString().equals(""))
{
answer1 = et1.getText().toString();
}
else
{
//No value entered
}