本文介绍了我的if/else语句始终在执行else(android)的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我正在制作一个可以帮助我解决问题的应用程序(很难解释它是什么),但它基本上是从EditText中获取一个4字母的String并在TextView中输出另一个String.但是,即使if语句为true,if/else语句也始终执行else.我将我的代码放在下面.
I am making an app to help me with something (It's quite hard to explain what it is) but it basically takes a 4 letter String from an EditText and outputs another String in the TextView. However the if/else statement is always executing the else even if the if statement is true. I will put my code below.
public class egkk extends ActionBarActivity {
EditText editText;
TextView textView;
Button button;
String aerodrome;
String firstTwo;
String firstThree;
String first;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_egkk);
button = (Button)findViewById(R.id.gen);
button.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View view) {
editText = (EditText)findViewById(R.id.egkk_edit);
textView = (TextView)findViewById(R.id.s_view);
aerodrome = editText.getText().toString().toLowerCase();
firstTwo = aerodrome.substring(0, 1);
firstThree = aerodrome.substring(0, 2);
first = aerodrome.substring(0, 0);
if (aerodrome.length() == 4){
if (aerodrome.equals("egkk")){
textView.setText("3750 - 3763: Gatwick APC general usage" + "3764 - 3767: Gatwick ADC circuit traffic and transits");
}
if (firstThree.equals("egj")){
textView.setText("Squawk is 7760 – 7775");
}
if (firstTwo.equals("eg")){
textView.setText("Sqauwk is 4301 - 4377");
}
if (firstTwo.equals("ei")){
textView.setText("Squawk is 4430 – 4477");
}
if (firstTwo.equals("lf")){
textView.setText("Squawk is 2201 – 2277");
}
if (firstTwo.equals("le")){
textView.setText("Squawk is 2201 – 2277");
}
if (firstTwo.equals("lp")){
textView.setText("Squawk is 2201 – 2277");
}
if (firstTwo.equals("fa")){
textView.setText("Squawk is 2201 – 2277");
}
if (firstTwo.equals("eh")){
textView.setText("Squawk is 7310 – 7677");
}
if (firstTwo.equals("eh")){
textView.setText("Squawk is 7310 – 7677");
}
if (first.equals("k")){
textView.setText("Squawk is 7610 – 7677");
}
if (first.equals("c")){
textView.setText("Squawk is 7610 – 7677");
}
if (firstTwo.equals("mu")){
textView.setText("Squawk is 7610 – 7677");
}
if (firstTwo.equals("my")){
textView.setText("Squawk is 7610 – 7677");
}
if (firstTwo.equals("mk")){
textView.setText("Squawk is 7610 – 7677");
}
if (firstTwo.equals("mt")){
textView.setText("Squawk is 7610 – 7677");
}
if (firstTwo.equals("md")){
textView.setText("Squawk is 7610 – 7677");
}
if (firstTwo.equals("tj")){
textView.setText("Squawk is 7610 – 7677");
}
if (firstTwo.equals("mu")){
textView.setText("Squawk is 7610 – 7677");
}
else {
textView.setText("Squawk is 1140 – 1177");
}
}
else {
textView.setText("Error");
}
}
});
}
@Override
public boolean onCreateOptionsMenu(Menu menu) {
// Inflate the menu; this adds items to the action bar if it is present.
getMenuInflater().inflate(R.menu.menu_egkk, menu);
return true;
}
@Override
public boolean onOptionsItemSelected(MenuItem item) {
// Handle action bar item clicks here. The action bar will
// automatically handle clicks on the Home/Up button, so long
// as you specify a parent activity in AndroidManifest.xml.
int id = item.getItemId();
//noinspection SimplifiableIfStatement
if (id == R.id.action_settings) {
return true;
}
return super.onOptionsItemSelected(item);
}
}
`
推荐答案
这是因为您的情况:
if (aerodrome.length() == 4){
从不匹配.这意味着,在这里,当您获取文本时,永远不会是长度为4 length
的字符串.
Never matches. This means, here, when you get the text, is never a String of 4 length
.
aerodrome = editText.getText().toString().toLowerCase();
这篇关于我的if/else语句始终在执行else(android)的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!