我的問題是這樣的: 我在EditText中輸入文本,然後顯示每個單獨的字母,但可以使用2個ImageView進行顯示。 我對ImageView沒有任何問題,問題是當你輸入單詞到for,試圖延遲顯示第一個字母並繼續與其他人,但它不起作用,只是發送給我的最後一個字母字輸入Android中的循環,單詞和圖像
代碼:
public class deletreo extends Activity {
TextView tv;
EditText etxt;
ImageView img,img2;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
requestWindowFeature(Window.FEATURE_NO_TITLE);
getWindow().setFlags(WindowManager.LayoutParams.FLAG_FULLSCREEN,
WindowManager.LayoutParams.FLAG_FULLSCREEN);
setContentView(R.layout.deletreo);
tv = new TextView(this);
etxt = (EditText)findViewById(R.id.text);
Button btn = (Button)findViewById(R.id.btn7);
btn.setOnClickListener(new OnClickListener() {
public void onClick(View v) {
letra();
}
});
}//fin bundle
public void letra() {
String t = "";
t = etxt.getText().toString();
String l = t.toLowerCase();
int p = l.length();
try{
for(int j = 0 ; j < p ; j++){
if(l.charAt(j) == 'a' || l.charAt(j) == 'A'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.aa);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_1);
}
if(l.charAt(j) == 'b' || l.charAt(j) == 'B'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.bb);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_2);
}
if(l.charAt(j) == 'c' || l.charAt(j) == 'C'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.cc);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_3);
}
if(l.charAt(j) == 'd' || l.charAt(j) == 'D'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.dd);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_4);
}
if(t.charAt(j) == 'e' || t.charAt(j) == 'E'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.ee);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_5);
}
if(t.charAt(j) == 'f' || t.charAt(j) == 'F'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.ff);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_6);
}
if(t.charAt(j) == 'g' || t.charAt(j) == 'G'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.gg);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_7);
}
if(t.charAt(j) == 'h' || t.charAt(j) == 'H'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.hh);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_8);
}
if(t.charAt(j) == 'i' || t.charAt(j) == 'I'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.ii);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_9);
}
if(t.charAt(j) == 'j' || t.charAt(j) == 'J'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.jj);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_10);
}
if(t.charAt(j) == 'k' || t.charAt(j) == 'K'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.kk);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_11);
}
if(t.charAt(j) == 'l' || t.charAt(j) == 'L'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.ll);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_12);
}
if(t.charAt(j) == 'm' || t.charAt(j) == 'M'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.mm);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_13);
}
if(t.charAt(j) == 'n' || t.charAt(j) == 'N'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.nn);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_14);
}
if(t.charAt(j) == 'ñ' || t.charAt(j) == 'Ñ'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.nini);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_15);
}
if(t.charAt(j) == 'o' || t.charAt(j) == 'O'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.oo);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_16);
}
if(t.charAt(j) == 'p' || t.charAt(j) == 'P'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.pp);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_17);
}
if(t.charAt(j) == 'q' || t.charAt(j) == 'Q'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.qq);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_18);
}
if(t.charAt(j) == 'r' || t.charAt(j) == 'R'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.rr);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_19);
}
if(t.charAt(j) == 's' || t.charAt(j) == 'S'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.ss);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_20);
}
if(t.charAt(j) == 't' || t.charAt(j) == 'T'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.tt);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_21);
}
if(t.charAt(j) == 'u' || t.charAt(j) == 'U'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.uu);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_22);
}
if(t.charAt(j) == 'v' || t.charAt(j) == 'V'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.vv);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_23);
}
if(t.charAt(j) == 'w' || t.charAt(j) == 'W'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.ww);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_24);
}
if(t.charAt(j) == 'x' || t.charAt(j) == 'X'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.xx);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_25);
}
if(t.charAt(j) == 'y' || t.charAt(j) == 'Y'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.yy);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_26);
}
if(t.charAt(j) == 'z' || t.charAt(j) == 'Z'){
img = (ImageView)findViewById(R.id.img);
img.setImageResource(R.drawable.zz);
img2 = (ImageView)findViewById(R.id.img2);
img2.setImageResource(R.drawable.image_27);
}
Thread.sleep(2000);
}//fin del for
}//fin try
catch (InterruptedException e) {
e.printStackTrace();
}
finally{}
}//fin letra();
}//fin
這是很多代碼。你可以使用調試器(甚至是日誌語句)來縮小問題的範圍嗎? –
對於您的巨大級聯'if/else'語句,您應該使用'switch'語句。它將更加清潔和高效。 – Michael
它是,只是在哪裏代碼,如果它只是比較在EditText中輸入的字母,我甚至沒有改變,通過切換到一些嚴重 – DiegoF