我认为这个问题非常简单。
事情是:我的TextView(位于TableLay中的TableRow中)离开可见屏幕。
屏幕截图应该澄清:
我试过
tvValue.setLinksClickable(true);
tvValue.setEllipsize(TruncateAt.MIDDLE);
1)不应用截断 2)显然,它们显示链接,但它们不可点击!
由于它是一个高度动态的东西,我需要在代码中进行所有初始化。 所以,这是代码:
public class StartUp extends Activity {
private TableLayout mTableLayout;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
final Context appContext = getApplicationContext();
mTableLayout = new TableLayout(appContext);
mTableLayout.setBackgroundColor(0xffffffff);
String[] h = new String[] { "email", "facebook", "youtube" };
String[] v = new String[] { "lol@xd.de", "http://www.facebook.com/010101010101010", "http://www.youtube.com/lalala" };
LayoutParams rowLayoutParams = new LayoutParams(LayoutParams.FILL_PARENT, LayoutParams.WRAP_CONTENT);
for (int i = 0; i < h.length; ++i) {
TableRow row = new TableRow(appContext);
row.setLayoutParams(rowLayoutParams);
TextView tvHead = new TextView(appContext);
tvHead.setTextColor(0xff336699);
tvHead.setPadding(0, 0, 10, 0);
tvHead.setText(h[i]);
TextView tvValue = new TextView(appContext);
tvValue.setTextColor(0xff191919);
tvValue.setLinksClickable(true);
tvValue.setText(v[i]);
tvValue.setEllipsize(TruncateAt.MIDDLE);
row.addView(tvHead);
row.addView(tvValue);
mTableLayout.addView(row);
}
setContentView(mTableLayout);
}
}
答案 0 :(得分:3)