InputMethodManager imm = (InputMethodManager)getSystemService(Context.INPUT_METHOD_SERVICE);
//隐藏软键盘
// imm.hideSoftInputFromWindow(tv.getWindowToken(), 0);
//显示软键盘
// imm.showSoftInputFromInputMethod(tv.getWindowToken(), 0);
//切换软键盘的显示与隐藏
imm.toggleSoftInputFromWindow(tv.getWindowToken(), 0, InputMethodManager.HIDE_NOT_ALWAYS);
//或者
// imm.toggleSoftInput(0, InputMethodManager.HIDE_NOT_ALWAYS);

1.android 中设置Activity启动时候是单例(singleTask)时候是，onActivityResult将失去应有的功能，当startActivityForResult执行时候将立即执行onActivityResult方法，原因待查，先mark...

自定义布局，继承view的时候，在构造函数中引入xml文件，这种写法是错误，它是view包下。不是wight包下，所以需继承wight包下的控件，然后在myBarView(Context context, AttributeSet attrs)带两个参数的构造函数下引入xml文件。
可能这种理解本身就是错误的。这是我自己找的原因，待查。

生成水印的过程。其实分为三个环节：第一，载入原始图片；第二，载入水印图片；第三，保存新的图片。
/**
* create the bitmap from a byte array
*
* @param src the bitmap object you want proecss
* @param watermark the water mark above the src
* @return return a bitmap object ,if paramter's length is 0,return null
*/
private Bitmap createBitmap( Bitmap src, Bitmap watermark )
{
String tag = "createBitmap";
Log.d( tag, "create a new bitmap" );
if( src == null )
{
return null;
}
int w = src.getWidth();
int h = src.getHeight();
int ww = watermark.getWidth();
int wh = watermark.getHeight();
//create the new blank bitmap
Bitmap newb = Bitmap.createBitmap( w, h, Config.ARGB_8888 );//创建一个新的和SRC长度宽度一样的位图
Canvas cv = new Canvas( newb );
//draw src into
cv.drawBitmap( src, 0, 0, null );//在 0，0坐标开始画入src
//draw watermark into
cv.drawBitmap( watermark, w - ww + 5, h - wh + 5, null );//在src的右下角画入水印
//save all clip
cv.save( Canvas.ALL_SAVE_FLAG );//保存
//store
cv.restore();//存储
return newb;

public class fanShe {
private String print(int s,String sss) {
System.out.println(s + sss);
return "今天周日";
}
}
反射
fanShe fanshe = new fanShe();
// 获取这个反射对象
Class<fanShe> tm_class = (Class<fanShe>) fanshe.getClass();
// 得到这个方法，后面传参的class类型,顺序要与方法一致
Method print = tm_class.getDeclaredMethod("print", int.class,String.class);
// 打开这个方法的访问权限，因为这个方法是私有的
print.setAccessible(true);
// 执行这个方法，ss是这个方法执行的返回值,Object里面传参与方法的类型一致
String returnss = (String) print.invoke(fanshe, new Object[] { 1,"天下第一" });
System.err.println(returnss);
打印结果
今天周日
1天下第一

计算距离的逻辑是从Android的提供的接口（Location.distanceBetween）中拔来的，应该是最精确的方法了。
EG:
public static double computeDistance(double lat1, double lon1,
double lat2, double lon2) {
// Based on http://www.ngs.noaa.gov/PUBS_LIB/inverse.pdf
// using the "Inverse Formula" (section 4)
int MAXITERS = 20;
// Convert lat/long to radians
lat1 *= Math.PI / 180.0;
lat2 *= Math.PI / 180.0;
lon1 *= Math.PI / 180.0;
lon2 *= Math.PI / 180.0;
double a = 6378137.0; // WGS84 major axis
double b = 6356752.3142; // WGS84 semi-major axis
double f = (a - b) / a;
double aSqMinusBSqOverBSq = (a * a - b * b) / (b * b);
double L = lon2 - lon1;
double A = 0.0;
double U1 = Math.atan((1.0 - f) * Math.tan(lat1));
double U2 = Math.atan((1.0 - f) * Math.tan(lat2));
double cosU1 = Math.cos(U1);
double cosU2 = Math.cos(U2);
double sinU1 = Math.sin(U1);
double sinU2 = Math.sin(U2);
double cosU1cosU2 = cosU1 * cosU2;
double sinU1sinU2 = sinU1 * sinU2;
double sigma = 0.0;
double deltaSigma = 0.0;
double cosSqAlpha = 0.0;
double cos2SM = 0.0;
double cosSigma = 0.0;
double sinSigma = 0.0;
double cosLambda = 0.0;
double sinLambda = 0.0;
double lambda = L; // initial guess
for (int iter = 0; iter < MAXITERS; iter++) {
double lambdaOrig = lambda;
cosLambda = Math.cos(lambda);
sinLambda = Math.sin(lambda);
double t1 = cosU2 * sinLambda;
double t2 = cosU1 * sinU2 - sinU1 * cosU2 * cosLambda;
double sinSqSigma = t1 * t1 + t2 * t2; // (14)
sinSigma = Math.sqrt(sinSqSigma);
cosSigma = sinU1sinU2 + cosU1cosU2 * cosLambda; // (15)
sigma = Math.atan2(sinSigma, cosSigma); // (16)
double sinAlpha = (sinSigma == 0) ? 0.0 :
cosU1cosU2 * sinLambda / sinSigma; // (17)
cosSqAlpha = 1.0 - sinAlpha * sinAlpha;
cos2SM = (cosSqAlpha == 0) ? 0.0 :
cosSigma - 2.0 * sinU1sinU2 / cosSqAlpha; // (18)
double uSquared = cosSqAlpha * aSqMinusBSqOverBSq; // defn
A = 1 + (uSquared / 16384.0) * // (3)
(4096.0 + uSquared *
(-768 + uSquared * (320.0 - 175.0 * uSquared)));
double B = (uSquared / 1024.0) * // (4)
(256.0 + uSquared *
(-128.0 + uSquared * (74.0 - 47.0 * uSquared)));
double C = (f / 16.0) *
cosSqAlpha *
(4.0 + f * (4.0 - 3.0 * cosSqAlpha)); // (10)
double cos2SMSq = cos2SM * cos2SM;
deltaSigma = B * sinSigma * // (6)
(cos2SM + (B / 4.0) *
(cosSigma * (-1.0 + 2.0 * cos2SMSq) -
(B / 6.0) * cos2SM *
(-3.0 + 4.0 * sinSigma * sinSigma) *
(-3.0 + 4.0 * cos2SMSq)));
lambda = L +
(1.0 - C) * f * sinAlpha *
(sigma + C * sinSigma *
(cos2SM + C * cosSigma *
(-1.0 + 2.0 * cos2SM * cos2SM))); // (11)
double delta = (lambda - lambdaOrig) / lambda;
if (Math.abs(delta) < 1.0e-12) {
break;
}
}
return b * A * (sigma - deltaSigma);
}

还有一个计算两个坐标间距离比较简单的方法，这里思考的地球是一个圆球。
public static double getDistance(double lat1,double longt1 , double lat2,double longt2
) {
double PI = 3.14159265358979323; // 圆周率
double R = 6371229; // 地球的半径
double x, y, distance;
x = (longt2 - longt1) * PI * R
* Math.cos(((lat1 + lat2) / 2) * PI / 180) / 180;
y = (lat2 - lat1) * PI * R / 180;
distance = Math.hypot(x, y);
return distance;
}