以下是計算一個音視頻文件,播放所需要帶寬的程序:
public int getBitrate(String path) {
File f = new File(path);
if (f.exists()) {
int mpegPackets = 0; int getpcr = 0; long pcr0 = 0; long pcr1; int pid = 0; try { FileInputStream in = new FileInputStream(f); while (true) { byte[] b = new byte[188]; int ret = in.read(b, 0, 188); if (ret == -1) break; mpegPackets++; if (b[0] != 0x47) { return -1;
} System.out.println(""); System.out.println(mpegPackets + " : "); for(int k=0; k<ret; k++){ System.out.print(Integer.toHexString(b[k]&0xFF) + " "); if(k%10 ==0){ System.out.println(""); } } if ((((b[3] >> 4) & 0x3) & 0x02) != 0) {//ptati判斷adaptation_field_control =='10' || ='11') //確定後面爲adaptation_field if (b[4] == 0) //adaptation_field_length = 0 則退出循環 continue; if ((b[5] & 0x10) != 0) { //對應 PCG_flag = 0 則表示此packet 不包含pcr 信息 if (getpcr == 10) { //當找到第11個包含pcr 信息的packet pid = ((b[1] & 0xff & 0x1f) << 8) | (b[2] & 0xff); int tmp; pcr0 = ((((b[6] & 0xff) << 24) + ((b[7] & 0xff) << 16) + ((b[8] & 0xff) << 8) + (b[9] & 0xff))); pcr0 <<= 1; if ((b[10] & 0xff & 0x80) != 0) pcr0 |= 1; //pcr0 爲 program_clock_reference_base (33bit) pcr0 *= 300; //OPCR(i) = OPCR_base(i)*300 + OPCR_ext(i) tmp = ((b[10] & 0xff & 1) << 8) + b[11]; //tmp 爲 program_clock_reference_extension (9bit) pcr0 += tmp; mpegPackets = 0; } if ((getpcr > 20) && pid == (((b[1] & 0xff & 0x1f) << 8) | (b[2] & 0xff))) { //從第21個包含pcr信息的packet的包並且pid與"第11個包含pcr 信息的packet的pid"相同 int tmp; pcr1 = ((((b[6] & 0xff) << 24) + ((b[7] & 0xff) << 16) + ((b[8] & 0xff) << 8) + (b[9] & 0xff))); pcr1 <<= 1; if ((b[10] & 0xff & 0x80) != 0) pcr1 |= 1; pcr1 *= 300; tmp = ((b[10] & 0xff & 1) << 8) + b[11] & 0xff; pcr1 += tmp; long PCRdelta = pcr1 - pcr0; double bitRate = ((40608 * mpegPackets * 1.0) / PCRdelta) * 1000000; //40608 = 188 * 216 return (int) bitRate; } getpcr++; } } } } catch (FileNotFoundException e) { e.printStackTrace(); } catch (IOException e) { e.printStackTrace(); } } else return 0; return 0;
}以上是根據pcr來計算播放所需要帶寬