我正在尝试实现一个Android应用程序来连接到开源软件Motion。目标是能够检查应用程序的状态并获取最后拍摄的图像。
我不是用Java编程，我的背景主要是C和Python。我对理解Android的UI部分没有任何实际问题，但我发现使用任何类型的字节缓冲区都非常痛苦。 Motion软件具有非常简单的HTTP API。在Java中打开URL连接很容易。默认页面的响应如下所示
Motion 3.2.12 Running [4] Threads
0
1
2
3
就我的目的而言，应用程序需要做的第一件事是解析线程数。在某些时候，我也可以从第一行检索版本号，但这在目前并不重要。
这是我的代码
package com.hydrogen18.motionsurveillanceviewer;
import java.io.BufferedInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.net.HttpURLConnection;
import java.net.URL;
import java.util.List;
public class MotionHttpApi {
String host;
int port = 80;
boolean secure = false;
int numberOfThreads = -1;
String getBaseUrl()
{
StringBuilder sb = new StringBuilder();
sb.append(secure ? "https://" : "http://");
sb.append(host);
sb.append(':');
sb.append(port);
return sb.toString();
}
public int getNumberOfCameras() throws IOException
{
if(numberOfThreads == -1)
{
retrieveSplash();
}
if(numberOfThreads == 1)
{
return 1;
}
return numberOfThreads - 1;
}
void retrieveSplash () throws IOException
{
URL url = new URL(getBaseUrl());
HttpURLConnection conn = (HttpURLConnection)url.openConnection();
if(conn.getResponseCode()!=HttpURLConnection.HTTP_OK)
{
throw new IOException("Got response code" + conn.getResponseCode());
}
try{
Byte[] buffer = new Byte[512];
byte[] sbuf = new byte[128];
int offset = 0;
InputStream in = new BufferedInputStream(conn.getInputStream());
boolean foundInfoString= false;
while( ! foundInfoString)
{
//Check to make sure we have not run out of space
if(offset == buffer.length)
{
throw new IOException("Response too large");
}
//Read into the smaller buffer since InputStream
//can't write to a Byte[]
final int result = in.read(sbuf,0,sbuf.length);
//Copy the data into the larger buffer
for(int i = 0; i < result;++i)
{
buffer[offset+i] = sbuf[i];
}
//Add to the offset
offset+=result;
//Wrap the array as a list
List<Byte> list = java.util.Arrays.asList(buffer);
//Find newline character
final int index = list.indexOf((byte) '\n');
//If the newline is present, extract the number of threads
if (index != -1)
{
//Find the number of threads
//Thread number is in the first lin like "[X]"
final int start = list.indexOf((byte)'[');
final int end = list.indexOf((byte)']');
//Sanity check the bounds
if(! (end > start))
{
throw new IOException("Couldn't locate number of threads");
}
//Create a string from the Byte[] array subset
StringBuilder sb = new StringBuilder();
for(int i = start+1; i != end; ++i)
{
final char c = (char) buffer[i].byteValue();
sb.append(c);
}
String numThreadsStr = sb.toString();
//Try and parse the string into a number
try
{
this.numberOfThreads = Integer.valueOf(numThreadsStr);
}catch(NumberFormatException e)
{
throw new IOException("Number of threads is NaN",e);
}
//No more values to extract
foundInfoString = true;
}
//If the InputStream got EOF and the into string has not been found
//Then an error has occurred.
if(result == -1 && ! foundInfoString )
{
throw new IOException("Never got info string");
}
}
}finally
{
//Close the connection
conn.disconnect();
}
}
public MotionHttpApi(String host,int port)
{
this.host = host;
this.port = port;
}
}
当您致电getNumberOfCameras()时，代码工作正常。但我认为我不能真正理解java的东西，因为retrieveSplash方法太复杂了。我可以在10行左右的C行或1行Python中做同样的事情。当然，必须有一种更健全的方法来操纵java中的字节？
我认为有一些样式问题，比如每当整数无法解析时我可能不应该抛出IOException。但这是一个单独的问题。
答案 0 :(得分：1)
按照Gautam Tandon的建议阅读第一行，然后使用正则表达式。 然后，您可以检查正则表达式是否匹配，甚至可以轻松提取数字。
正则表达式'可以在http://txt2re.com创建。我已经为你做过了。 该页面甚至可以创建Java，Pyhton，C等文件供您使用。
// URL that generated this code:
// http://txt2re.com/index-java.php3?s=Motion%203.2.12%20Running%20[4]%20Threads&-7&-19&-5&-20&-1&2&-22&-21&-62&-63&15
import java.util.regex.*;
class Main
{
public static void main(String[] args)
{
String txt="Motion 3.2.12 Running [4] Threads";
String re1="(Motion)"; // Word 1
String re2="( )"; // White Space 1
String re3="(3\\.2\\.12)"; // MMDDYY 1
String re4="( )"; // White Space 2
String re5="(Running)"; // Word 2
String re6="( )"; // White Space 3
String re7="(\\[)"; // Any Single Character 1
String re8="(\\d+)"; // Integer Number 1
String re9="(\\])"; // Any Single Character 2
String re10="( )"; // White Space 4
String re11="((?:[a-z][a-z]+))"; // Word 3
Pattern p = Pattern.compile(re1+re2+re3+re4+re5+re6+re7+re8+re9+re10+re11,Pattern.CASE_INSENSITIVE | Pattern.DOTALL);
Matcher m = p.matcher(txt);
if (m.find())
{
String word1=m.group(1);
String ws1=m.group(2);
String mmddyy1=m.group(3);
String ws2=m.group(4);
String word2=m.group(5);
String ws3=m.group(6);
String c1=m.group(7);
String int1=m.group(8);
String c2=m.group(9);
String ws4=m.group(10);
String word3=m.group(11);
System.out.print("("+word1.toString()+")"+"("+ws1.toString()+")"+"("+mmddyy1.toString()+")"+"("+ws2.toString()+")"+"("+word2.toString()+")"+"("+ws3.toString()+")"+"("+c1.toString()+")"+"("+int1.toString()+")"+"("+c2.toString()+")"+"("+ws4.toString()+")"+"("+word3.toString()+")"+"\n");
}
}
}
//-----
// This code is for use with Sun's Java VM - see http://java.sun.com/ for downloads.
//
// Paste the code into a new java application or a file called 'Main.java'
//
// Compile and run in Unix using:
// # javac Main.java
// # java Main
//
String int1=m.group(8);为您提供所需的整数。当然，您可以简化上面的代码。这是现在详细说明的方式。
答案 1 :(得分：0)
您可以使用BufferedReader大大简化retrieveSplash方法。这是一个更简单的函数版本：
void retrieveSplash_simpler() throws IOException {
URL url = new URL(getBaseUrl());
HttpURLConnection conn = (HttpURLConnection)url.openConnection();
// open the connection
conn.connect();
// create a buffered reader to read the input stream line by line
BufferedReader reader = new BufferedReader(new InputStreamReader(conn.getInputStream()));
// find number of threads
String firstLine = reader.readLine();
int x = firstLine.indexOf("[");
int y = firstLine.indexOf("]");
if (x > 0 && y > 0 && x < y) {
try {
numberOfThreads = Integer.parseInt(firstLine.substring(x+1, y));
} catch (NumberFormatException nfe) {
// disconnect and throw exception
conn.disconnect();
throw new IOException("Couldn't locate number of threads");
}
} else {
// disconnect and throw exception
conn.disconnect();
throw new IOException("Couldn't locate number of threads");
}
// disconnect
conn.disconnect();
}
我通过在适当的位置使用try / catch / finally块来进一步清理上面的方法，这样我就不必复制“conn.disconnect（）”了。但是我没有这样做，以保持简单（尝试/捕获/最终有时变得棘手......）。