从服务器上解压缩文件，HTTP [英] Unzip file from server http

问题描述

我要解压缩从服务器接收到一个文件有问题。我发送一个请求，和服务器显示我code字符的文本：•

有了这个$ C C I接收来自服务器的zip，但我需要这个文本文件解压的，并显示在EdText。

 公共无效postLoginData（）{    //创建一个新的HttpClient和邮政头
    HttpClient的HttpClient的=新DefaultHttpClient（）;
    HttpPost httppost =新HttpPost（
            HTTP：//本地主机：80 / MOBILE / MBSERVER.V25？）;
    尝试{
        字符串VENDED_VEN =2，EM preS_CFG =1，VERSAO_CFG =100，功能=GetMARCAS，REQUERY = NULL;        COMANDO =（EditText上）findViewById（R.id.comando）;
        //字符串PLS = comando.getText（）的toString（）。        清单＆LT;＆的NameValuePair GT;值=新的ArrayList＆LT;＆的NameValuePair GT;（6）;        值=新的ArrayList＆LT;＆的NameValuePair GT;（）;
        values​​.add（新BasicNameValuePair（EM preS，EM preS_CFG））;
        values​​.add（新BasicNameValuePair（USUARI，VENDED_VEN））;
        values​​.add（新BasicNameValuePair（VERSAO，VERSAO_CFG））;
        values​​.add（新BasicNameValuePair（ORIGEM，AD））;
        values​​.add（新BasicNameValuePair（FUNCAO功能））;
        values​​.add（新BasicNameValuePair（RQUERY，REQUERY））;
        httppost.setEntity（新UrlEn codedFormEntity（值））;
        //执行HTTP POST请求
        Log.w（日志，执行HTTP POST请求）;
        HTT presponse响应= httpclient.execute（httppost）;
        Log.w（日志，值：+值）;        字符串str = inputStreamToString（response.getEntity（）的getContent（））
                的ToString（）;        Log.w（日志，STR）;
        如果（str.toString（）。equalsIgnoreCase（真））{
            Log.w（日志，TRUE）;
            result.setText（登录成功）;        }其他{
            Log.w（LOG，假）;
            result.setText（STR）;
        }
    }赶上（ClientProtocolException E）{
        e.printStackTrace（）;
    }赶上（IOException异常五）{
        e.printStackTrace（）;
    }
}私人的StringBuilder inputStreamToString（InputStream为）{
    串线=;
    StringBuilder的总=新的StringBuilder（）;    //环绕式InputStream的一个BufferedReader
    RD的BufferedReader =新的BufferedReader（新的InputStreamReader（是））;
    //读取响应，直到结束
    尝试{
        而（（行= rd.readLine（））！= NULL）{
            total.append（线）;
        }
    }赶上（IOException异常五）{
        e.printStackTrace（）;
    }
    //返回满弦
    总回报;
}@覆盖
公共无效的onClick（查看视图）{
    如果（查看== OK）{
        postLoginData（）;
    }
}
}
 

解决方案

如果这是一个标准的zip文件，可以使用 java.util.zip 包。

下面是解压即中有一个文件夹归档，并将其写入到文件中的一个例子。

 的FileInputStream zipInputStream =新的FileInputStream（新文件（cacheDir，zipFileName））;
FileOutputStream中datOutputStream =新的FileOutputStream（新文件（cacheDir，datFileName））;//解压缩和处理ZIP文件
ZipInputStream ZIS =新ZipInputStream（zipInputStream）;
ZipEntry的泽= NULL;
//通过循环存档
而（（ZE = zis.getNextEntry（））！= NULL）{
    如果（ze.getName（）。的toString（）。等于（MyFolder中/ MYFILE.DAT））{//改变这一切你的文件夹/文件被命名为存档内
        而（量（bufferLength = zis.read（缓冲液，0，1023））！=  -  1）{
            datOutputStream.write（缓冲液，0，BufferLength中）;
        }
     }
     zis.closeEntry（）;
}
 

I have a problem to unzip a file received from server. I send a request, and server show me a text with code characters :S

With this code i receive a zip from server, but i need this text file unziped, and show in a EdText.

public void postLoginData() {

    // Create a new HttpClient and Post Header
    HttpClient httpclient = new DefaultHttpClient();


    HttpPost httppost = new HttpPost(
            "http://localhost:80/MOBILE/MBSERVER.V25?");
    try {
        String VENDED_VEN = "2", EMPRES_CFG = "1", VERSAO_CFG = "100", function = "GetMARCAS", REQUERY = null;

        comando = (EditText) findViewById(R.id.comando);
        // String PLS = comando.getText().toString();

        List<NameValuePair> values = new ArrayList<NameValuePair>(6);

        values = new ArrayList<NameValuePair>();
        values.add(new BasicNameValuePair("EMPRES", EMPRES_CFG));
        values.add(new BasicNameValuePair("USUARI", VENDED_VEN));
        values.add(new BasicNameValuePair("VERSAO", VERSAO_CFG));
        values.add(new BasicNameValuePair("ORIGEM", "AD"));
        values.add(new BasicNameValuePair("FUNCAO", function));
        values.add(new BasicNameValuePair("RQUERY", REQUERY));


        httppost.setEntity(new UrlEncodedFormEntity(values));
        // Execute HTTP Post Request
        Log.w("LOG", "Execute HTTP Post Request");
        HttpResponse response = httpclient.execute(httppost);
        Log.w("LOG", "VALUES:"+values);

        String str = inputStreamToString(response.getEntity().getContent())
                .toString();



        Log.w("LOG", str);
        if (str.toString().equalsIgnoreCase("true")) {
            Log.w("LOG", "TRUE");
            result.setText("Login successful");

        } else {
            Log.w("LOG", "FALSE");
            result.setText(str);
        }
    } catch (ClientProtocolException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }
}

private StringBuilder inputStreamToString(InputStream is) {
    String line = "";
    StringBuilder total = new StringBuilder();

    // Wrap a BufferedReader around the InputStream
    BufferedReader rd = new BufferedReader(new InputStreamReader(is));
    // Read response until the end
    try {
        while ((line = rd.readLine()) != null) {
            total.append(line);
        }
    } catch (IOException e) {
        e.printStackTrace();
    }
    // Return full string
    return total;
}

@Override
public void onClick(View view) {
    if (view == ok) {
        postLoginData();
    }
}
}    

解决方案

If it's a standard zip file, you can use the java.util.zip package.

Here's an example of unzipping an archive that has a folder in it, and writing it to a file.

FileInputStream zipInputStream = new FileInputStream(new File(cacheDir, zipFileName));
FileOutputStream datOutputStream = new FileOutputStream(new File(cacheDir, datFileName));

// unzip and process ZIP file
ZipInputStream zis = new ZipInputStream(zipInputStream);
ZipEntry ze = null;
// loop through archive
while ((ze = zis.getNextEntry()) != null) {
    if (ze.getName().toString().equals("myfolder/myfile.dat")) { // change this to whatever your folder/file is named inside the archive
        while ((bufferLength = zis.read(buffer, 0, 1023)) != -1) {
            datOutputStream.write(buffer, 0, bufferLength);
        }
     }
     zis.closeEntry();
}

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