Files of zip type

Question

Hi friends,
sender Legacy system is sending files in .Zip format.
how the adapter is going to process those files?
what configurations r to be needed in adapter configuration?
& also for processing of .pdf files what configurations should be done?
thanks in advance.
shiva.

Answer 1

Hi Shiva,
the adapter can process the file. At mapping time you need to dezip the msg. An example in java:
import java.io.*;
import java.util.Map;
import java.util.zip.*;
import com.sap.aii.mapping.api.StreamTransformation;
import com.sap.aii.mapping.api.StreamTransformationException;
public class DecompressMapping implements StreamTransformation {
     /* (non-Javadoc)
      * @see com.sap.aii.mapping.api.StreamTransformation#setParameter(java.util.Map)
     public void setParameter(Map arg0) {
          // TODO Auto-generated method stub
     /* (non-Javadoc)
      * @see com.sap.aii.mapping.api.StreamTransformation#execute(java.io.InputStream, java.io.OutputStream)
     public void execute(InputStream arg0, OutputStream arg1)
          throws StreamTransformationException {
               try {
          int len;
          byte[] buf = new byte[8192];
          boolean first_iter = true;
          ByteArrayOutputStream baos = new ByteArrayOutputStream();
          while((len = arg0.read(buf) ) != -1)
               if (first_iter)
                 if ((buf[0] != 'P') || (buf[1] != 'K'))
                      arg1.write(buf, 0, len);
                    while((len = arg0.read(buf) ) != -1)
                         arg1.write(buf, 0, len);                           
                      return;     
                 else
                      first_iter = false;
               baos.write(buf, 0, len);
          byte[] in_data = baos.toByteArray();
          ZipInputStream zis = new ZipInputStream(new ByteArrayInputStream(in_data));
               ZipEntry ze = zis.getNextEntry();
               if (ze != null)
                    while((len = zis.read(buf) ) != -1)
                         arg1.write(buf, 0, len);
          } catch (IOException e) {
               e.printStackTrace();
Regards,
Udo