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I'm stored the content of the image as a blob in the database , the content type in the photo entity is a byte array (byte[]) で動的な写真を表示したい。私は2つの異なる方法を試しましたが、どちらも役に立たなかった.

1)

 public StreamedContent getImage() throws IOException {
  byte[] img=photo.getContent();
 InputStream in = new ByteArrayInputStream(img);
 StreamedContent  image = new DefaultStreamedContent(in,"image/jpeg");
  return image;
}

2)

public StreamedContent getImage() throws IOException {
  byte[] img=photo.getContent();
 InputStream in = new ByteArrayInputStream(img);
 BufferedImage bufferedImg = ImageIO.read(in);
 ByteArrayOutputStream os = new ByteArrayOutputStream();
 ImageIO.write(bufferedImg, "jpeg", os);

 StreamedContent   image = new DefaultStreamedContent(new ByteArrayInputStream(os.toByteArray()),"image/jpeg");

  return image;
}

およびビューページで：

  <p:graphicImage value="#{controller.image}"/>

だから、誰かがそれを機能させるのを手伝ってくれませんか!!

4

1 に答える 1

2

まず、データベースから BLOB を取得する必要があります。

public StreamedContent getImage() throws SQLException {
    Blob imgBlob;
    try
    {
     //Select the blob from your database here. Use a PreparedStatement (I'll call it stmt here) for this.
     ...
     ResultSet res = stmt.executeQuery();
     res.next();
     imgBlob = res.getBlob("columnName");
    }
    catch(SQLException e){ throw e; }
    byte[] img=imgBlob.getBytes(1, imgBlob.length()); //Get the blob as a byte array
    InputStream in = new ByteArrayInputStream(img);
    StreamedContent  image = new DefaultStreamedContent(in,"image/jpeg");
    return image;
}

Blob クラスの詳細。

于 2012-10-08T00:06:27.090 に答える