12

我有 PostgreSQL 数据库，并且有数据类型为“bytea”的列“image”。我无法修改列或数据库配置。JPA 注释 POJO 包含跟随映射

@Column(name="image")
private byte[] image;

返回的数据格式如下（这只是一个示例）

WF5ClN6RlpLZ0hJTUdNQ1FJWmkwcFVGSUdNQ0lDWUE5TUEvanRFeElwK2x0M2tBQUFBQVNVVk9SSzVDWUlJPQo=

当我将此数据写入文件 (.jpeg) 时，照片查看器显示“这是已损坏的文件”。我也知道实际的图像字节数据看起来与上面的示例不同。我阅读了一些博客，其中提到 PostgreSQL 将十六进制转换应用于 bytea 数据。如何在有或没有 JPA 的情况下将其恢复为原始数据？

数据库 - PostgresSQL 版本 9.5.1

司机

<dependency>
    <groupId>org.postgresql</groupId>
    <artifactId>postgresql</artifactId>
    <version>9.4-1205-jdbc41</version>
</dependency>
4

5 回答 5

10

尝试用注释你的实体@Lob

@Lob
@Column(name="image")
private byte[] image;

如果您使用的是休眠实现，您也可以@Type(type="org.hibernate.type.BinaryType")在列中添加。

@Lob
@Column(name="image")
@Type(type="org.hibernate.type.BinaryType")
private byte[] image;
于 2016-02-19T14:36:48.920 回答
6

图像实体

package com.example;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;

@Entity
public class ImageEntity {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    @Column(name="image")
    private byte[] image;

    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

    public byte[] getImage() {
        return image;
    }

    public void setImage(byte[] image) {
        this.image = image;
    }
}

图像存储库

package com.example;

import org.springframework.data.jpa.repository.JpaRepository;
import org.springframework.stereotype.Repository;

@Repository
public interface ImageRepository extends JpaRepository<ImageEntity, Long> {
}

测试

package com.example;

import java.io.ByteArrayOutputStream;
import java.io.IOException;
import java.io.InputStream;

import javax.annotation.Resource;

import org.junit.Test;
import org.junit.runner.RunWith;
import org.springframework.boot.test.SpringApplicationConfiguration;
import org.springframework.test.context.junit4.SpringJUnit4ClassRunner;

import junit.framework.TestCase;

@RunWith(SpringJUnit4ClassRunner.class)
@SpringApplicationConfiguration(classes = TestApplication.class)
public class ImageDaoTest {

    @Resource
    private ImageRepository imageRepository;

    @Test
    public void testImage() throws IOException {

        // Read an image from disk. Assume test.png exists
        ByteArrayOutputStream out = new ByteArrayOutputStream();

        try (InputStream in = getClass().getResourceAsStream("test.png")) {
            int length;
            byte[] buffer = new byte[1024];
            while ((length = in.read(buffer)) != -1) out.write(buffer, 0, length);
        }

        byte[] image = out.toByteArray();

        // Store image to DB
        ImageEntity imageEntiry = new ImageEntity();
        imageEntiry.setImage(image);
        long imageEntiryId = imageRepository.save(imageEntiry).getId();

        // Retrieve image from DB
        ImageEntity resultImageEntiry = imageRepository.findOne(imageEntiryId);
        byte[] resultImage = resultImageEntiry.getImage();

        // Compare retrieved image with source image by byte to byte comparison
        for (int i = 0; i < resultImage.length; i++) {
            TestCase.assertEquals(image[i], resultImage[i]);
        }

    }

}

它适用于带有 9.4.1207.jre7 jdbc 驱动程序的 Postgres 9.5.0-1。

于 2016-02-19T15:25:42.997 回答
5

返回的数据看起来好像是 base64 编码的。在写入文件之前，您必须将其解码回二进制数据。

有关解码的更多信息，请查看此处

于 2016-02-19T13:02:21.240 回答
2

我正在添加可能对其他人有用的完整代码（跳过 try/catch），

String base64EncryptedImage = new String(image);
decoded = org.apache.commons.codec.binary.Base64.decodeBase64(base64EncryptedImage);
ImageOutputStream out = new FileImageOutputStream(new File("D://abc.png"));
out.write(decoded);
out.close();
于 2016-02-22T08:54:06.040 回答
0

插入图像，您将使用：

  //Get the Large Object Manager to perform operations with
    LargeObjectManager lobj = ((org.postgresql.PGConnection)conn).getLargeObjectAPI();

    // Create a new large object
    int oid = lobj.create(LargeObjectManager.READ | LargeObjectManager.WRITE);

    // Open the large object for writing
    LargeObject obj = lobj.open(oid, LargeObjectManager.WRITE);

    // Now open the file
    File file = new File("myimage.gif");
    FileInputStream fis = new FileInputStream(file);

    // Copy the data from the file to the large object
    byte buf[] = new byte[2048];
    int s, tl = 0;
    while ((s = fis.read(buf, 0, 2048)) > 0) {
        obj.write(buf, 0, s);
        tl += s;
    }

    // Close the large object
    obj.close();

// Now insert the row into imageslo
PreparedStatement ps = conn.prepareStatement("INSERT INTO imageslo VALUES (?, ?)");
ps.setString(1, file.getName());
ps.setInt(2, oid);
ps.executeUpdate();
ps.close();
fis.close();

// Finally, commit the transaction.
conn.commit();

从大对象中检索图像：

// All LargeObject API calls must be within a transaction block
conn.setAutoCommit(false);

// Get the Large Object Manager to perform operations with
LargeObjectManager lobj = ((org.postgresql.PGConnection)conn).getLargeObjectAPI();

PreparedStatement ps = conn.prepareStatement("SELECT imgoid FROM imageslo WHERE imgname = ?");
ps.setString(1, "myimage.gif");
ResultSet rs = ps.executeQuery();
while (rs.next()) {
    // Open the large object for reading
    int oid = rs.getInt(1);
    LargeObject obj = lobj.open(oid, LargeObjectManager.READ);

    // Read the data
    byte buf[] = new byte[obj.size()];
    obj.read(buf, 0, obj.size());
    // Do something with the data read here

    // Close the object
    obj.close();
}
rs.close();
ps.close();

// Finally, commit the transaction.
conn.commit();
于 2016-02-19T12:49:46.857 回答