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java - 如何在Java Rest API中获取InputStream文件的真实路径

转载 作者:行者123 更新时间:2023-12-02 00:03:14 24 4
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我有一个 Java Rest API Post 方法,它发送图像 (InputStream) 作为参数,并且我必须将其保存在 Oracle 中的 blob 列中。

我需要获取此InputStream的完整路径(真实路径)以将此图像保存在数据库中。我的代码如下。

@POST
@Path("/upload")
@Consumes(MediaType.MULTIPART_FORM_DATA)
public Response uploadFile(
@FormDataParam("file") InputStream uploadedInputStream,
@FormDataParam("file") FormDataContentDisposition fileDetail) {
String UPLOAD_FOLDER = "c:/uploadedFiles/"; // my rest api does not have this file, how to get at runtime?
String uploadedFileLocation = UPLOAD_FOLDER + fileDetail.getFileName(); // this line is ok

我想做这样的事情:

String UPLOAD_FOLDER = uploadedInputStream.getRealPathName();

String UPLOAD_FOLDER = fileDetail.getRealPathName();

最佳答案

我通过将输入流转换为字节数组解决了这个问题。我将 byte[] 转发给数据库持久化方法。我的代码如下:

public byte[] toByteArray(InputStream is) throws IOException{
ByteArrayOutputStream baus = new ByteArrayOutputStream();
byte[] buffer = new byte[1024];
int len;
while((len= is.read(buffer)) != -1){
baus.write(buffer, 0, len);
}
return baus.toByteArray();
}

@POST
@Path("/upload")
@Consumes(MediaType.MULTIPART_FORM_DATA)
public Response uploadFile(
@FormDataParam("file") InputStream uploadedInputStream,
@FormDataParam("file") FormDataContentDisposition fileDetail) {
...
byte[] b = toByteArray(uploadedInputStream);
business.saveUploadedFileInDatabase(b);
...
}

public boolean uploadFile(byte[] b) throws SQLException, IOException{
...
PreparedStatement ps = conn.prepareStatement("INSERT INTO TABLE_IMAGE_TEST (CODE, IMAGE) VALUES (?, ?)");
pstmt.setLong(1, 1L);
pstmt.setBytes(2, b);
pstmt.execute();
...
}

关于java - 如何在Java Rest API中获取InputStream文件的真实路径,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/58170204/

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