在Groovy中，如何正确从httpservletrequest获取文件

Question

我正在用Groovy脚本编写一个REST API，该API将从客户端接收文件上传。 REST API将通过HttpservletRequest接收文件。 我试图通过获取其输入流来从httpservletrequest获取文件，然后将其转换为文件以保存到适当的文件夹。 我的代码如下：

RestApiResponse doHandle(HttpServletRequest request, RestApiResponseBuilder apiResponseBuilder, RestAPIContext context) {
    InputStream inputStream = request.getInputStream()              
    def file = new File(tempFolder + "//" + fileName)
    
    FileOutputStream outputStream = null
    try
    {
        outputStream = new FileOutputStream(file, false)
        int read;
        byte[] bytes = new byte[DEFAULT_BUFFER_SIZE];
        while ((read = inputStream.read(bytes)) != -1) {
            outputStream.write(bytes, 0, read);
        }
    }
    finally {
        if (outputStream != null) {
            outputStream.close();
        }
    }
    inputStream.close();

    // the rest of the code
}

创建文件，但所有文件都是损坏的。 当我尝试用记事本打开它们时，一开始，所有这些都与以下相似：

-----------------------------134303111730200325402357640857
Content-Disposition: form-data; name="pbUpload1"; filename="Book1.xlsx"
Content-Type: application/vnd.openxmlformats-officedocument.spreadsheetml.sheet

我做错了吗？如何正确获取文件？

Answer 1

找到了使用多植物的解决方案

import org.apache.commons.fileupload.MultipartStream
import org.apache.commons.io.FileUtils

InputStream inputStream = request.getInputStream()
//file << inputStream;  
String fileName = "";
final String CD = "Content-Disposition: "
MultipartStream multipartStream =  new MultipartStream(inputStream, boundary);

//Block below line because it always return false for some reason
// but should be used as stated in document
//boolean nextPart = multipartStream.skipPreamble();

//Block below line as in my case, the part I need is at the first part
// or maybe I should use it and break after successfully get the file name
//while(nextPart) {
String[] headers = multipartStream.readHeaders().split("\\r\\n")
ContentDisposition cd = null
for (String h in headers) {
    if (h.startsWith(CD)) {
        cd = new ContentDisposition(h.substring(CD.length()));
        fileName = cd.getParameter("filename");         }
}             
def file = new File(tempFolder + "//" + fileName)
ByteArrayOutputStream output = new ByteArrayOutputStream(1024)
try
{
    multipartStream.readBodyData(output)
    FileUtils.writeByteArrayToFile(file, output.toByteArray());
}
finally {
    if (output != null) {
        output.flush();
        output.close(); 
    }
}
inputStream.close();