将大文件(大于1GB)上传到grails3

问题描述

将大文件(> 1GB)上载到grails，我只需要通过流访问它，而无需将整个文件保存到磁盘或RAM.但是，如何访问上传流?我尝试了一个拦截器:

Uploading a large file (>1GB) to a grails, I only need to access it via stream, no need to save the entire file to disk or RAM. However how can I access the upload stream? I tried with an Interceptor:

class UploadLargeFileInterceptor {

int order = HIGHEST_PRECEDENCE

UploadLargeFileInterceptor() {
    match(controller:"apiv2", action:"uploadLarge")
}

boolean before() {

    log.error('before')
    log.error(request.getInputStream().text.length() as String)
    true
}

boolean after() {

    log.error('after')
    true
}

void afterView() {
    // no-op
}
}

，但是流长度始终为0，并且在控制器中有一个多部分文件，我试图避免该文件，因为它将存储整个文件.有什么想法吗?

but stream length is always 0 and in the controller there is a multipart file which I am trying to avoid because it will store the whole file. Any ideas?

推荐答案

无需拦截器.您可以在控制器内部访问用户上传的文件流，如下所示:

No need for an interceptor. You can access user uploaded file stream like this inside controller:

MultipartFile uploadedFile = params.filename as MultipartFile
println "file original name" + uploadedFile.originalFilename
println "file content type" + uploadedFile.contentType
println "file size" + uploadedFile.size //consider as stream length
InputStream inputStream = uploadedFile.getInputStream() //access the input stream

还要确保在 application.groovy 中允许这样的最大大小:

Also make sure to allow max size in application.groovy like this:

grails.controllers.upload.maxFileSize = 1572864000 //1500MB (X MB * 1024 * 1024)
grails.controllers.upload.maxRequestSize = 1572864000 //1500MB

或者，如果您正在像 nginx 这样的网络服务器上落后，请调整该请求&超时配置也是如此.

Or if you are running behind on webserver like nginx, tune that request & timeout config also.

使用上面的代码和 -Xms512m -Xmx1G 堆大小，我上传了1.25Gb文件，没有问题.如果要上传到文件存储/云存储，请确保lib使用流:

With the above code and -Xms512m -Xmx1G heap size I uploaded 1.25Gb file with no issue. If you are uploading to file storage / cloud storage, make sure that lib uses stream like:

文件: org.apache.commons.io.FileUtils.copyToFile(inputStream，new File('< file path>'))Azure: storageBlob.upload(inputStream，length)

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