CODEWITHSUNDEEP
javaamazon-web-servicesamazon-s3aws-lambda
hermt2
hermt2

Reputation: 874

Is it possible to create a file in my S3 lambda function?

I am creating a java function for AWS Lambda that brings in a file from AWS S3 like this:

InputStream videoObjectStream = awsS3Video.getObjectContent();

I am also utilizing FFmpegFrameGrabber, which requires that I specify a file path whenever I create a new frameGrabber, i.e.: FFmpegFrameGrabber frameGrabber = new FFmpegFrameGrabber(filePath)

I am trying to convert the InputStream into a temporary file in my Lambda function, but it is not allowing me to create a file. Here is my code to convert the videoObjectStream into a file:

byte[] inputBuffer = null;

        try {
            inputBuffer = IOUtils.toByteArray(videoObjectStream);
        } catch (IOException e1) {
            // TODO Auto-generated catch block
            e1.printStackTrace();
        }

        System.out.println("The length of the byte array is " + inputBuffer.length);


        try {
            FileOutputStream videoOS = new FileOutputStream(videoDetails.get("videoFileKey"), false);
            videoOS.write(inputBuffer);
            videoOS.close();
        } catch (FileNotFoundException e1) {
            // TODO Auto-generated catch block
            e1.printStackTrace();
        } catch (IOException ex) {
            ex.printStackTrace();
        }

        File tempVideoFile = new File(videoDetails.get("videoFileKey"));

        if (tempVideoFile.exists()) {
            System.out.println("The file exists");
        } else {
            System.out.println("The file does not exist");
        }

Then, I get the following stack trace, saying that this is a read-only file system:

java.io.FileNotFoundException: currentPath1490660005410.mp4 (Read-only file system)
    at java.io.FileOutputStream.open0(Native Method)
    at java.io.FileOutputStream.open(FileOutputStream.java:270)
    at java.io.FileOutputStream.<init>(FileOutputStream.java:213)
    at java.io.FileOutputStream.<init>(FileOutputStream.java:133)
    at com.amazonaws.lambda.LambdaFunctionHandler.convertVideo(LambdaFunctionHandler.java:67)
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
    at java.lang.reflect.Method.invoke(Method.java:498)
    at lambdainternal.EventHandlerLoader$PojoMethodRequestHandler.handleRequest(EventHandlerLoader.java:456)
    at lambdainternal.EventHandlerLoader$PojoHandlerAsStreamHandler.handleRequest(EventHandlerLoader.java:375)
    at lambdainternal.EventHandlerLoader$2.call(EventHandlerLoader.java:1139)
    at lambdainternal.AWSLambda$2.call(AWSLambda.java:94)
    at lambdainternal.AWSLambda.startRuntime(AWSLambda.java:290)
    at lambdainternal.AWSLambda.<clinit>(AWSLambda.java:57)
    at java.lang.Class.forName0(Native Method)
    at java.lang.Class.forName(Class.java:348)
    at lambdainternal.LambdaRTEntry.main(LambdaRTEntry.java:94)

Is there any way around this? I need to manipulate the video data but cannot without making it into a file first. Any suggestions are welcome. Thank you.

Upvotes: 15

Views: 12615

Answers (1)

Mark B
Mark B

Reputation: 200476

You have to create the file in /tmp. That's the only location you are allowed to write to in the Lambda environment.

Upvotes: 32

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