How to read a file in AWS Lambda Function written

Question

I have written an AWS Lambda Handler as below :

package com.lambda;
import com.amazonaws.services.lambda.runtime.Context;
import com.amazonaws.services.lambda.runtime.LambdaLogger;
import com.amazonaws.services.lambda.runtime.RequestStreamHandler;
import java.io.*;

public class TestDetailsHandler implements RequestStreamHandler {


    public void  handleRequest(InputStream input,OutputStream output,Context context){

        // Get Lambda Logger
        LambdaLogger logger = context.getLogger();

        // Receive the input from Inputstream throw exception if any

        File starting = new File(System.getProperty("user.dir"));
            System.out.println("Source Location" + starting);

           File cityFile = new File(starting + "City.db");
        FileInputStream fis = null;

        try {
            fis = new FileInputStream(cityFile);

            System.out.println("Total file size to read (in bytes) : "
                    + fis.available());

            int content;
            while ((content = fis.read()) != -1) {
                // convert to char and display it
                System.out.print((char) content);
            }

        } catch (IOException e) {
            e.printStackTrace();
        } finally {
            try {
                if (fis != null)
                    fis.close();
            } catch (IOException ex) {
                ex.printStackTrace();
            }
        }
    }

}

Its read a file : City.db , available in resources folder, even I kept to everywhere see below :

But it showing following message on execution of this lambda function :

START RequestId: 5216ea47-fc43-11e5-96d5-83c1dcdad75d Version: $LATEST
Source Location/
java.io.FileNotFoundException: /city.db (No such file or directory)
    at java.io.FileInputStream.open0(Native Method)
    at java.io.FileInputStream.open(FileInputStream.java:195)
    at java.io.FileInputStream.<init>(FileInputStream.java:138)
    at com.lambda.TestDetailsHandler.handleRequest(TestDetailsHandler.java:26)
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
    at java.lang.reflect.Method.invoke(Method.java:497)
    at lambdainternal.EventHandlerLoader$StreamMethodRequestHandler.handleRequest(EventHandlerLoader.java:511)
    at lambdainternal.EventHandlerLoader$2.call(EventHandlerLoader.java:972)
    at lambdainternal.AWSLambda.startRuntime(AWSLambda.java:231)
    at lambdainternal.AWSLambda.<clinit>(AWSLambda.java:59)
    at java.lang.Class.forName0(Native Method)
    at java.lang.Class.forName(Class.java:348)
    at lambdainternal.LambdaRTEntry.main(LambdaRTEntry.java:93)
END RequestId: 5216ea47-fc43-11e5-96d5-83c1dcdad75d
REPORT RequestId: 5216ea47-fc43-11e5-96d5-83c1dcdad75d  Duration: 58.02 ms  Billed Duration: 100 ms     Memory Size: 1024 MB    Max Memory Used: 50 MB  

Contents of the Pom.xml file :

<?xml version="1.0" encoding="UTF-8"?>
<project xmlns="http://maven.apache.org/POM/4.0.0"
         xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
    <modelVersion>4.0.0</modelVersion>

    <groupId>com.lambda</groupId>
    <artifactId>testdetails</artifactId>
    <version>1.0-SNAPSHOT</version>
    <packaging>jar</packaging>
    <name>test-handler</name>

    <dependencies>
        <dependency>
            <groupId>com.amazonaws</groupId>
            <artifactId>aws-lambda-java-core</artifactId>
            <version>1.1.0</version>
        </dependency>

        <dependency>
            <groupId>junit</groupId>
            <artifactId>junit</artifactId>
            <version>4.11</version>
            <scope>test</scope>
        </dependency>



        </dependencies>

    <build>
        <plugins>
            <plugin>
                <groupId>org.apache.maven.plugins</groupId>
                <artifactId>maven-shade-plugin</artifactId>
                <version>2.3</version>
                <configuration>
                    <createDependencyReducedPom>false</createDependencyReducedPom>
                </configuration>
                <executions>
                    <execution>
                        <phase>package</phase>
                        <goals>
                            <goal>shade</goal>
                        </goals>
                    </execution>
                </executions>
            </plugin>
        </plugins>
    </build>
</project>

I have used various ways to keep file here and there , but at the end its not working. May you please let me know what is wrong here ?

However in my another project where I have kept xyz.properties file in resources folder and reading from a PropertyManager file, its working fine. When I tested it on my system its working fine, but on AWS Lambda function it doesn't work.

Answer 1

If the file in located under resources directory, then the following solution should work:

String fileName = "resources/config.json";

Path path = Paths.get(this.getClass().getResource("/").toURI());
Path resourceLocation = path.resolve(fileName);

try(InputStream configStream = Files.newInputStream(resourceLocation)) {
            //use your file stream as you need.
}

Here the most important part is "resources/config.json", it must not be "/resources/config.json", because the file location is /var/task/resources/config.json in lambda, I checked.

Hope this helps who still face problem in reading file in aws lambda.

Answer 2

This is how I did it, let's say this is how your project structure looks like -

And you want to read the file config.properties which is inside the project-dir/resources directory.

The code for reading the content of the file would be -

InputStream input = null;
try {
    Path path = Paths.get(PropertyUtility.class.getResource("/").toURI());

    // The path for config file in Lambda Instance -
    String resourceLoc = path + "/resources/config.properties";

    input = new FileInputStream(resourceLoc);
} catch(Exception e) {
    // Do whatever
}

If you are following this project structure and using this code, then it will work in AWS Lambda.

PropertyUtility is just a utility class that I have created to read the contents of the config file. The PropertyUtility class looks like this -

As you can see in the above code, the path of the config file is different in the local system and in Lambda Instance.

In your local machine, PropertyUtility.class.getResource("/") points to bin, that is why you have to do path.getParent(), to point it to the project-directory which is HelloLambda in this example.

For the Lambda Instance, PropertyUtility.class.getResource("/") points directly to the project-directory.

Answer 3

I have made following changes in my code and now its works perfect :

Majorly changed following two lines :

ClassLoader classLoader = getClass().getClassLoader();

File cityFile = new File(classLoader.getResource("City.db").getFile());

package com.lambda;


import com.amazonaws.services.lambda.runtime.Context;
import com.amazonaws.services.lambda.runtime.LambdaLogger;
import com.amazonaws.services.lambda.runtime.RequestStreamHandler;
import java.io.*;

public class TestDetailsHandler implements RequestStreamHandler {


    public void  handleRequest(InputStream input,OutputStream output,Context context){

        // Get Lambda Logger
        LambdaLogger logger = context.getLogger();

        // Receive the input from Inputstream throw exception if any

        ClassLoader classLoader = getClass().getClassLoader();

        File cityFile = new File(classLoader.getResource("City.db").getFile());
        FileInputStream fis = null;

        try {
            fis = new FileInputStream(cityFile);

            System.out.println("Total file size to read (in bytes) : "
                    + fis.available());

            int content;
            while ((content = fis.read()) != -1) {
                // convert to char and display it
                System.out.print((char) content);
            }

        } catch (IOException e) {
            e.printStackTrace();
        } finally {
            try {
                if (fis != null)
                    fis.close();
            } catch (IOException ex) {
                ex.printStackTrace();
            }
        }
    }

Answer 4

If the file is located under resources folder, you can use it directly in lambda by using something like the following code:

final BufferedReader br = new BufferedReader(new FileReader("/flows/cancellation/MessageArray.json"));

I wanted to read a json file, you can have different use case, but the code works.