SAM 调用不会采用本地环境变量答案

【问题标题】：SAM invoke won't take local env varsSAM 调用不会采用本地环境变量
【发布时间】：2023-04-06 11:11:01
【问题描述】：

我有一个带有基本端点的示例 SAM 应用程序。我只想通过以下方式在本地运行它：

sam local invoke -e events/event-post-item.json putItemFunction --profile myprofile -n local.json

local.json如下：

{
    "getAllItems": {
        "SAMPLE_TABLE": "mywebservices-SampleTable-1BS18COYN2SHV"
    },
    "getById": {
        "SAMPLE_TABLE": "mywebservices-SampleTable-1BS18COYN2SHV"
    },
    "putItem": {
        "SAMPLE_TABLE": "mywebservices-SampleTable-1BS18COYN2SHV"
    }
}

下面是putItemFunction的代码

// Create clients and set shared const values outside of the handler

// Create a DocumentClient that represents the query to add an item
const dynamodb = require('aws-sdk/clients/dynamodb');

const docClient = new dynamodb.DocumentClient();

// Get the DynamoDB table name from environment variables
const tableName = process.env.SAMPLE_TABLE;


/**
 * A simple example includes a HTTP post method to add one item to a DynamoDB table.
 */
exports.putItemHandler = async (event) => {
    
    const { body, httpMethod, path } = event;
    if (httpMethod !== 'POST') {
        throw new Error(`postMethod only accepts POST method, you tried: ${httpMethod} method.`);
    }
    // All log statements are written to CloudWatch by default. For more information, see
    // https://docs.aws.amazon.com/lambda/latest/dg/nodejs-prog-model-logging.html
    console.log('received:', JSON.stringify(event));

    // Get id and name from the body of the request
    const { id, name } = JSON.parse(body);

    // Creates a new item, or replaces an old item with a new item
    // https://docs.aws.amazon.com/AWSJavaScriptSDK/latest/AWS/DynamoDB/DocumentClient.html#put-property
    const params = {
        TableName: tableName,
        Item: { id, name },
    };
    await docClient.put(params).promise();

    const response = {
        statusCode: 200,
        body,
    };

    console.log(`response from: ${path} statusCode: ${response.statusCode} body: ${response.body}`);
    return response;
};

我运行此程序，但出现“找不到资源”错误。我已确保个人资料详细信息正确无误。

问题在于处理程序中的这一行：const tableName = process.env.SAMPLE_TABLE;

如果我在这里硬编码表名，它可以正常工作。否则该函数总是产生tableName 值“SampleTable”...

它应该采用我提供的环境变量的值。不是“SampleTable”...我做错了什么？

【问题讨论】：

标签： amazon-web-services aws-lambda amazon-dynamodb

【解决方案1】：

local environment filelocal.json 中的键是 lambda 函数名称。

"putItemFunction:"，在你的情况下不是"putItem:"。

【讨论】：