Node.JS 仅返回 MySQL“WHERE IN”子句的部分记录

Posted 2023-02-19

【中文标题】Node.JS 仅返回 MySQL“WHERE IN”子句的部分记录

【英文标题】：Node.JS is returning only a part of records for the MySQL "WHERE IN" clause

【发布时间】：2021-12-08 04:26:29

【问题描述】：

我正在尝试执行以下GET 函数。

http://127.0.0.1:3000/sportfolioimages/getbyportfoliolist?portfolioid_list=69,70,71

我的代码在 Node.js 中，请查看下方。

const mysql = require('mysql2');
const errorCodes = require('source/error-codes');
const PropertiesReader = require('properties-reader');

const prop = PropertiesReader('properties.properties');

const con = mysql.createConnection(
  host: prop.get('server.host'),
  user: prop.get("server.username"),
  password: prop.get("server.password"),
  port: prop.get("server.port"),
  database: prop.get("server.dbname")
);


exports.getSellerPortfolioItemImagesByPortfolioList = (event, context, callback) => 

  const params = event.queryStringParameters;


  if (!params || portfolioid_list == null) 
    context.callbackWaitsForEmptyEventLoop = false;
    var response = errorCodes.missing_parameters;
    callback(null, response)
  
  else 

    const portfolioid_list = event.queryStringParameters.portfolioid_list;
    context.callbackWaitsForEmptyEventLoop = false;

    const sql = "SELECT * FROM peresia.seller_portfolio_item_images WHERE idseller_portfolio_item IN (?)";
      con.execute(sql, [portfolioid_list], function (err, result) 
      console.log(sql);
        if (err) 
          console.log(err);
          var response = errorCodes.internal_server_error;
          callback(null, response);
        
        else 
          var response = 
            "statusCode": 200,
            "headers": 
              "Content-Type": "application/json"
            ,
            "body": JSON.stringify(result),
            "isBase64Encoded": false
          ;
          callback(null, response)
        
      );


  
;

我的代码总是返回我调用中值列表的第一个值。由于我的值列表是69,70,71，它始终只返回匹配69 的记录，并且即使数据库中有记录，70 和71 也不会返回任何记录。

举个例子，下面是我执行上述GET函数时得到的结果。

[
    
        "idseller_portfolio_item_images": 22,
        "idseller_portfolio_item": 69,
        "image_url": "https://database.com/portfolio/IMG_20211020_114254-1634730049335.jpg"
    ,
    
        "idseller_portfolio_item_images": 23,
        "idseller_portfolio_item": 69,
        "image_url": "https://database.com/portfolio/IMG_20211020_114254-1634730049335.jpg"
    ,
    
        "idseller_portfolio_item_images": 31,
        "idseller_portfolio_item": 69,
        "image_url": "https://peresia3.s3.us-east-2.amazonaws.com/portfolio/IMG_20211020_114254-1634730049335.jpg"
    ,
    
        "idseller_portfolio_item_images": 32,
        "idseller_portfolio_item": 69,
        "image_url": "https://database/portfolio/IMG_20211020_114254-1634730049335.jpg"
    
]

如果我直接在数据库中运行 MySQL 代码，我将能够毫无问题地获取完整的记录集。

为什么会这样，我该如何解决？

【参考方案1】：

根据this，您将需要这样的东西

const portfolioid_list = [69,70,71];

const sql = "SELECT * FROM peresia.seller_portfolio_item_images WHERE idseller_portfolio_item IN (?,?,?)";
      con.execute(sql, portfolioid_list, function (err, result)  ...

这样您就可以动态地构建您的查询。 reference

const portfolioid_list = event.queryStringParameters.portfolioid_list.split(",");  //converts to a list ["69", "70", "71"]
    
portfolioidValsPlaceHolders=Array(portfolioid_list.length).fill("?").join();
const sql = "SELECT * FROM peresia.seller_portfolio_item_images WHERE idseller_portfolio_item IN ("+portfolioidValsPlaceHolders+")"; 
      con.execute(sql, portfolioid_list, function (err, result)  ...    

【讨论】：

好的，但是我应该如何知道列表的确切长度？似乎是一个错误。我现在将检查 git。

您可以将查询参数字符串解析为一个列表，然后使用列表的长度您可以构建“？”的编号在sql语句中需要。只是一个建议

你能帮我解决这个问题吗？我对 Node.js 知之甚少

const portfolioid_list = event.queryStringParameters.portfolioid_list.split(","); //converts to a list ["69", "70", "71"] let sql = "SELECT * FROM peresia.seller_portfolio_item_images WHERE idseller_portfolio_item IN ("; sql += "?,".repeat(portfolioid_list.length); //adding the "?" sql = sql.substring(0, sql.length - 1); //remove the last unwanted "," sql += ")"; //now the sql is complete 可能有更优雅的方式来做到这一点