Question

我正在使用sql数据库提供服务。我确实结合了两个查询来创建单个查询，因为查询存储在不同的表中。但是它在我试图执行最终查询的代码的最后部分给出了错误。

代码如下：

const mysql = require('mysql');
const express = require('express');
var app = express();
const bodyParser = require('body-parser');
app.use(bodyParser.json());
app.use(bodyParser.urlencoded({ extended: true })); 



var mysqlConnection = mysql.createConnection({
    host: 'localhost',
    user: 'root',
    password: 'root',
    database: 'abc',

});




mysqlConnection.connect((err) => {
    if (!err)
        console.info('DB connection Succeed...!!');
    else
        console.error('DB connection Error \n Error :' + JSON.stringify(err, undefined, 2));
});

app.listen(4001, ()=> console.log('Express server is running on port 4001 '));



app.use((req, res, next) => {
    res.header("Access-Control-Allow-Origin", "*");
    res.header("Access-Control-Allow-Headers", "Origin, X-Requested-With, Content-Type, Accept, Authorization");
    if (req.method === 'OPTIONS'){
        res.header("Access-Control-Allow-Methods", "POST, PUT, PATCH, DELETE, GET");
        return res.status(200).json({});
    }
    if (req.method === 'PATCH'){
        res.header("Access-Control-Allow-Methods", "POST, PUT, PATCH, DELETE, GET");
        return res.status(200).json({});
    }
    next();
});


app.get('/drawChart',(req,res)=>{

    var query1 = null;
    var query2 = null;
    var finalquery = null;
    var finquery=null;
    mysqlConnection.query('SELECT * FROM t_table WHERE t_code= ?', 70, function',function(err, result){
        if (err) 
            return res.json({})
        query1=result[0].T_QUERY;
    });

    mysqlConnection.query('SELECT * FROM t_table_attr_val WHERE Table_ATTR_TYPE="GROUPBY" and t_CODE=?', 70, ', function(err, result){
        if (err) 
            return res.json({})
        query2=result[0].T_ATTR TYPE;
        finalquery=query1 + ' Group by ';
        finalquery +=query2;
        console.log(finalquery)
     });

     mysqlConnection.query(finalquery,function(err,result){
         if(err)
             return res.json({})
         console.log("result is "+ result);
     })
});

这是错误消息

Cannot use 'in' operator to search for 'typeCast' in null
    at Connection.query 

(C:\Users\ASUS\node_modules\mysql\lib\Connection.js:186:47)

我无法弄清楚代码出了什么问题。

Node.js：不能使用'in'运算符在null中搜索'typeCast'

0 个答案: