Question

我一直试图让我的网站调用此API函数，但它调用的是不正确的查询，即使它在我的文件中正确显示。服务器端js代码如下：

app.post('/createPhysician/', function(req, res) {
    console.log("below is the req body for createPhysician");
    console.log(req.body);

    var createPromise = interact.createPhysician(
        req.body.firstName,
        req.body.lastName,
        req.body.yearNum,
        req.body.position,
        req.body.isAttending,
        req.body.highRiskTrained);

    createPromise.then(function(createResponse) {
        res.json("successful"); // returns the physicianID for the createUsers
    }).catch(function(err) {
        console.log(err);
        console.log(req.body);
        res.json("Terrible job you botched it");
    }); 
});

调用查询的交互文件如下： var mysql = require（'mysql'）;

var con = mysql.createConnection({
    host: "localhost",
    user: "root",
    password: "password",
    database: "dbName"
});

con.connect(function(err) {
    if (err) {
        console.error('Error:- ' + err.stack);
        return;
    }
    console.log('Connected Id:- ' + con.threadId);
});


createPhysician: function(physicianID,
    firstName,
    lastName,
    yearNum,
    position,
    isAttending,
    highRiskTrained)  {

    var qry = "insert into Physician values (firstName, lastName, yearNum, position, isAttending, highRiskTrained \
    ('"+firstName+"', '"+lastName+"', "+yearNum+", '"+position+"', '"+isAttending+"', '"+highRiskTrained+"');"
    return runQuery(qry);
}

当我在邮递员中使用以下数据运行它时，我收到的错误甚至不再出现在我的查询中。

API正文：

{
    "firstName": "Jane",
    "lastName": "Doe",
    "yearNum": 1,
    "position": "coder",
    "isAttending": 1,
    "highRiskTrained": 0
}

邮递员错误信息：

{
    "code": "ER_BAD_FIELD_ERROR",
    "errno": 1054,
    "sqlMessage": "Unknown column 'undefined' in 'field list'",
    "sqlState": "42S22",
    "index": 0,
    "sql": "insert into Physician (physicianID, firstName, lastName, yearNum, position, isAttending, highRiskTrained)             values (undefined, 'jane', 'doe', 1, 'coder', '1', '0');"
}

下面是我的runQuery代码：

function runQuery(queryString) {
    return new Promise(function(resolve, reject) {
        con.query(queryString, function (err, result, fields){
            if (err) reject(err);
            resolve(result);
        });
    });
}

Answer 1

我认为您的问题是由SQL语法引起的。试试这个：

var qry = "insert into Physician (firstName, lastName, yearNum, position, isAttending, highRiskTrained) values ('"+firstName+"', '"+lastName+"', "+yearNum+", '"+position+"', '"+isAttending+"', '"+highRiskTrained+"');";

API代码未运行正确的查询

1 个答案: