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Javascript MYSQL节点问题_Javascript_Mysql_Node.js_Web - Fatal编程技术网

Javascript MYSQL节点问题

Javascript MYSQL节点问题,javascript,mysql,node.js,web,Javascript,Mysql,Node.js,Web,运行下面的代码会抛出一个MYSQL错误,我已经检查了语法,它似乎是正确的,但不确定抛出错误的原因 app.post('/:roomname/create',function(request, response){ let roomID = ""+request.params.roomname+""; console.log(roomID); let insertRoom = "INSERT INTO chatrooms (roomname) VALUES (?)"; conn.query(

运行下面的代码会抛出一个MYSQL错误,我已经检查了语法,它似乎是正确的,但不确定抛出错误的原因

app.post('/:roomname/create',function(request, response){
let roomID =  ""+request.params.roomname+"";
console.log(roomID);
let insertRoom = "INSERT INTO chatrooms (roomname) VALUES (?)";


conn.query(insertRoom,[roomID], function(error, result) {
  if (err) throw err;
console.log("1 record inserted");
});
错误:

node_modules/mysql/lib/protocol/Parser.js:80
    throw err; // Rethrow non-MySQL errors
    ^

ReferenceError:err未定义

您的处理错误名称和回调错误名称不相同! 您必须在回调和处理块中定义相同的错误名称。 请检查并使用这只鹬。使用后,如果有任何错误发生,请让我在评论中知道

app.post('/:roomname/create',function(request, response){
let roomID =  ""+request.params.roomname+"";
console.log(roomID);
let insertRoom = "INSERT INTO chatrooms (roomname) VALUES (?)";


conn.query(insertRoom,[roomID], function(err, result) {
  if (err) throw err;
console.log("1 record inserted");
});
在回调函数中,您正在写入“error”,但在中,如果您正在检查“err”。在回调函数中,将其更改为

conn.query(insertRoom,[roomID], function(err, result) {

您可以发布实际使用的JavaScript吗?
“插入房间(roomname)值(“+roomID+”)”。在@MattPengelly发布的variableNew代码周围加上单引号
conn.query(insertRoom,[roomID], function(err, result) {