我正在尝试创建一个应用程序来转换温度。我在我的temp.js类中进行POST调用，然后它应该调用我的ajax.js类来处理信息并执行计算，以返回我需要的结果。然而，它从来没有进入我的ajax调用，所以我现在被卡住了。如果有人有任何建议或答案，请告诉我。temp.js
const express = require('express');
const path = require('path');
const router = express.Router();
const app = express();
const bodyParser = require("body-parser");
app.use(bodyParser.json({extended: false}));
router.get("/",function(req, res){
res.sendFile(path.join(__dirname,"/tempConverter.html"));
});
router.post("/ajaxTemp", function(req, res){
console.log(req.body.data);
console.log(req.headers['content-type']);
var body = req.body;
res.send(JSON.stringify(body));
});
module.exports = router;
ajax.js
$(document).ready(function(){
$("#tempForm").submit(function(e){
e.preventDefault();
var tempInput = $("#tempInput").val();
var fromOp = $("#fromOp").val();
var toOp = $("#toOp").val();
if(fromOp !== toOp){
if(fromOp === "Fahrenheit"){
//F -> C
if(toOp === "Celsius"){
var result = (tempInput - 32)*(5/9);
}
//F -> K
else{
var result = (tempInput - 32)*(5/9)+273.15;
}
}else if(fromOp === "Celsius"){
//C -> F
if(toOp === "Fahrenheit"){
var result = (tempInput + 32)*(9/5)+273.15;
}
//C -> K
else{
var result = tempInput + 273.15;
}
}else{
//K -> C
if(toOp === "Celsius"){
var result = tempInput - 273.15;
}
//K -> F
else{
var result = (tempInput + 32)*(9/5)-273.15;
}
}
}else{
alert("Cannot convert Same Degree");
}
$.ajax({
url : "/ajaxTemp",
data : {
result
},
type: "POST",
contentType: "application/x-www-form-urlencoded",
success : function(res){
console.log("made it here");
$("#tempForm").append("<br/>"+res);
}, error: function(err){
console.log(err);
}
})
})
});
html文件：
<html>
<head>
</head>
<body>
<h1>
Lets convert some Temperatures
</h1>
<form id="tempForm" action="tempConverter/ajaxTemp" method="POST">
<input id="tempInput" type="number">
<select id="fromOp">
<option id="fromF">Farenheit</option>
<option id="fromC">Celsius</option>
<option id="fromK">Kelvin</option>
</select>
Convert to:
<select id="toOp">
<option id="toF">Farenheit</option>
<option id="toC">Celsius</option>
<option id="toK">Kelvin</option>
</select>
<input type="submit" value="Submit">
</form>
</body>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.5.1/jquery.min.js"></script>
</html>
上面的代码只是我正在创建的路线之一。我有一个应用程序输入的主类：
app.js
const express = require('express');
const bodyParser = require('body-parser');
const path = require('path');
const app = express();
app.use(bodyParser.urlencoded({ extended: false }));
app.use(bodyParser.json());
app.listen(8080);
app.use("/tempConverter",require('./assets/Temperature/temp.js'));
您的html运行良好。您的快递API出错。你忘了要求表达，并且遗漏了一些语法。
const express = require('express');
const router = express.Router();
const app = express();
// Remember to update ajax url at ajax.js file
const port = 8080;
// Your routes are here ...
router.get(...)
router.post(...)
app.use('/', router);
// If your run server local.
// Your api url will be http://localhost:8080
app.listen(port, () => console.log(`Example app listening on port ${port}!`));
下面这行是错误的，因为您的路由将是/tempConverter/tempConverter而不是/tempConverter
app.use("/tempConverter",require('./assets/Temperature/temp.js'));
您的代码应该修改如下。
// Write as this for best practice
const router = require('./assets/Temperature/temp.js');
app.use("/", router);
// This line must be after app.use
app.listen(8080);