I am learning Node.js at the moment, and i am trying to create a shopping list application, and i am trying to implement a search route, which checks if the query matches the val, here is the code
const express = require('express'); const bodyParser = require('body-parser'); const app = express(); const port = 3000; //Array List let list = ['Fish', 'Lettuce', 'Chicken']; //Set view engine to pug app.set('view engine', 'pug'); //use bodyParser app.get('/', function(request, response){ response.render('list', {list}); }); app.get('/search', function(request, response){ return list.map(function(val){ if(request.query.search === val){ return response.send('Yup you got ' + val); } response.send('Not found') }); }); app.get('/new-item', function(request, response){ response.render('new'); }); app.post('/add-item', function(request, response){ let add = response.send(request.body); list.push(add); }); app.listen(port, function(){ console.log('Listening on Port 3000'); }); Now the problem is with the /search route's if conditional, i know the reason i am getting the error is because i can't use the response.send twice, i am looking for a way to send either response, based on if the condition is met. Any help is appreciated Thanks