app.js
app.get('/results',showResult)
var express = require('express')
var n = req.query.query;
mysql_crawl.query('SELECT prod_name, full_price FROM `xxx` WHERE MATCH(data_index) AGAINST("'+n+'")', function(error, product_data) {
res.render('result.html', {result: n, related: product_data: product_data})
});
result.html
<form action= '/results' class="input-group input-group-lg" id="search">
<input type="text" class="input-group form-control" name="query" value='<%= result %>'/>
<span class="input-group-btn">
<button class="btn btn-secondary" type='button'>Search</button>
</span>
</form>
When I search, I get a URL like example.com/results?query=keyword, but I want a results page at example.com/s/keyword.
Both pages should be the same results. If I change the keyword in the URL, it should show search results for that keyword.
So my question is, how do I create the example.com/s/keyword page to duplicate the results at example.com/results?query=keyword?
app.get('/s/price/:sea_query', function (req, res) {
var n = req.params.sea_query;
};
I can use this to get data from url, thanks :)
You can create new route as
app.get('/s/:keyword', function (req, res) {
var n = req.params.keyword;
// do rest of the task
};
Instead of repeating same code in two route, you can create a function then call the same function in both routes.