url-parameters

问题 I am facing an issue on getting the value of tagid from my URL: localhost:8888/p?tagid=1234 . Help me out to correct my controller code. I am not able to get the tagid value. My code is as follows: app.js : var express = require(\'express\'), http = require(\'http\'), path = require(\'path\'); var app = express(); var controller = require(\'./controller\')({ app: app }); // all environments app.configure(function() { app.set(\'port\', process.env.PORT || 8888); app.use(express.json()); app

问题 When the user accesses this URL running on my flask app, I want the web service to be able to handle the parameters specified after the question mark: http://10.1.1.1:5000/login?username=alex&password=pw1 #I just want to be able to manipulate the parameters @app.route(\'/login\', methods=[\'GET\', \'POST\']) def login(): username = request.form[\'username\'] print(username) password = request.form[\'password\'] print(password) 回答1: Use request.args to get parsed contents of query string: from

问题 I have a string like this: abc=foo&def=%5Basf%5D&xyz=5 How can I convert it into a JavaScript object like this? { abc: \'foo\', def: \'[asf]\', xyz: 5 } 回答1: Edit This edit improves and explains the answer based on the comments. var search = location.search.substring(1); JSON.parse('{"' + decodeURI(search).replace(/"/g, '\\"').replace(/&/g, '","').replace(/=/g,'":"') + '"}') Example Parse abc=foo&def=%5Basf%5D&xyz=5 in five steps: decodeURI: abc=foo&def=[asf]&xyz=5 Escape quotes: same, as

问题 I\'m looking for a jQuery plugin that can get URL parameters, and support this search string without outputting the JavaScript error: \"malformed URI sequence\". If there isn\'t a jQuery plugin that supports this, I need to know how to modify it to support this. ?search=%E6%F8%E5 The value of the URL parameter, when decoded, should be: æøå (the characters are Norwegian). I don\'t have access to the server, so I can\'t modify anything on it. 回答1: function getURLParameter(name) { return