In Dart I am working on a web game. In this I would need a function where I can get data from the URL, in the same way that you would get it in PHP. How would I do this?
Say I, for example, append the following to my URL when I load my web game: ?id=15&randomNumber=3.14. How can I get them in Dart either as a raw string (preferred) or in some other format?
You can use the Uri class from dart:core
https://api.dartlang.org/apidocs/channels/stable/dartdoc-viewer/dart-core.Uri
ie:
main() {
print(Uri.base.toString()); // http://localhost:8082/game.html?id=15&randomNumber=3.14
print(Uri.base.query); // id=15&randomNumber=3.14
print(Uri.base.queryParameters['randomNumber']); // 3.14
}
import 'dart:html';
void doWork(){
var uri = Uri.dataFromString(window.location.href); //converts string to a uri
Map<String, String> params = uri.queryParameters; // query parameters automatically populated
var param1 = params['param1']; // return value of parameter "param1" from uri
var param2 = params['param2'];
print(jsonEncode(params)); //can use returned parameters to encode as json
}
Import 'dart:html' then you can use window.location...
Related
This is my code:
#Get('/get-presigned-url/:extension?')
async getS3PresignedUrl(#Param('extension') extension?: string): Promise<AwsTempPostUrlDto> {
return await this.aws.getTemporaryPostUrl(extension);
}
But the extension in the swagger is required.
How can I set url params as optional
I think you should use #Query instead of #Param
Is there a way to implement GraphQL in flutter?
I was trying making the API call with the query and variables objects in a JSON object.
type '_InternalLinkedHashMap' is not a subtype of type 'String' in type cast
I have been using graphql_flutter package for a few weeks now and it seems to work well enough. Here is an example:
import 'package:graphql_flutter/graphql_flutter.dart' show Client, InMemoryCache;
...
Future<dynamic> post(
String body, {
Map<String, dynamic> variables,
}) async {
final Client client = Client(
endPoint: endpoint,
cache: new InMemoryCache(),
);
final Future<Map<String, dynamic>> result =
client.query(query: body, variables: variables);
return result;
}
To use just give it the graphql and any variables. i.e. a delete mutation may look like
String deleteMutation =
'''mutation deleteItem(\$itemId: ID!) {
deleteItem(input: { itemId: \$itemId}) {
itemId
}
}'''.replaceAll('\n', ' ');
await post(deleteMutation , variables: <String, dynamic>{'itemId': itemId});
This is updated and working solution of #aqwert
import 'package:graphql_flutter/graphql_flutter.dart';
...
HttpLink link = HttpLink(uri: /*your url here*/); // you can also use headers for authorization etc.
GraphQLClient client = GraphQLClient(link: link as Link, cache: InMemoryCache());
QueryOptions query = QueryOptions(
document:
r'''
mutation deleteItem($id: String!) {
deleteItem(callId: $id)
}
''',
variables: {'id' : id}
);
var result = await client.query(query);
Is there a function to do urlencoding in Dart? I am doing a AJAX call using XMLHttpRequest object and I need the url to be url encoded.
I did a search on dartlang.org, but it didn't turn up any results.
var uri = 'http://example.org/api?foo=some message';
var encoded = Uri.encodeFull(uri);
assert(encoded == 'http://example.org/api?foo=some%20message');
var decoded = Uri.decodeFull(encoded);
assert(uri == decoded);
http://www.dartlang.org/docs/dart-up-and-running/contents/ch03.html#ch03-uri
Update: There is now support for encode/decode URI in the Dart Uri class
Dart's URI code is placed in a separate library called dart:uri (so it can be shared between both dart:html and dart:io). It looks like it currently does not include a urlencode function so your best alternative, for now, is probably to use this Dart implementation of JavaScript's encodeUriComponent.
Uri.encodeComponent(url); // To encode url
Uri.decodeComponent(encodedUrl); // To decode url
I wrote this small function to convert a Map into a URL encoded string, which may be what you're looking for.
String encodeMap(Map data) {
return data.keys.map((key) => "${Uri.encodeComponent(key)}=${Uri.encodeComponent(data[key])}").join("&");
}
I dont' think there is yet. Check out http://unpythonic.blogspot.com/2011/11/oauth20-and-jsonp-with-dartin-web.html and the encodeComponent method.
Note, it's lacking some characters too, it needs to be expanded. Dart really should have this built in and easy to get to. It may have it in fact, but I didn't find it.
Safe Url Encoding in flutter
Ex.
String url = 'http://example.org/';
String postDataKey = "requestParam="
String postData = 'hdfhghdf+fdfbjdfjjndf'
In Case of get request :
Uri.encodeComponent(url+postDataKey+postData);
In Case of Post Data Request use flutter_inappwebview library
var data = postDataKey + Uri.encodeComponent(postData);
webViewController.postUrl(url: Uri.parse(url), postData: utf8.encode(data));
Uri.encodeComponent() is correct, Uri.encodeFull() has a bug, see below example:
void main() {
print('$text\n');
var coded = Uri.encodeFull(text);
print(coded);
print('\n');
coded = Uri.encodeComponent(text);
print(coded);
}
var text = '#2020-02-29T142022Z_1523651918_RC2EAF9OOHDB_RT.jpg';
I have a page with links. These links all end in the same way. For example www.site.com/fruit/apples, www.site.com/fruit/bananas, www.site.com/fruit/oranges, etc. I want all these links to call the same Dart app and have the app do some processing and then redirect you wherever you need to go (the bananas page vs. the oranges page). This way, I avoid having an actual HTML file for every single fruit. I can instead have a single landing template that gets populated with variable fruit data.
The part I'm hung up on is passing the url into the Dart app so it can do the handling. I understand main() cannot receive arguments, so what's another way?
You can use the route package to handle the URL's for you.
For example:
import 'package:route/client.dart';
final fruitUrl = new UrlPattern(r'/fruit/(\w+)');
main() {
var router = new Router()
..addHandler(fruitUrl, showFruit)
..listen();
}
void showFruit(String path) {
var fruit = fruitUrl.parse(req.path)[0];
// Display the page according to the fruit type
}
If you don't need to handle actual routes, and you just want to handle any query parameters passed of the form ?fruit=apple you don't have to use the routes package and can instead manually parse the URL:
Map params = {};
// If arguments were provided, decode them into params map
if(window.location.search.length > 1) {
// Break all arguments into form: fruit=apple
List<String> queryPairs = window.location.search.substring(1).split('&');
for(String queryPair in queryPairs) {
// Add arguments into params map: key=fruit, value=apple
List<String> queryPairList = queryPair.split('=');
params[queryPairList[0]] = queryPairList[1];
}
}
// Handle the proper action based on the fruit
switch(params['fruit']) {
case 'apple':
// ...
break;
// ...
case 'orange':
// ...
break;
}
In my action I have Request property. I want to use this request to get response from another URL (from another website). Is it possible?
To request a external webpage from C# code and retrieve it as a string. You can use the following snippet of code:
try
{
var request = WebRequest.Create("http://www.steelcm.com/");
var response = request.GetResponse();
var responseStream = response.GetResponseStream();
using(var reader = new StreamReader(responseStream))
{
// Convert stream object to string
string myWebPage = reader.ReadToEnd();
}
}catch(Exception e)
{
// TODO: Handle error exception
}
This will return the HTML in the string variable myWebPage. Also you will need to include the following 2 libraries:
using System.Net;
using System.IO;
Of course you can manipulate the request URL dependant on a users input.