request

Axios vs Request

前提是你 提交于 2020-05-27 03:59:05
问题 I am doing a post request to an URL with some formdata.... I am interested in capturing the "command":"insert" part which is in the response.. when I make a post to an url using AXIOS. I dont get this "command":"insert" part axios.post('https://www.localgov.ie/en/views/ajax', { validation_date_from: "10/10/2017", view_name : "bcsm_search_results", view_display_id : "notice_search_pane", view_path : "bcms/search" }).then(function(response){ console.log( response.data) console.log("------------

C++ Http Request with POCO

六月ゝ 毕业季﹏ 提交于 2020-05-24 21:04:41
问题 I'm wondering how I can do a request to a URL (e.g. download a picture and save it) with POCO in C++? I got this little code so far #include <iostream> #include <string> #include "multiplication.h" #include <vector> #include <HTTPRequest.h> using std::cout; using std::cin; using std::getline; using namespace Poco; using namespace Net; int main() { HTTPRequest *test = new HTTPRequest("HTTP_GET", "http://www.example.com", "HTTP/1.1"); } 回答1: Normally POCO has a great advantage to be very simple

How to call function after async requests finish in SwiftUI?

倖福魔咒の 提交于 2020-05-16 08:59:40
问题 I have a function that calls 2 types of api requests to get a bunch of data i need in my app. In the function I make a request for locations, then for each location in the response I make a different request to get details of that specific location. (ex. if request 1 returns 20 locations, my second request is called 20 times, once for each location) My function code here: func requestAndCombineGData(location: CLLocation, radius: Int) { // Clears map of markers self.mapView.clear() // Calls

Submit a POST form in Cypress and navigate to the resulting page

别来无恙 提交于 2020-05-15 08:53:05
问题 I'm having issues with Cypress loading the response body when I utilize the cy.request() command. In our application, when a form is filled out and submitted, it POSTs, and the response body is the new page. When I'm trying to do in Cypress is programmatically fill out the form. So I set up a cy.request() command, with the body filled with the form fields, which is the same as what happens when you fill it out manually. When I run the command, I can view the console and see that the correct

Submit a POST form in Cypress and navigate to the resulting page

ⅰ亾dé卋堺 提交于 2020-05-15 08:53:04
问题 I'm having issues with Cypress loading the response body when I utilize the cy.request() command. In our application, when a form is filled out and submitted, it POSTs, and the response body is the new page. When I'm trying to do in Cypress is programmatically fill out the form. So I set up a cy.request() command, with the body filled with the form fields, which is the same as what happens when you fill it out manually. When I run the command, I can view the console and see that the correct

How to change request url before making request in scrapy?

荒凉一梦 提交于 2020-05-14 19:56:07
问题 I need to modify my request url before a response is downloaded. But I'm not able to change it. Even after modifying the request url using request.replace(url=new_url) , the process_response prints the non-modified url. Here's the code of the middleware: def process_request(self, request, spider): original_url = request.url new_url= original_url + "hello%20world" print request.url # This prints the original request url request=request.replace(url=new_url) print request.url # This prints the

Set a body for WebClient when making a Post Request

与世无争的帅哥 提交于 2020-05-12 20:36:31
问题 So I have an api that I want to call to. The first call is an ahoy call and in the body of the request I need to send the ship_type, piratename and my piratepass. I then want to read the response which has my treasure booty that i will use for later. I'm able to do this with web request. but i feel like there is a better way to do it with webclient. (way I currently do it in webrequest) //Credentials for the Pirate api string piratename = "IvanTheTerrible"; string piratepass= "YARRRRRRRR";

PHP GuzzleHttp. How to make a post request with params?

一世执手 提交于 2020-05-09 17:43:08
问题 How to make a post request with GuzzleHttp( version 5.0 ). I am trying to do the following: $client = new \GuzzleHttp\Client(); $client->post( 'http://www.example.com/user/create', array( 'email' => 'test@gmail.com', 'name' => 'Test user', 'password' => 'testpassword' ) ); But I am getting the error: PHP Fatal error: Uncaught exception 'InvalidArgumentException' with the message 'No method can handle the email config key' 回答1: Try this $client = new \GuzzleHttp\Client(); $client->post( 'http:

How to make an XMLHttpRequest that loads an HTML file into the div?

£可爱£侵袭症+ 提交于 2020-05-08 18:57:30
问题 I'm trying to make an XMLHttpRequest that loads HTML from an external file and inserts the content of the file into the div . When I run the function, it inserts the HTML in all of the body which is not adequate. My code: --------------------------> HTML <-------------------------- <!DOCTYPE html> <html> <head> <meta charset="utf-8"> <script src="shit.js" charset="utf-8"></script> <link rel="stylesheet" href="index.css"> <title>Test</title> </head> <body> <button type="button" name="button"

How to make an XMLHttpRequest that loads an HTML file into the div?

心已入冬 提交于 2020-05-08 18:57:10
问题 I'm trying to make an XMLHttpRequest that loads HTML from an external file and inserts the content of the file into the div . When I run the function, it inserts the HTML in all of the body which is not adequate. My code: --------------------------> HTML <-------------------------- <!DOCTYPE html> <html> <head> <meta charset="utf-8"> <script src="shit.js" charset="utf-8"></script> <link rel="stylesheet" href="index.css"> <title>Test</title> </head> <body> <button type="button" name="button"