fetch url in javascript
fetch('https://api.github.com/users/kipngetich33')
.then(response => response.json())
.then(formatedResponse => console.log(formatedResponse))
//NB please replace url with your ownfetch url in javascript
fetch('https://api.github.com/users/kipngetich33')
.then(response => response.json())
.then(formatedResponse => console.log(formatedResponse))
//NB please replace url with your ownfetch api in javascript
// GET Request.
fetch('https://jsonplaceholder.typicode.com/users')
// Handle success
.then(response => response.json()) // convert to json
.then(json => console.log(json)) //print data to console
.catch(err => console.log('Request Failed', err)); // Catch errorsfetch and then javascript send parameters
const url = 'https://randomuser.me/api';
// The data we are going to send in our request
let data = {
name: 'Sara'
}
// The parameters we are gonna pass to the fetch function
let fetchData = {
method: 'POST',
body: data,
headers: new Headers()
}
fetch(url, fetchData)
.then(function() {
// Handle response you get from the server
});fetch suntax
fetch('http://example.com/movies.json')
.then((response) => {
return response.json();
})
.then((data) => {
console.log(data);
});fetch html from url
fetch('/about').then(function (response) {
// The API call was successful!
return response.text();
}).then(function (html) {
// Convert the HTML string into a document object
var parser = new DOMParser();
var doc = parser.parseFromString(html, 'text/html');
// Get the image file
var img = doc.querySelector('img');
console.log(img);
}).catch(function (err) {
// There was an error
console.warn('Something went wrong.', err);
});Copyright © 2021 Codeinu
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