I have an arrays=[John; Alex; Mark], I wanna to show the elements of this array one by one by 3 second delay.
for (var i=0; i<=3; i++)
{
setTimeout(function(){x.innerHTML=s[i]},3000)
}
It seems very simple problem, but I can't figure out.
setTimeout starts with a lower case s3000 * i, not 3000 or they'll all fire at oncei, not the values it had as you went through the loop.This works, and completely avoids the loop variable issue:
var s = ['John', 'Mark', 'Alex'];
var i = 0;
(function loop() {
x.innerHTML = s[i];
if (++i < s.length) {
setTimeout(loop, 3000); // call myself in 3 seconds time if required
}
})(); // above function expression is called immediately to start it off
Note how it uses "pseudo-recursion" to trigger the next iteration 3000ms after the completion of the previous iteration. This is preferable to having n outstanding timers all waiting at the same time.
http://jsfiddle.net/rlemon/mHQVz/1/
I got to tinkering... albeit this is probably not the best solution it was fun.
var x = document.getElementById('x'),
s = ['John', 'Mark', 'Alex'];
(function loop() {
s.length && (x.innerHTML = s.shift(), setTimeout(loop, 3000));
})();
Alnitak's solution is alot better. However they both would work (his is just more readable and less hacky also does not destroy the array).
this will also help:
const fruits = ['apple', 'banana', 'mango', 'guava'];
let index = 0;
const primtMe = (value, i) => {
if (i < fruits.length) {
setTimeout(() => {
console.log(i + ' value = ' + value)
primtMe(fruits[i + 1], i + 1)
}, 3000);
} else {
return;
}
}
primtMe(fruits[index], index)
Try
var s=['John', 'Alex', 'Mark'];
var x = document.getElementById('x');
function display(i){
if(i >= s.length){
i = 0;
}
x.innerHTML = s[i];
setTimeout(function(){
display(i + 1)
}, 2000)
}
display(0)
Demo: Fiddle
If you do not use closure, you will end up with i being undefined. This is because in each iteration you are overriding what i is. By the time it finishes, it will be undefined. Using a closure will preserve i.
On another note, it's kind of pointless to hard code in values (i.e. i<3) when you can just check for length. This way, if s ever changes, you for loop will still grab everything.
var s = ['john','mark','brian'];
for (var i = 0; i < s.length; i++) {
(function(i) {
setTimeout(function() {
x.innerHTML = s[i];
}, 3000*(i+1));
})(i);
}
Your code won't work, since you set four timeouts of 2000 milliseconds (i.e. 2 seconds) at a time. You'd better use closure that sets three timeouts (by number of elements in array) with 3000 milliseconds of delay. It can be done with the following code (note that setTimeout is written from the small letter):
var s = ["John", "Alex", "Mark"];
for (var i = 0; i < s.length; i++) {
(function(i) {
setTimeout(function() {
x.innerHTML = s[i];
}, 3000 * i);
})(i);
}
You can use setInterval to show elements one by one after 3 seconds delay:
s=["John", "Alex", "Mark"];
var i = 0;
var id = setInterval(function(){
if(i > s.length) {
clearInterval(id);
}
else{
console.log(s[i++]);
}
}, 3000);
Try this without pseudo-recursion
var arr = [10,20,30,40]; // your array
var i = 0;
var interval = 2000 // 2 sec, you can add your time required
function callInterval() {
// set a variable for setInterval
var time = setInterval(()=>{
console.log('['+arr[i]+','+i+']');
i++;
if(i==arr.length){
this.clearInterval(time);// clear the interval after the index
}
}, interval);
}
callInterval();I solved this problem by using setInterval. Here is my solution:
function delayPrint(arr, delay) {
let index = -1;
const interval = setInterval(function () {
if (++index === arr.length) {
clearInterval(interval);
return;
}
console.log(arr[index]);
}, delay);
}
let s = ['John', 'Alex', 'Mark'];
delayPrint(s, 3000);
// output:
// John after 3s
// Alex after 3s
// Mark after 3s