Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Call function on scroll only once

I got a question regarding a function that will be called if the object is within my screen. But when the object is within my screen the function is been called and a alert is been fired. But if I close the alert and scroll further down the event is called again. I do not want that. How can I solve that?

Working example

My code so far:

<div id="wrapper">
    scroll down to see the div
</div>
<div id="tester"></div>

JS

$(window).on('scroll',function() {
    if (checkVisible($('#tester'))) {
        alert("Visible!!!")        
    } else {
        // do nothing 
    }
});

function checkVisible( elm, eval ) {
    eval = eval || "object visible";
    var viewportHeight = $(window).height(), // Viewport Height
        scrolltop = $(window).scrollTop(), // Scroll Top
        y = $(elm).offset().top,
        elementHeight = $(elm).height();   

    if (eval == "object visible") return ((y < (viewportHeight + scrolltop)) && (y > (scrolltop - elementHeight)));
    if (eval == "above") return ((y < (viewportHeight + scrolltop)));
}

What I want is that the If function only will be called 1 time and not on every scroll if the object is visible.

like image 242
Rotan075 Avatar asked Aug 21 '15 07:08

Rotan075


2 Answers

Try using .one():

$(window).one('scroll',function() {
   // Stuff
});

Or, unlink the event inside:

$(window).on('scroll',function() {
   // After Stuff
   $(window).off('scroll');
});

Guess you might need this code:

// this will check if element is in viewport

function checkVisible(elm, eval) {
  eval = eval || "object visible";
  var viewportHeight = $(window).height(), // Viewport Height
    scrolltop = $(window).scrollTop(), // Scroll Top
    y = $(elm).offset().top,
    elementHeight = $(elm).height();

  if (eval == "object visible") return y < viewportHeight + scrolltop && y > scrolltop - elementHeight;
  if (eval == "above") return y < viewportHeight + scrolltop;
}

$(window).on("scroll", function () {
  if (checkVisible($("#tester"))) {
    alert("Visible!!!");
    $(window).off("scroll");
  } else {
    // do nothing
  }
});

Fiddle: http://jsfiddle.net/c68nz3q6/

like image 94
Praveen Kumar Purushothaman Avatar answered Nov 19 '22 22:11

Praveen Kumar Purushothaman


Let's try to solve your problem in javascript as all of the answers here are in jquery. You can use global variable as browser retains its value as long as the page is not refreshed. All those variables declared outside the function are called global variables and can be accessed by any function.

window.onscroll = myScroll;
var counter = 0; // Global Variable
function myScroll(){
   var val = document.getElementById("value");
   val.innerHTML = 'pageYOffset = ' + window.pageYOffset;
   if(counter == 0){ // if counter is 1, it will not execute
     if(window.pageYOffset > 300){
        alert('You have scrolled to second div');
        counter++; // increment the counter by 1, new value = 1
     }
   }
  }
#wrapper,#tester {
   width: 300px;
  height: 300px;
  border: 1px solid black;
  padding: 10px;
  }
#wrapper p {
  text-align: center;
  }
#tester {
  border: 1px solid crimson;
  }
#value {
   position: fixed;
  left: auto;
  right: 40px;
  top: 10px;
  }
<p id = "value"></p>
<div id="wrapper">
    <p>scroll down to div to see the alert</p>
</div>
<div id="tester"></div>
like image 39
Amir Saleem Avatar answered Nov 19 '22 23:11

Amir Saleem