CODEWITHSUNDEEP
ajaxjquery
fabrizio
fabrizio

Reputation: 259

block double request while ajax is pending

I really need to understand how block the requests ajax while is pending. Let me explain an example:

I have an input type called "share" and when I write something and press enter ajax take the result and it insert inside mysql. If I do this pressing a lot of time the enter button it take more results like how many times I pressed enter even if there is a control in php. I think is a problem of the request ajax because the control php I'm sure is right. Thank you so much.

this is the code of example:

form

<input type="text" class="share">

Js

$.fn.enterKey = function (fnc) { // function for check when i press enter
    return this.each(function () {
        $(this).keypress(function (ev) {
            var keycode = (ev.keyCode ? ev.keyCode : ev.which);
            if (keycode == '13') {
                fnc.call(this, ev);
            }
        })
    })
}

$('.share').enterKey(function() { // ajax request
    var comment = $(this);
    if (comment.val() != "") {
        $.ajax({
            type: "POST",
            url:  add_comment.urlRemove,
            data: "text=" + comment.val(),
            success: function(html) {
                comment.val('');
            },
            error: function(){
                alert('Error on ajax call');
            }
        }); 
    } 
    else {
        return false;
    }
});

Upvotes: 1

Views: 893

Answers (1)

Karl Anderson
Karl Anderson

Reputation: 34846

You can set a variable to determine whether or not the request has already been sent, like this:

var alreadySent = false;

$('.share').enterKey(function() {
    var comment = $(this);
    if (comment.val() != "") {   
        if(!requestSent) {
            // Set variable to true
            alreadySent = true;

            $.ajax({
                type: "POST",
                url:  add_comment.urlRemove,
                data: "text=" + comment.val(),
                success: function(html) {
                    // Set variable to false
                    alreadySent = false;
                    comment.val('');
                },
                error: function(){
                    // Set variable to false
                    alreadySent = false;
                    alert('Error on ajax call');
                }
           }); 
        }
    }
});

Upvotes: 1

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