Issue with Firefox not recognizing keydown events for the backspace key

I am currently developing a terminal emulator and have encountered an issue with capturing the backspace key in Firefox. While I am able to capture the first backspace press and delete the last character in the input prompt, the problem arises when trying to remove multiple characters.

For more information, please visit the official website:

To see the replication of the issue, you can view it here: http://jsfiddle.net/BgtsE/1/

Below is the JavaScript code snippet:

function handleKeys(e){
    var evt = e || window.event;
    var key = evt.charCode || evt.keyCode;
    if(evt.type == "keydown")
    {
        curr_key = key;
        if(key == 8)
        {
            evt.preventDefault();
            if(0 < $('body').text().length)
                $('body').text($('body').text().slice(0,-1));
        }
    }
    else if(evt.type == "keypress")
    {
        if(97 <= key && key <= 122)
        {
            if(curr_key != key)
                $('body').append(String.fromCharCode(key));
        }
        else
            $('body').append(String.fromCharCode(key));
    }
}
$(function(){
    $('html').live({
        keydown:function(e){
            handleKeys(e);
        },
        keypress:function(e){
            handleKeys(e);
        }
    })
})​

Answer №1

Check out this solution here: http://jsfiddle.net/NBZG8/1/

To ensure compatibility with both Chrome and Firefox, make sure to handle the backspace key in both keydown and keypress events.

function handleKeys(e){
    var evt = e || window.event;
    var key = evt.charCode || evt.keyCode;

    if (evt.type == "keydown") {
        curr_key = key;
        if(key == 8 && !$.browser.mozilla) {
            backspaceHandler(evt);
        }
    } else if (evt.type == "keypress") {
        if (key == 8) {
            backspaceHandler(evt);
        } else if (97 <= key && key <= 122) {
            if(curr_key != key) {
                $('body').append(String.fromCharCode(key));
            }
        } else {
            $('body').append(String.fromCharCode(key));
        }
    }
}

function backspaceHandler(evt) {
    evt.preventDefault();
    if(0 < $('body').text().length) {
        $('body').text($('body').text().slice(0,-1));
    }  
};

$(function(){
    $('html').live({
        keydown : handleKeys,
        keypress : handleKeys
    })
})​

Answer №2

When using Firefox Windows 17.0.1, I encountered an issue where any value returned by $("selector").text() had an additional newline character at the end. This caused problems when trying to use substring in my code:

<html>
    <head>
        <title>test</title>
        <script src="jquery.js"></script>
        <script>
            $("document").ready(function(){
                console.log("body text seems to have a new line character");
                console.log(($('body').text()[5]=="\n"));
            });

            function handleKeys(e){
                var evt = e || window.event;
                var key = evt.charCode || evt.keyCode;
                if(evt.type == "keydown")
                {
                    curr_key = key;
                    if(key == 8)
                    {
                        evt.preventDefault();
                        if(0 < $('body').text().length)
                            //$('body').text($('body').text().slice(0,-2));
                             $('body').text($('body').text().substring(0,$('body').text().length-1)); 
                    }
                }
                else if(evt.type == "keypress")
                {
                    if(97 <= key && key <= 122)
                    {
                        if(curr_key != key)
                            $('body').append(String.fromCharCode(key));
                    }
                    else
                        $('body').append(String.fromCharCode(key));
                }
            }
            $(function(){
                $('html').live({
                    keydown:function(e){
                        handleKeys(e);
                    },
                    keypress:function(e){
                        handleKeys(e);
                    }
                })
            })
        </script>
    </head>
    <body>12345</body>
</html>

Answer №3

Encountering a similar issue with keypress in Mozilla prompted me to seek solutions. I'm grateful for this discussion as it helped resolve my problem, and now I intend to share my code in case others face a similar challenge.
In my scenario, I attempted to automatically add spaces when the user inputs two numbers. Unfortunately, this functionality did not work in Firefox, leading me to tweak my code accordingly:

$(function() {

    $('#field1, #field2').on('keypress',function(event) {
        event = event || window.event;
        var charCode = event.keyCode || event.which,
            lgstring = $(this).val().length,
            trimstring;
        if(charCode === 8) {
            event.returnValue = false;
            if(event.preventDefault)
                event.preventDefault();
            if(0 < $(this).val().length) {
                $(this).val($(this).val().slice(0,-1));
            }  
        }
        else if(((charCode > 31) && (charCode < 48 || charCode > 57)) || lgstring >= 14) {
            event.returnValue = false;
            if(event.preventDefault)
                event.preventDefault();
        }
        else {
            trimstring = $(this).val().replace(/ /g,"");
            if((lgstring !== 0) && (trimstring.length % 2) === 0 ) {
                $(this).val($(this).val() + ' ');
            }
        }
    });

});

It's worth noting that while Mozilla treats backspace as a keypress, Chrome does not exhibit the same behavior.

Pardon any language issues, as English is not my first language.

Answer №4

$('#id').keypress(function(e) {

if(e.charCode > 0 || e.keyCode === 8){
 if(e.keyCode === 8){
   return true;
 }else if((e.charCode !== 0) && ((e.charCode  > 57 && e.charCode  < 65)){
   return false;
   }
 }else if((e.keyCode !== 0) && ((e.keyCode  > 57 && e.keyCode  < 65)){
  return false;
 }
});

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