I'm trying to use a Regex to only allow letters, '-', and the backspace on input fields. I need to be able to use the left and right arrow keys and the delete key. I was attempting to do this with:
$("#input").keypress(function (e) {
var regex = new RegExp("^[a-zA-Z-\b]+$");
var key = String.fromCharCode(!e.charCode ? e.which : e.charCode);
if((!(e.keyCode == 37 || e.keyCode == 39 || e.keyCode == 46))){
if(!regex.test(key)){
return false;
}
}
});
where I would check if the key is not the left arrow (37), right arrow (39), or the delete key (46), and then check if the key is not in the regex.
This works on IE and Chrome, however on Firefox the key codes 37 and 39 also correspond to ' and % respectively.
The key codes work on all of the browsers, the issue is that [on firefox only] if I allow for the arrows and delete keys it also allows for the ' and % characters.
I was actually able to figure it out. For IE and Chrome the keyCode and Which values are the same. On Firefox the keyCode value of the left arrow is 37, but the which value is 0.
I was able to solve it doing this:
$("#input").keypress(function (e) {
var regex = new RegExp("^[a-zA-Z-\b]+$");
var key = String.fromCharCode(!e.charCode ? e.which : e.charCode);
if(!((e.keyCode == 37 && e.which == 0) || (e.keyCode == 39 && e.which == 0) || (e.keyCode == 46 && e.which == 0))){
if(!regex.test(key)){
return false;
}
}
});
Related
I have a little curiosity to understand, I've created a responsive site And I used 2 scripts to disable the numeric keys in an input field; The first does not allow the numbers to be entered,The second, a warning message appears that warns you that you can only type alpha characters, I would like to understand why on smartphones only work on firefox while on pc work perfectly:
This script disables the numeric keys:
enter code here
function Check(e) {
var keyCode = (e.keyCode ? e.keyCode : e.which);
if (keyCode > 47 && keyCode < 58) {
e.preventDefault();
}
}
And this makes the message appear:
$(document).ready(function(){
$(".inputTextBox").keypress(function(event){
var inputValue = event.which;
// allow letters and whitespaces only.
if(!(inputValue >= 65 && inputValue <= 123) && (inputValue != 32 && inputValue != 0)) {
event.preventDefault();
document.getElementById('warning07').style.display='block';
document.getElementById('07').innerHTML = "Inserire solo lettere!";
}
console.log(inputValue);
});
});
Can you explain me? Thanks in advance.
Good morning,
I facing a issue on the IE 10 where my keypress still can enter '%' but the FF and Chrome no such issue.
I found out that the key 37 is the [ left arrow ] which match with '%' in ASCII.
My sample code as below:
$('#refId').bind("keypress", function(event) {
// allow letters, numbers and keypad numbers ONLY
var key = event.charCode;
if((key >= 48 && key <= 57) ||
(key >= 65 && key <= 90) ||
(key >= 97 && key <= 122)){
return true;
}
//allow backspace, tab, left arrows, right arrow, delete
key = event.keyCode;
if(key == 8 ||
key == 9 ||
key == 37 ||
key == 39 ||
key == 46){
return true;
}
return false;
});
Can give me some idea how to fix this?
Thanks.
-fsloke
Use var key = event.which; instead and join the if-statements.
The event.which property normalizes event.keyCode and event.charCode.
It is recommended to watch event.which for keyboard key input.
- https://api.jquery.com/event.which/
$('#refId').on("keydown", function(event) {
// allow letters, numbers and keypad numbers ONLY
var key = event.which;
if((key >= 48 && key <= 57) ||
(key >= 65 && key <= 90) ||
(key >= 97 && key <= 122) ||
key == 8 ||
key == 9 ||
key == 37 ||
key == 39 ||
key == 46) {
return true;
}
return false;
});
I have to disable some symbols from html input.
e.which is not working properly on Firefox. Firefox disables backspace and ect.
Here is JS Fiddle:
var code = window.event ? event.keyCode : e.which;
event.keyCode works on firefox, but does not work with String.fromCharCode(code).
jQuery normalizes e.which, so you don't have to worry about this at all.
Also, it's a lot easier to just listen for the correct keycodes, there's no reason to convert the keycode to a character just to filter it out with indexOf ?
$('#foo').keydown(function(e) {
var code = e.which;
if (code == 8 || code == 13) return true; // backspace and enter
if (code < 48 || code > 57 || code == 188 || code == 190) return false;
});
FIDDLE
To keep most keys active, and just mainly disable characters, you could filter like this
$('#foo').keydown(function(e) {
var key = e.which;
if (!e.shiftKey && !e.altKey && !e.ctrlKey &&
key >= 48 && key <= 57 || // numbers
key >= 96 && key <= 105 || // Numeric keypad
key == 190 || key == 188 || key == 109 || key == 110 || // comma, period and minus, . on keypad
key == 8 || key == 9 || key == 13 || // Backspace and Tab and EnterEnd
key == 35 || key == 36 || // Home and
key == 37 || key == 39 || // left and right arrows
key == 46 || key == 45) // Del and Ins
return true;
return false;
});
FIDDLE
You've got two errors in your script:
you called the event parameter event but referred to e.which.
2. you have to call evt.preventDefault() for preventing the typed character to appear.
The latter point is wrong when adding jQuery event handlers. 'Normal' DOM handlers require preventDefault(), see also this comment.
→ jsFiddle
$('#foo').keypress(function(evt) {
var code = window.event ? event.keyCode : evt.which;
var chr = String.fromCharCode(code);
if ("0123456789.,".indexOf(chr) < 0) {
return false;
}
});
I'm validating my input field using key code to enter (a-z, A-Z, _). The javascript code looks something like this:
function checkForSpecialCharacters(event) {
var keycode;
keycode = event.keyCode ? event.keyCode : event.which;
if ((keycode >= 48 && keycode <= 57) || (keycode >= 65 && keycode <= 90) || (keycode == 9)
|| (keycode == 95)||(keycode == 8) || (keycode >= 97 && keycode <= 122)) {
return true;
}
else {
return false;
}
return true;
}
This works pretty fine in Chrome, IEs, but in firefox it prevents arrow key also.
I've gone through What are the ascii values of up down left right?. But In my case I need to prevent to enter all the special characters in input field.
The give solution in the above link does not fulfill my requirement.
Kindly reply with your positive response.
Thanks.
Got the solution of this problem. Just use event.which. Here which is a property of the event object. It contains the key code of the key which was pressed to trigger the event (eg: keydown, keyup etc.).
So just get the value fo the key pressed using event.which and check in if condition.
In my case in Firefox browser I was getting 0 for arrow keys. so I just added on more or condition like:
if ((keycode >= 48 && keycode <= 57) || (keycode >= 65 && keycode <= 90) || (keycode == 9)
|| (event.which == 0) || (keycode == 95)||(keycode == 8) || (keycode >= 97 && keycode <= 122)) {
return true;}
Hope it will help other pal.
Thanks
I am wanting to restrict the input characters for a text box to [a-z0-9_-]. However whenever if do this buttons like backspace and the arrow keys don't work. I have found some attempts on this website and others but either they don't work properly on all browsers or they use a black list. For example the W3Schools website example black lists numbers. Is there a way to use white list (the one above) and still allow keys like backspace, arrows, home, end etc? Or do I have to add everyone of the key codes that match the keys I want to allow? I do something like this (this is shortened for simplicity).
EDIT - Added code
<input type="text" onkeypress="return checkInput();">
function checkInput(){
return /[a-z0-9_-]/gi.test(String.fromCharCode(window.event.keyCode));
}
Just change the regex in the example to something like this:
numcheck = /[^a-z0-9_-]/;
Or better yet, avoid the double negative with:
numcheck = /[a-z0-9_-]/;
return numcheck.test(keychar);
Then you can look up the keycodes of backspace, etc. and check for them too:
if (keychar === 8) return true;
...
Or even put them in your regex:
numcheck = /[a-z0-9_\x08-]/;
You haven't provided any code samples, so it's hard to be specific in a response, but as a general strategy, try this: instead of trying to whitelist characters that can be input while they are being typed in, validate the contents of the text box after every key stroke to make sure that it still contains valid characters. If it doesn't, remove the last character entered.
This approach will allow special keys like backspace, etc., while at the same time achieve what it sounds like you are really after: a valid value in the text box.
Yes you can limit the input of characters. For example create a function that checks what is going on, return true if everything is OK and false if not:
// return true for 1234567890A-Za-z - _
function InputCheck(e) {
if ((e.shiftKey && e.keyCode == 45) || e.which != 8 && e.which != 0 && (e.which < 48 || e.which > 57)) {
if (e.which == 45 || e.which == 95 || (e.which >= 65 && e.which <= 90) || (e.which >= 97 && e.which <= 122))
return true;
return false;
}
return true;
}
once you have the function, hook it into you input (this is with jQuery):
$('#InputID').keypress(InputCheck);
You can make as complicated a check as you want, for example this will allow for USD money values:
function InputCheck(e) {
if ((e.shiftKey && e.keyCode == 45) || e.which != 8 && e.which != 0 && (e.which < 48 || e.which > 57) && e.which != 46 && e.which != 36) {
return false;
}
// . = 46
// $ = 36
var text = $(this).val();
// Dollar sign first char only
if (e.which == 36 && text.length != 0) {
return false;
}
// Only one decimal point
if (e.which == 46 && text.indexOf('.') != -1) {
return false;
}
// Only 2 numbers after decimal
if (text.indexOf('.') != -1 && (text.length - text.indexOf('.')) > 2) {
return false;
}
return true;
}
You can press any key you like, as long as you keep the value from including anything
not in the white-list.
inputelement.onkeyup=function(e){
e=e || window.event;
var who=e.target || e.srcElement;
who.value= who.value.replace(/[^\w-]+/g,'');
}
Add this code to onkeypress event.
var code;
document.all ? code = e.keyCode : code = e.which;
return ((code > 64 && code < 91) || (code > 96 && code < 123) || code == 8 || code == 32 || (code >= 48 && code <= 57));
For browser compatibility, You can add
var k = e.keyCode == 0 ? e.charCode : e.keyCode;