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I don't want to allowed only the special character like < > and &

What I have tried:

and here's the code that i search on the internet but this code did not accept all of the special character.
HTML
<pre><html xmlns="http://www.w3.org/1999/xhtml">
<head>
    <title></title>
    <style type="text/css">
        body
        {
            font-size: 9pt;
            font-family: Arial;
        }
    </style>
</head>
<body>
    Alphanumeric value:
    <input type="text" id="text1" onkeypress="return IsAlphaNumeric(event);" ondrop="return false;"
         />
    <span id="error" style="color: Red; display: none">* Special Characters not allowed</span>
    <script type="text/javascript">
        var specialKeys = new Array();
        specialKeys.push(8); //Backspace
        specialKeys.push(9); //Tab
        specialKeys.push(46); //Delete
        specialKeys.push(36); //Home
        specialKeys.push(35); //End
        specialKeys.push(37); //Left
        specialKeys.push(39); //Right
        function IsAlphaNumeric(e) {
            var keyCode = e.keyCode == 1 ? e.charCode : e.keyCode;
            var ret = ((keyCode >= 48 && keyCode <= 57) || (keyCode >= 65 && keyCode <= 90) || (keyCode >= 97 && keyCode <= 122) || (specialKeys.indexOf(e.keyCode) != -1 && e.charCode != e.keyCode));
            document.getElementById("error").style.display = ret ? "none" : "inline";
            return ret;
        }
    </script>
</body>
</html>
Posted
Updated 25-Oct-17 5:00am
v2

1 solution

Try this function instead:
JavaScript
function IsAlphaNumeric(e) {
        var keyCode = e.keyCode == 1 ? e.charCode : e.keyCode;
        var ret = ((keyCode > 31 && keyCode < 127) && keyCode != 38 && keyCode != 60 && keyCode != 62);
            document.getElementById("error").style.display = ret ? "none" : "inline";
            return ret;
}
It will allow any character with ASCII code[^] from 32 to 126 apart from < and > and &

YOu can easily add or exclude others.

Also, just to add: do not rely on JavaScript (client side) validation for user input. You must always run the same checks in code (VB, C#, whatever) on the server as well.
 
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v2
Comments
Member 13427032 27-Oct-17 8:07am    
Thank you so much this code is working and i will be adding a validation code in c#. Thanks again
Member 13427032 3-Nov-17 15:05pm    
hi sir good day, i add keyCode == 13 but the enter still is not working. and i check on the ASCII code the value of the enter is not there. can you help me about the ENTER key
A_Griffin 3-Nov-17 15:13pm    
what do you expect to happen when you press enter? IT's a single line text box...
Member 13427032 3-Nov-17 15:58pm    
i have a textbox that is multiline and when i push the enter button it shows the error message (Special Characters not allowed) can you help me when pressing the enter it will add a new line on the textbox. Please
A_Griffin 3-Nov-17 16:20pm    
Just change the one line to:
var ret = (keyCode == 13) || (((keyCode > 31 && keyCode < 127) && keyCode != 38 && keyCode != 60 && keyCode != 62));

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

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