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JavaScript Regular Expressions in JavaScript Validating a Form Form Validation

Posted by Kelinia Johnson
Kelinia Johnson
10,189 Points

I need help! Quickly!!! In this challenge you'll write a validator for a hexadecimal string used to set a CSS color.

I don't know how to write this string!

app.js 
// Type inside this function
function isValidHex(text) {
  const hexRegEx = /^(#)$ [a-f] 6 $/i.test(text);
}

const hex = document.getElementById("hex");
const body = document.getElementsByTagName("body")[0];

hex.addEventListener("input", e => {
  const text = e.target.value;
  const valid = isValidHex(text);
  if (valid) {
    body.style.backgroundColor = "rgb(176, 208, 168)";
  } else {
    body.style.backgroundColor = "rgb(189, 86, 86)";
  }
});
index.html 
<!DOCTYPE html>
<html>

<head>
    <title>DOM Manipulation</title>
</head>
<link rel="stylesheet" href="style.css" />

<body>
    <div id="content">
        <p>Enter a valid hex value below to make the screen turn green.</p>
        <input type="text" id="hex">
    </div>
    <script src="app.js"></script>
</body>

</html>

here's another attempt:

var hexRegEx = /"#"+[a-z]$/.test(text)

1 Answer

jb30
jb30
44,806 Points
jb30
jb30
44,806 Points

You have /^(#)$ [a-f] 6 $/i. I think that the $ character should be at the end of your regular expression, before the /i, but not following the (#). You might not want a space between (#) and [a-f], unless you are expecting your input to have it. If you want a value from [a-f] to repeat 6 times, you could try [a-f]{6} instead of having a space followed by the character 6 followed by another space followed by the end of the line.

Instead of limiting your hexadecimal numbers to using the characters a, b, c, d, e, f, or their uppercase equivalents, you might also allow them to use the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9, by changing [a-f] to [\da-f] or [a-f\d].

If you know that all of your colors have six hexadecimal digits, then those changes should be sufficient. However, hexadecimal colors such as #aabbcc can also be written as #abc, which your code would claim to be invalid. If you want to allow both three digit and six digit hexadecimal numbers, try /^#([\da-f]{3}){1,2}$/i.

Kelinia Johnson
Kelinia Johnson
10,189 Points

it says i need to limit my regex to meet the requirements