I've created a palindrome checker. I want to take it one step further and show
the letters being compared as it is being checked.
HTML:
<p id="typing"></p>
<input type="text" id="textBox" onkeyup="pal(this.value);" value="" />
<div id="response"></div>
<hr>
<div id="palindromeRun"></div>
JS:
To do this, I run
the recursive check... Then if it is a palindrome, I run colorLetters(), which I'm trying to highlight in
green each letter as it is being checked. Right now it is just rewriting palindromeRun's HTML with
the first letter. I know this is because of
the way I'm resetting its HTML.
I don't know how to just grab
the first and last letter, change only those letters' css, then increment i and j on
the next setTimeout loop.
var timeout2 = null;
function pal (input) {
var str = input.replace(/\s/g, '');
var str2 = str.replace(/\W/, '');
if (checkPal(str2, 0, str2.length-1)) {
$("#textBox").css({"color" : "
green"});
$("#response").html(input + " is a palindrome");
$("#palindromeRun").html(input);
colorLetters(str2, 0, str2.length-1);
}
else {
$("#textBox").css({"color" : "red"});
$("#response").html(input + " is not a palindrome");
}
if (input.length <= 0) {
$("#response").html("");
$("#textBox").css({"color" : "black"});
}
}
function checkPal (input, i, j) {
if (input.length <= 1) {
return false;
}
if (i === j || ((j-i) == 1 && input.charAt(i) === input.charAt(j))) {
return true;
}
else {
if (input.charAt(i).toLowerCase() === input.charAt(j).toLowerCase()) {
return checkPal(input, ++i, --j);
}
else {
return false;
}
}
}
function colorLetters(myinput, i, j) {
if (timeout2 == null) {
timeout2 = setTimeout(function () {
console.log("called");
var firstLetter = $("#palindromeRun").html(myinput.charAt(i))
var secondLetter = $("#palindromeRun").html(myinput.charAt(j))
$(firstLetter).css({"color" : "red"});
$(secondLetter).css({"color" : "
green"});
i++;
j++;
timeout2=null;
}, 1000);
}
}
Secondary: If possible, I'd just like to have it colors
the letters as
the user is typing... I realize this will require a setTimeout on each keyup, but also am not sure how to write this.
