jQuery don't work inside PHP echo - php

Question

Value of $qty is in my database. I've read few similar questions and I already added $(document).ready but it's still not working.
echo "<div class='sp-quantity'>
<div class='sp-minus fff'><a class='ddd' href='#' data-multi='-1'>-</a></div>
<div class='col-xs-3 sp-input'>
<input type='text' class='quntity-input form-control qty' pid='$bookid' id='qty-$bookid' value='$qty'/>
</div>
<div class='sp-plus fff'><a class='ddd' href='#' data-multi='1'>+</a></div>
</div>";
echo "<script>";
echo "$(document).ready(function(){";
echo "$('.ddd').on('click', function() {";
echo "var $button = $(this);";
echo "var $input = $button.closest('.sp-quantity').find('input.quntity-input');";
echo "$input.val(function(i, value) {";
echo "return +value + (1 * +$button.data('multi'));";
echo "});";
echo "});";
echo "});";
echo "</script>";

Answer 1

You're getting your PHP & Javascript variable syntax confused. PHP uses $ as the start of every variable name. Javascript doesn't, plus jquery uses $ in a special way. On top of that, PHP does automatic string substitution of anything starting with a $ that can be a valid variable name as long as the string uses " instead of '. So the result of your code is:
<script>
$(document).ready(function(){
$('.ddd').on('click', function() {
var $button = $(this);
var $input = $button.closest('.sp-quantity').find('input.quntity-input');
$input.val(function(i, value) {
return +value + (1 * +$button.data('multi'));
});
});
});
</script>
$button and $input will be substituted with the contents of the PHP variables of those names. I suspect that's not what you want - though I have done things like that at times myself. What I think you want is:
<script>
$(document).ready(function(){
$('.ddd').on('click', function() {
var button = $(this);
var input = button.closest('.sp-quantity').find('input.quntity-input');
input.val(function(i, value) {
return +value + (1 * + button.data('multi'));
});
});
});
</script>
Two other suggestions:
1 - If you have absolutely no PHP values within a bunch of Javascript (or plain HTML) echo lines, you can exit out of PHP code mode by using ?> and not use echo and avoid any issues of string substitution.
2 - Personally, I prefer using {$string} for my variables inside strings as it makes the substitution more obvious and sometimes less ambiguous.