Is there a way to find number of items (li
tag) inside a ul
which has it's float: set to left. Assume i got loads of folders which are visually represented by li
tags. Due to floating behavior as soon as the li
does not fit in single row they will be pushed down giving it Rows and columns look.my questions is
Using jQuery ..etc Is there a way to determine the number of li
for each row
<ul>
<li>A</li>
<li>B</li>
<li>C</li>
<li>D</li>
<li>E</li>
<li>F</li>
<li>G</li>
<li>H</li>
<li>I</li>
<li>J</li>
</ul>
ul,li{margin:0 0;padding:0 0;}
ul li{display:inline;float:left}
For my resolution li's upto F makes one row and the rest makes another like below. I would like a jquery way to get number of elements ( which is 6 in my case ) in row 1.
A B C D E F
G H I J
I am adding keyboard events to navigate to each folder(li) using left,right,top,down. Left right is as simple as highlighting next and previous li's. Top and down keys need to go to next row and above row. Now you will get it :)
We can do this by checking the number of LIs with their top (y-axis) positions matching that of their previous siblings.
Demo: http://jsfiddle.net/KgBpx/1/
Updated demo: http://jsfiddle.net/KgBpx/2/
Note: recalculation done on window resize as well to fully demonstrate the logic.
Code:
function calculateLIsInRow() {
var lisInRow = 0;
$('ul li').each(function() {
if($(this).prev().length > 0) {
if($(this).position().top != $(this).prev().position().top) return false;
lisInRow++;
}
else {
lisInRow++;
}
});
console.log('No: of lis in a row = ' + lisInRow);
}
calculateLIsInRow();
$(window).resize(calculateLIsInRow);
NOTE
The count of LIs will be same for all rows except maybe the last if the total number of LIs is not a multiple of lisInRow. The number of LIs in the last row can easily be found out by doing: $('ul li').length % lisInRow
Using filter() makes for smaller amount of code.
function filterByTop(top) {
return function(i) { return $(this).offset().top == top; };
}
var el = $('li:first');
var rowSz = $('li').filter(filterByTop(el.offset().top)).length;
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