Get Two Lowest Values From A Table With Jquery

Question

I have a table:

Solution 1:

First of all, check the Fiddle Demo

Consider your HTML, here is my css & js solution:

javascript

var tds = $('td.value');
var arr = $.makeArray(tds);
arr.sort(function(a,b){
    returnparseInt($(b).text()) - parseInt($(a).text());
});

$(arr[arr.length-2]).addClass('lowest');
$(arr[arr.length-1]).addClass('lowest');

css

td.lowest {background-color:lightgreen;}

Solution 2:

var ok = $('table tr td:last-child').map(function () {
    return $(this).text();
}).get().sort(function (x, y) {
    return x - y
});

alert(ok[0] + ' - is the smallest');
alert(ok[1] + ' - is the 2nd smallest');

http://jsfiddle.net/263Ps/

Solution 3:

Try

var map = {}, array = [];
var $tds = $('table tr').slice(1).find('td:last-child').each(function () {
    var value = +$(this).text();
    if (map[value]) {
        map[value].push(this);
    } else {
        map[value] = [this];
        array.push(value);
    }
})
array.sort(function(a1, a2){
    return a1 - a2;
});
$tds.filter(function () {
    return $.inArray(+$(this).text(), array) < 2;
}).css('color', 'red')

Demo: Fiddle

Solution 4:

var values = [],
    tds = $("table tr").find("td:last");

tds.each(function(){ 
    $(this).data("val",$(this).text());
    values.push(parseInt($(this).text()))
})
values.sort(function(x,y){return x > y});
tds.each(function(){
    if($(this).data("val") == values[0] || $(this).data("val") == values[1])
       $(this).css("color","red");
})

values[0] and values[1] will be smallest and second smallest values respectively.

Solution 5:

This should do the trick:-

var values = $('td:last-child').sort(function(a, b) {
   return $(a).text() - $(b).text();
}).slice(0, 2).css('color', 'red').map(function() {
   return $(this).text();
}).get();

So, sort each td:last-child by its text() value (low to high), then use slice() to get the first two elements. Apply the css() to these elements, then use map() to return their text(), which will be assigned to var values

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