Reputation: 1135
I want to filter (show/hide) elements by class. I want to have lets say 5 filters working together.
<select id='filter4'>...</select>
<select id='filter5'>...</select>
<select id='filter6'>...</select>
<select id='filter7'>...</select>
and
<select id='filter6' multiple>...</select>
jquery:
jQuery(document).ready(function ($) {
$("select").on("change", function () {
var filterVal = $("select#filter4").val();
var filterVal2 = $("select#filter5").val();
var filterVal3 = $("select#filter6").val();
var filterVal4 = $("select#filter7").val();
var filterVal5 = $("select#filter8").val();
var filterSelector = "";
var filter2Selector = "";
var filter3Selector = "";
var filter4Selector = "";
var filter5Selector = "";
if (filterVal == "all" && filterVal2 == "all" && filterVal3 == "all" && filterVal4 == "all" && filterVal5 === null) {
//show all items
$(".item").fadeIn("fast");
} else {
if (filterVal != "all") {
filterSelector = "." + filterVal;
}
if (filterVal2 != "all") {
filter2Selector = "." + filterVal2;
}
if (filterVal3 != "all") {
filter3Selector = "." + filterVal3;
}
if (filterVal4 != "all") {
filter4Selector = "." + filterVal4;
}
if (filterVal5 !== null) {
filter5Selector = "." + filterVal5;
}
$(".item").hide();
$(filterSelector + filter2Selector + filter3Selector + filter4Selector + filter5Selector).fadeIn("fast");
}
});
});
See everything in this jsfiddle !
The first four single filters work with any problem. The fifth multi-select element does not work (filter) properly when more than one option is selected.
EDIT: It was needed to replace commas with dots (multiple css selectors) with help of g modifier (/ /g) which is global match:
$(filter4Selector + filter5Selector.replace(/,/g, ".")).fadeIn("fast");
see http://jsfiddle.net/mahish/dum7n/ .
Answer below offers different code which works as well!
Upvotes: 0
Views: 4802
Reputation: 9326
Cleaned up your code and fixed the multiple queries. Note that:
.class1.class2.class3
means class1 && class2 && class 3
.class1, .class2, .class3
means class1 || class2 || class 3
That was cause of earlier confusion in the comments.
jQuery(document).ready(function ($) {
var values = [7, 8];
$("select").on("change", function () {
var showAll = true,
show = [],
joined;
$.each(values, function (index, id) {
var $el = $('#filter' + id),
multi = $el.attr('multiple'),
val = $el.val();
if (multi) {
if (val !== null) {
showAll = false;
$.each(val, function (i, v) {
show.push( v );
});
}
} else {
if (val != 'all') {
showAll = false;
show.push( val );
}
}
});
console.log(showAll);
console.log(show);
if (showAll) {
//show all items
$(".item").fadeIn("fast");
} else {
joined = '.' + show.join('.');
console.log( joined );
$(".item").hide();
$( joined ).fadeIn("fast");
}
});
$.each(values, function (index, id) {
$('#filter' + id).chosen();
});
});
Upvotes: 2
Reputation: 22405
Separate the selectors with commas and that should work (google jQuery multiple selectors for more information)
Upvotes: 0