Reputation: 301
I'm a bit of a noob, but managed to make something still with the help of one JavaScript snippet I found here (need to dig it up soon, but for now, my problem:)
I'm making this website: manu.mymaterial.org which I wish would sort the images the same way with dropdown menus like it works now with the checkboxes. I have tried and tried but nothing seems to work.
I have this prototype for you for fiddling out the problem, if you have the time. I would appreciate help immensely :)
I would like to be able to get same results with the dropdown menus (that don't work now) like with the checkboxes (that do work now)
https://codepen.io/manujarvinen/pen/BaRzGVa
Please tell me if I'm not clear enough or some information is missing. I'll be glad to get that info for you.
Cheers!
Manu
EDIT: The answer to this question is awesome! But even more polished answer came in here in my second question: Filter accurately with BOTH checkboxes AND dropdown menus at the same time - Excellent!
Image of the manu.mymaterial.org website, just to give you some context:
And this is how I would like it to be:
var $filterCheckboxes = $('input[type="checkbox"]');
var filterFunc = function() {
var selectedFilters = [];
$filterCheckboxes.filter(':checked').each(function() {
var v = this.value;
if (selectedFilters.indexOf(v) === -1)
selectedFilters.push(v);
});
$('.animal' && '.filterDiv')
.hide()
.filter(
function(_,a) {
var itemCat = $(a).data('category').split(' ');
return selectedFilters.every(
function(c){
return itemCat.indexOf(c) > -1;
})
})
.show();
}
$filterCheckboxes.on('change', filterFunc);
body {
width: 100%;
text-align: center;
background-color: black;
color: white;
}
.grid {
width: 300px;
margin: 50px auto;
display: grid;
grid-template-columns: 1fr 1fr 1fr;
}
.filterDiv {
width: 100px;
height: 100px;
}
<!-- Help needed in this URL: https://stackoverflow.com/q/68315184/4383420 -->
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
</head>
<div class=grid>
<select>
<option value="showAll">show all</option>
<option value="violet">violet</option>
</select>
<select>
<option value="showAll">show all</option>
<option value="blue">blue</option>
</select>
<select>
<option value="showAll">show all</option>
<option value="yellow">yellow</option>
</select>
</div>
<div class=grid>
<label>violet
<input type="checkbox" value="violet" />
<span class="checkmark"></span>
</label>
<label>blue
<input type="checkbox" value="blue" />
<span class="checkmark"></span>
</label>
<label>yellow
<input type="checkbox" value="yellow" />
<span class="checkmark"></span>
</label>
</div>
<div class=grid>
<div class="filterDiv" data-category="violet blue" style="background-color: blue"></div>
<div class="filterDiv" data-category="violet red" style="background-color: red"></div>
<div class="filterDiv" data-category="yellow" style="background-color: yellow"></div>
</div>
Upvotes: 1
Views: 287
Reputation: 9893
You expect menu works same as checkbox so you don't need to <option value="showAll">show all</option>
simply you can get all selected
option and doing like filterFunc
Here is working sample:
var $filterCheckboxes = $('input[type="checkbox"]');
var $filtermenues = $('.grid1');
filterfuncAnother = function () {
var selectedFilters = [];
$filtermenues.find(":selected").each(function () {
debugger
var v = this.value;
if (selectedFilters.indexOf(v) === -1 && v)
selectedFilters.push(v);
});
$('.animal' && '.filterDiv')
.hide()
.filter(
function (_, a) {
var itemCat = $(a).data('category').split(' ');
if (itemCat.indexOf("showAll") > -1)
return;
return selectedFilters.every(
function (c) {
return itemCat.indexOf(c) > -1;
})
})
.show();
}
var filterFunc = function () {
var selectedFilters = [];
debugger
$filterCheckboxes.filter(':checked').each(function () {
var v = this.value;
if (selectedFilters.indexOf(v) === -1)
selectedFilters.push(v);
});
$('.animal' && '.filterDiv')
.hide()
.filter(
function (_, a) {
var itemCat = $(a).data('category').split(' ');
return selectedFilters.every(
function (c) {
return itemCat.indexOf(c) > -1;
})
})
.show();
}
$filterCheckboxes.on('change', filterFunc);
$('select').on('change', filterfuncAnother);
body {
width: 100%;
text-align: center;
background-color: black;
color: white;
}
.grid {
width: 300px;
margin: 50px auto;
display: grid;
grid-template-columns: 1fr 1fr 1fr;
}
.filterDiv {
width: 100px;
height: 100px;
}
<!-- Help needed in this URL: https://stackoverflow.com/q/68315184/4383420 -->
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class=grid1>
<select>
<option value="">--</option>
<!--<option value="showAll">show all</option>-->
<option value="violet">violet</option>
</select>
<select>
<option value="">--</option>
<!--<option value="showAll">show all</option>-->
<option value="blue">blue</option>
</select>
<select>
<option value="">--</option>
<!--<option value="showAll">show all</option>-->
<option value="yellow">yellow</option>
</select>
</div>
<div class=grid>
<label>violet
<input type="checkbox" value="violet" />
<span class="checkmark"></span>
</label>
<label>blue
<input type="checkbox" value="blue" />
<span class="checkmark"></span>
</label>
<label>yellow
<input type="checkbox" value="yellow" />
<span class="checkmark"></span>
</label>
</div>
<div class=grid>
<div class="filterDiv" data-category="violet blue" style="background-color: blue"></div>
<div class="filterDiv" data-category="violet red" style="background-color: red"></div>
<div class="filterDiv" data-category="yellow" style="background-color: yellow"></div>
</div>
Upvotes: 1