How can I move selected option in multiselect up a

Question

I have select

<select multiple id="select2"> 
     <option value="1">1</option> 
     <option value="2">2</option> 
     <option value="3">3</option>
</select>

And two buttons

<input type="button" value="Up" onclick="up()">
<input type="button" value="Down" onclick="down()">

How can I move selected option in multiselect up and down by buttons using jquery?

Answer 1

if you do not use jquery

function moveUp(){
    var select = document.getElementById("columnOrder");
    var options = select && select.options;
    var selected = [];

    for (var i = 0, iLen = options.length; i < iLen; i++) {
        if (options[i].selected) {
            selected.push(options[i]);
        }
    }

    for (i = 0, iLen = selected.length; i < iLen; i++) {
        var index = selected[i].index;

        if(index == 0){
            break;
        }

        var temp = selected[i].text;
        selected[i].text = options[index - 1].text;
        options[index - 1].text = temp;

        temp = selected[i].value;
        selected[i].value = options[index - 1].value;
        options[index - 1].value = temp;

        selected[i].selected = false;
        options[index - 1].selected = true;
    }
}

function moveDown(){
    var select = document.getElementById("columnOrder");
    var options = select && select.options;
    var selected = [];

    for (var i = 0, iLen = options.length; i < iLen; i++) {
        if (options[i].selected) {
            selected.push(options[i]);
        }
    }

    for (i = selected.length - 1, iLen = 0; i >= iLen; i--) {
        var index = selected[i].index;

        if(index == (options.length - 1)){
            break;
        }

        var temp = selected[i].text;
        selected[i].text = options[index + 1].text;
        options[index + 1].text = temp;

        temp = selected[i].value;
        selected[i].value = options[index + 1].value;
        options[index + 1].value = temp;

        selected[i].selected = false;
        options[index + 1].selected = true;
    }
}

Answer 2

<script type="text/javascript">
  $(document).ready(function() {
    $('li').click(function() {
      // the clicked LI
      var clicked = $(this);

      // all the LIs above the clicked one
      var previousAll = clicked.prevAll();

      // only proceed if it's not already on top (no previous siblings)
      if(previousAll.length > 0) {
        // top LI
        var top = $(previousAll[previousAll.length - 1]);

        // immediately previous LI
        var previous = $(previousAll[0]);

        // how far up do we need to move the clicked LI?
        var moveUp = clicked.attr('offsetTop') - top.attr('offsetTop');

        // how far down do we need to move the previous siblings?
        var moveDown = (clicked.offset().top + clicked.outerHeight()) - (previous.offset().top + previous.outerHeight());

        // let's move stuff
        clicked.css('position', 'relative');
        previousAll.css('position', 'relative');
        clicked.animate({'top': -moveUp});
        previousAll.animate({'top': moveDown}, {complete: function() {
          // rearrange the DOM and restore positioning when we're done moving
          clicked.parent().prepend(clicked);
          clicked.css({'position': 'static', 'top': 0});
          previousAll.css({'position': 'static', 'top': 0}); 
        }});
      }
    });
  });
</script>

<ul>
 <li><a>Hank</a></li>
 <li><a>Alice</a></li>
 <li><a>Tom</a></li>
 <li><a>Ashlee</a></li>
</ul>

Answer 3

Update: Fixed code for multiple options selected, based on @patrick dw's suggestion.

$(document).ready(function(){
    $('input[type="button"]').click(function(){
        var $op = $('#select2 option:selected'),
            $this = $(this);
        if($op.length){
            ($this.val() == 'Up') ? 
                $op.first().prev().before($op) : 
                $op.last().next().after($op);
        }
    });
});

Test it here »

No need to use inline onclick="" event listeners. jQuery takes full control of separating presentation from functionality.

Answer 4

function up() {
    var selected = $("#select2").find(":selected");
    var before = selected.prev();
    if (before.length > 0)
        selected.detach().insertBefore(before);
}

function down() {
    var selected = $("#select2").find(":selected");
    var next = selected.next();
    if (next.length > 0)
        selected.detach().insertAfter(next);
}

Answer 5

I created a jquery plugin for this: https://github.com/UziTech/jquery.moveSelected.js

usage:

$("button#up").click(function(){
  $("select").moveSelectedUp();
});
$("button#down").click(function(){
  $("select").moveSelectedDown();
});

http://jsfiddle.net/UziTech/qr5qfhgg/

Answer 6

here is the same idea as the previously-posted non-jquery example, but with some strategic code re-use. My need was to have the buttons always operate on the same select element, named "cols". You could put "sel" as a parameter of the moveUp() and moveDown() functions if you want something more generic.

function moveUp() {
    var sel = document.getElementById("cols");
    var i1=0, i2=1;
    while (i2 < sel.options.length) {
        swapIf(sel,i1++,i2++);
    }
}
function moveDown() {
    var sel = document.getElementById("cols");
    var i1=sel.options.length-1, i2=i1-1;
    while (i1 > 0) {
        swapIf(sel,i1--,i2--);
    }
}
var swapVar = '';
function swapIf(sel,i1,i2) {
    if ( ! sel[i1].selected && sel[i2].selected) {
        swapVar = sel[i2].text;
        sel[i2].text = sel[i1].text;
        sel[i1].text = swapVar;
        swapVar = sel[i2].value;
        sel[i2].value = sel[i1].value;
        sel[i1].value = swapVar;
        sel[i1].selected = true;
        sel[i2].selected = false;
    }
}