我有两个使用最后版本jQuery Chosen plugin的相同选择。我想要实现的是如果你在第一个选择中选择选项,在第二个选择中禁用此选项,反之亦然。我尝试使用jQuery prop方法解决它,但它不起作用。
这是代码:
<select data-placeholder= "Keyword" multiple style="width:300px;" name="keywordContainSelect" multiple class="selectKeyword">
<option value=""> </option>
<option value="1">apartments</option>
<option value="2">edward</option>
<option value="3">shutters</option>
<option value="4">sprtsmen</option>
<option value="5">dinner</option>
<option value="6">bachelor</option>
<option value="7">remember</option>
<option value="8">pleasant</option>
<option value="9">connection</option>
<option value="10">instrument</option>
<option value="11">preference</option>
<option value="12">valley</option>
<option value="13">dashwoods</option>
<option value="14">marriage</option>
<option value="15">literature</option>
<option value="16">imprudence</option>
<option value="17">cottage</option>
<option value="18">ferrars</option>
<option value="19">gentleman</option>
<option value="20">sweetness</option>
<option value="21">barton</option>
<option value="22">provision</option>
<option value="23">account</option>
<option value="24">daughter</option>
<option value="25">goodness</option>
<option value="26">songs</option>
<option value="27">view</option>
<option value="28">assure</option>
<option value="29">extremity</option>
</select>
<script>
$("select[name=keywordContainSelect]").chosen();
</script>
<select data-placeholder= "Keyword" multiple style="width:300px;" name="keywordNotContainSelect" multiple class="selectKeyword">
<option value=""> </option>
<option value="1">apartments</option>
<option value="2">edward</option>
<option value="3">shutters</option>
<option value="4">sprtsmen</option>
<option value="5">dinner</option>
<option value="6">bachelor</option>
<option value="7">remember</option>
<option value="8">pleasant</option>
<option value="9">connection</option>
<option value="10">instrument</option>
<option value="11">preference</option>
<option value="12">valley</option>
<option value="13">dashwoods</option>
<option value="14">marriage</option>
<option value="15">literature</option>
<option value="16">imprudence</option>
<option value="17">cottage</option>
<option value="18">ferrars</option>
<option value="19">gentleman</option>
<option value="20">sweetness</option>
<option value="21">barton</option>
<option value="22">provision</option>
<option value="23">account</option>
<option value="24">daughter</option>
<option value="25">goodness</option>
<option value="26">songs</option>
<option value="27">view</option>
<option value="28">assure</option>
<option value="29">extremity</option>
</select>
<script>
$("select[name=keywordNotContainSelect]").chosen();
$("select[name=keywordContainSelect]").on('change', function(evt, params) {
var valSel = params.selected;
var valDeSel = params.deselected;
if (valSel > 0){
$('select[name=keywordNotContainSelect] option[value='+valSel+']').prop("disabled", true);
}
if (valDeSel > 0)
$('select[name=keywordNotContainSelect] option[value='+valSel+']').prop("disabled", false);
});
$("select[name=keywordNotContainSelect]").on('change', function(evt, params) {
var valSel = params.selected;
var valDeSel = params.deselected;
if (valSel > 0)
$('select[name=keywordContainSelect] option[value='+valSel+']').prop("disabled", true);
if (valDeSel > 0)
$('select[name=keywordContainSelect] option[value='+valSel+']').prop("disabled", false);
});
</script>
同样的问题和解决方案是here,但该解决方案不适用于新版本的jQuery Chosen插件。
答案 0 :(得分:2)
您在http://jsfiddle.net/aksvuy9q/2/
$(".selectKeyword").chosen().change( function() {
var selectedValue = $(this).find('option:selected').val();
$(".selectKeyword").find('option[value="'+ selectedValue +'"]:not(:selected)').attr('disabled','disabled');
$(".selectKeyword").trigger("chosen:updated");
});