我有3个下拉值,包含Select
,One
,Two
,Three
,以下是HTML。
<select class="dpdown" id="box1">
<option value="Select">Select</option>
<option value="One">One</option>
<option value="Two">Tow</option>
<option value="Three">Three</option>
</select>
<select class="dpdown" id="box2">
<option value="Select">Select</option>
<option value="One">One</option>
<option value="Two">Tow</option>
<option value="Three">Three</option>
</select>
<select class="dpdown" id="box3">
<option value="Select">Select</option>
<option value="One">One</option>
<option value="Two">Tow</option>
<option value="Three">Three</option>
</select>
条件
One
选择#box1
,那么应该从所有其他框中禁用One
,并且从Two
中选择#box1
再次,然后从其他框中禁用Two
并启用One
。即最后一个值被禁用而不是所有值。修改
$(document).ready(function(){
$('#box1').data('pre', $(this).val()); // added this line to get the pre value.
$("select").change(function(e){
var id = $(this).attr("id");
var before_change = $(this).data('pre');
alert(before_change); // alert shows blank values
});
});
我想用jQuery做这件事。请帮忙。
答案 0 :(得分:0)
尝试这样的事情:
var $selects = $('.dpdown').change(function() {
var val = $(this).val();
$selects
.children().prop('disabled', false)
.end().not(this)
.find('[value="' + val + '"]').prop('disabled', true);
});
&#13;
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<select class="dpdown" id="box1">
<option value="Select">Select</option>
<option value="One">One</option>
<option value="Two">Two</option>
<option value="Three">Three</option>
</select>
<select class="dpdown" id="box2">
<option value="Select">Select</option>
<option value="One">One</option>
<option value="Two">Two</option>
<option value="Three">Three</option>
</select>
<select class="dpdown" id="box3">
<option value="Select">Select</option>
<option value="One">One</option>
<option value="Two">Two</option>
<option value="Three">Three</option>
</select>
&#13;
我们的想法是将更改事件绑定到所有选择框。然后,当事件发生时,您启用所有选择框中的所有选项,然后您找到具有来自其他选择(当前this
除外)的匹配值的选项并禁用它们。
答案 1 :(得分:0)
你编码简单:
<script>
$('select').on('change', function() {
$("select option").removeAttr('disabled');
$("select option[value=" + $(this).val() + "]").attr('disabled','disabled')
});
</script>
答案 2 :(得分:0)
像这样:
$(".dpdown").on("change", function() {
var sel = $(this).val();
$(".dpdown").find("option").prop("disabled", false);
$(".dpdown").not(this).find("option[value='" + sel + "']").prop("disabled", true);
});
&#13;
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<select class="dpdown" id="box1">
<option value="Select">Select</option>
<option value="One">One</option>
<option value="Two">Tow</option>
<option value="Three">Three</option>
</select>
<select class="dpdown" id="box2">
<option value="Select">Select</option>
<option value="One">One</option>
<option value="Two">Tow</option>
<option value="Three">Three</option>
</select>
<select class="dpdown" id="box3">
<option value="Select">Select</option>
<option value="One">One</option>
<option value="Two">Tow</option>
<option value="Three">Three</option>
</select>
&#13;
答案 3 :(得分:0)
尝试这样的事情:
<强>已更新强>
var $selects = $(".dpdown").change(function() {
var selectedValues = [];
$.each($selects, function(index, select) {
var value = $(select).val();
if (value != "Select") {
selectedValues.push(value);
}
});
$selects
.find("option")
.prop("disabled", false);
$.each(selectedValues, function(index, value) {
$selects
.find("option[value='" + value +"']")
.not(":selected")
.prop("disabled", true);
});
});