javascript 使用 jQuery 删除值后刷新/重新加载下拉列表？
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Refresh/reload a dropdown list after removing values with jQuery?
提问by DreamTeK
The following code removes dropdown list values based on text entered in a textbox.
以下代码根据在文本框中输入的文本删除下拉列表值。
FIDDLE DEMO
小提琴演示
JQUERY
查询
var ddl = ('#ddl'),
txt = ('#txt');
$(txt).change(function() {
var x = $(txt).val(),
li = $(ddl).html();
if (x.length != 0) {
$(ddl).html(li);
$(ddl + ' :not([value="' + x + '"])').remove();
}
})<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input id="txt" type="text" placeholder="Enter a number...">
<select id="ddl">
<option value="0" selected="selected">Select...</option>
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
<option value="5">5</option>
</select>QUESTION
问题
How to restore dropdownlist values back to initial state when a new value is entered into the textbox?
在文本框中输入新值时，如何将下拉列表值恢复到初始状态？
I have been attempting to use:
我一直在尝试使用：
$(ddl).selectmenu("refresh");
but this stops the script from working as expected.
但这会阻止脚本按预期工作。
回答by leopik
Like so
像这样
...
$(ddl).html(li);
$(ddl + ' :not([value="' + x + '"])').hide();
} else {
$(ddl + ' :not([value="' + x + '"])').show();
}
...
Instead of removing the item completely, you simply hide. When you empty the input field, re-show all the items.
您只需隐藏即可，而不是完全删除该项目。当您清空输入字段时，重新显示所有项目。
回答by Pedro Ferreira
You could try something like this:
你可以尝试这样的事情：
var ddl = ('#ddl'),
txt = ('#txt');
$(txt).change(function () {
var x = $(txt).val(),
li = $(ddl).html();
$(ddl + ' option').show();
if (x.length != 0) {
$(ddl).html(li);
$(ddl + ' option:not([value="' + x + '"])').hide();
$(ddl + ' option[value="' + x + '"]').prop('selected',true);
}
});
