我正在使用jQuery,这是我的代码到目前为止:
$('.CategoryInput').each(function() {
$(this).click(function() {
var CategoryID = $( this ).attr( "category" );
var Clicked = $( this ).attr( "clicked" );
if(Clicked == 0){
$("#Category" + CategoryID).css({backgroundColor: '#fc3b66'});
$("#Category" + CategoryID).find( ".state" ).css({color: '#fff'});
$(".subcat" + CategoryID).css({backgroundColor: '#fc3b66'});
$(".subcat" + CategoryID).find( ".state" ).css({color: '#fff'});
$(".SubCategoryInput" + CategoryID).prop('checked', true);
$(".SubCategoryInput" + CategoryID).attr('clicked','1');
$(this).attr('clicked','1');
} else {
$("#Category" + CategoryID).css({backgroundColor: '#fdfdfd'});
$("#Category" + CategoryID).find( ".state" ).css({color: '#6a6d76'});
$(".subcat" + CategoryID).css({backgroundColor: '#fdfdfd'});
$(".subcat" + CategoryID).find( ".state" ).css({color: '#6a6d76'});
$(".SubCategoryInput" + CategoryID).prop('checked', false);
$(".SubCategoryInput" + CategoryID).attr('clicked','0');
$(this).attr('clicked','0');
}
});
});
这是我的HTML:
<div class="container">
<h1>Selecciona los productos que elaboras:</h1>
<?PHP
$r = mysql_query("SELECT * FROM `Category` WHERE `subCategory`='0' ORDER by Position ASC");
while($rowi = mysql_fetch_array($r))
{
$id = $rowi['id'];
$Name = $rowi['Name'];
$Description = $rowi['Description'];
?>
<div class="row">
<div class="maintag" id="Category<?PHP echo $id;?>">
<label class="state" for="categories"><?PHP echo $Name;?></label>
<input type="checkbox" class="CategoryInput" name="categories" category="<?PHP echo $id;?>" clicked="0">
(<?PHP echo $Description;?>)
</div>
</div>
<div class="row">
<?PHP
$r1 = mysql_query("SELECT * FROM `Category` WHERE `subCategory`='$id' ORDER by Position ASC");
while($rowi1 = mysql_fetch_array($r1))
{
$id1 = $rowi1['id'];
$Name1 = $rowi1['Name'];
?>
<div class="col-md-4" style="padding-left:0px;">
<div id="subtag<?PHP echo $id1;?>" class="subcat<?PHP echo $id;?> supersub">
<label class="state" for="categories"><?PHP echo $Name1;?></label>
<input type="checkbox" name="subcategory" class="SubCategoryInput<?PHP echo $id;?>" SubCategory="<?PHP echo $id1;?>" clicked="0">
</div>
</div>
<?PHP } ?>
</div>
<?PHP } ?>
</div>
我有几个复选框,我用作类别.这些类别具有子类别,因此此代码只是突出显示所选的复选框.如果所选复选框来自母类别,则会突出显示所有子类别的所有复选框.还将它们设置为已选中.
我到目前为止所取得的一切.现在我被困在如何获取所选复选框的id并将其添加到如下所示的变量中:
var SelectedCategoryIDs = '12,435,123,124,56,34,23';
因此,当我单击一个母类别时,它将向SelectedCategoryIDs添加所有子类别的ID.此外,当我取消选中已经检查过的母类别时,它将从SelectedCategoryID中删除所有与所有子类别对应的ID.
你能帮我实现吗?
我希望我解释得很好,如果您有任何疑问,请告诉我.
最佳答案 只需创建一个数组,然后用逗号分隔加入它:
var selectedArray = [];
$('.CategoryInput').each(function() {
$(this).click(function() {
selectedArray.push($(this).attr('id'));
});
});
/* and then join: */
var selectedString = selectedArray.join(","); // you'll give all ids comma separated