1

我有一个在我的网站上重复的功能

jQuery

$(function() {

//check the enabled state on load
if(!$(".notifyOwner").is(':checked')){
    $(".ownerDay").attr("disabled", "disabled");
     $(".ownerDaytype").attr("disabled", "disabled");
}

//toggle the enabled state when the checkbox is clicked
$(".notifyOwner").click(function() {

    if($(this).is(":checked")) {
   $(".ownerDay").removeAttr("disabled");
   $(".ownerDaytype").removeAttr("disabled");
   $(".ownerActive").removeClass("disabled");
} else {
     $(".ownerDay").attr("disabled", "disabled");
   $(".ownerDaytype").attr("disabled", "disabled");
    $(".ownerActive").addClass("disabled");
}



});

});

我试图将其作为可重用的代码,如下所示

function selectToggle(obj, dayclass, daytypeclass, txtactive){
var $event = $(obj);

if(!($event).is(":checked"))
{
    $(dayclass).attr("disabled", "disabled");
    $(daytypeclass).attr("disabled", "disabled");
}

 $event.click(function() {

    if($event.is(":checked"))
    {
     $(dayclass).removeAttr("disabled");
   $(daytypeclass).removeAttr("disabled");
   $(txtactive).removeClass("disabled");
} else {
     $(dayclass).attr("disabled", "disabled");
   $(daytypeclass).attr("disabled", "disabled");
    $(txtactive).addClass("disabled");
}
 });
}

并将其称为 onclick="selectToggle(this,'ownerDay','ownerDaytype','ownerActive')"

但它不起作用。我在这个脚本中的错误在哪里。

4

1 回答 1

1

您不会将 jquery select 称为类。即“ownerDay”在您的通话中应该是“.ownerDay”。其他人也一样。

于 2013-07-31T07:04:10.110 回答