Question

如何删除列表类型具有特定的属性名称，我试图从动态变量中删除具有属性名称的特定li。如何解决这个问题

var Attribute = 2;
<ul>
<li name="1">as</li>
<li name="2">asd</li>
<li name="3">asdf</li>
</select>
$("ul li").each(function () {
if($(this).hasAttr("name",Attribute){
$(this).remove();
}
else{
alert("no attribute present");
}
});

Answer 1

这有两个部分：

查找li：
```
var li = $('li[name="' + Attribute + '"]');
```
注意字符串连接。它生成一个像li[name="2"]这样的字符串。在您的示例中，我们不需要围绕属性值"，但如果它包含空格或其他字符，我们最好包含它们。

这是一个CSS选择器。关于他们的更多信息here。
使用remove删除它：
```
li.remove();
```

当然可以合并：

$('li[name="' + Attribute + '"]').remove();

直播示例：

＆＃13;

setTimeout(function() {
    var Attribute = 2;
    $('li[name="' + Attribute + '"]').remove();
}, 800);

＆＃13;

<ul>
<li name="1">as</li>
<li name="2">asd</li>
<li name="3">asdf</li>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

＆＃13;

Answer 2

您可以使用with alldates as ( SELECT convert(date,thedate) alldate FROM dbo.ExplodeDates( '2017-06-10' , '2017-06-10') as d ), empshift as ( select ed.Pk,ed.EmployeeName,sd.ShiftName,sd.Intime,sd.OutTime from EmployeeDetails ed left outer join EmployeeShiftDetails esd on (ed.Pk = esd.EmployeePk) left outer join ShiftDetails sd on (sd.Pk = esd.ShiftPk ) ) ,biometric1 as ( select min(punchtime) intime,max(punchtime) outtime,convert(date,punchtime) dates , empshift.EmployeeName,empshift.ShiftName,Intime ShiftInTime,OutTime ShiftOutTime from BioMetricPunchDetails bpd inner join empshift on (empshift.Pk =bpd.EmployeePk ) where CONVERT(date,punchtime) between '2017-06-10' and '2017-06-0' group by convert(date,punchtime), empshift.EmployeeName,empshift.ShiftName,Intime,OutTime ), fnl as ( select convert(date,intime) dates,intime, case when (intime = outtime ) then NULL else outtime end as outtime ,ShiftName,DATEDIFF(MINUTE,intime,outtime) duration,ShiftInTime,convert(time,intime) InTimeHrMin ,datediff(MINUTE,ShiftInTime, convert(time,intime)) lateby from biometric1 ), fnl1 as (select * from alldates left outer join fnl on (alldates.alldate = fnl.dates)) select ed.EmployeeName,fnl1.* from EmployeeDetails ed cross join fnl1检查属性。 attr()为您提供的属性值。

＆＃13;

attr()

＆＃13;

var Attribute = 2;

$("ul li").each(function () {

  if($(this).attr('name') == Attribute){
     $(this).remove();
   }
   else{
     //alert("no attribute present");
   }
});

＆＃13;

Answer 3

尝试使用简单的直接选择器$("ul").find('li[name="'+Attribute+'"]').remove()

＆＃13;

var Attribute = 2;
$("ul").find('li[name="'+Attribute+'"]').remove()

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<ul>
  <li name="1">as</li>
  <li name="2">asd</li>
  <li name="3">asdf</li>
</ul>

＆＃13;

如何从列表类型中删除特定属性名称

3 个答案: