Question

这是我的html（由ajax呈现）

<div class="filteredItem" guildid="1">
  <div class="filteredItem_image" style="background-image: url(http://visitbrabant.nl.domainpreview.nl/dev/files/images/guilds/1_g1_1_1.jpg)" onclick="DoNav('?action=item&amp;sid=1')"></div>
  <div class="filteredItem_right">
     <div class="filteredItem_name">Park Hilaria</div>
     <div class="filteredItem_city">Eindhoven</div>
     <div class="filteredItem_duration">undefined</div>
     <div class="filteredItem_short">De Kennedylaan in Eindhoven weer in Park Hilaria. Het tijdelijke pretpark is veel meer dan alleen een kermis. Het is zomers topvermaak voor alle leeftijden.</div>
     <div class="addToList" onclick="addToList(1)"></div>
  </div>

这是onclick addtolist函数：

function addToList (GuildId){
$.getJSON('http://visitbrabant.nl.domainpreview.nl/dev/vb/inc/api/set.useritem.ajax/?gId='+GuildId+'&pId=0&eId='+eId, function( data ) {
    if(data.status == 'error' && data.error =='user is not logged in'){
        alert('TODO:gebruiker is niet ingelogd, opslaan in de cookie');
    }else{
        console.log(data.planned);
        console.log($('.filteredItem[guildid='+2+'] .addToList'));
        if(data.planned == 1){
            $('.filteredItem[guildid='+GuildId+'] .addToList').addClass('inlist');
        }else{
            $('.filteredItem[guildid='+GuildId+'] .addToList').removeClass('inlist');
        }
    }
});
}

执行以下行时：

console.log($('.filteredItem[guildid='+2+'] .addToList'));

它在控制台中返回以下规则：

div.addToList, prevObject: n.fn.init[1], context: document, selector: ".filteredItem[guildid=2] .addToList", jquery: "1.11.0", constructor: function…]

当我在浏览器控制台中调用它时：

console.log($('.filteredItem[guildid='+2+'] .addToList'));

它返回相同但我称之为：

$('.filteredItem[guildid='+2+'] .addToList')

（在浏览器控制台中）IT工作很精细。

抱歉我的英语不好，我希望你能帮助我...

Answer 1

在将日志提交到列表后尝试阅读日志：

 function addToList (GuildId){
$.getJSON('http://visitbrabant.nl.domainpreview.nl/dev/vb/inc/api/set.useritem.ajax/?gId='+GuildId+'&pId=0&eId='+eId, function( data ) {
    if(data.status == 'error' && data.error =='user is not logged in'){
        alert('TODO:gebruiker is niet ingelogd, opslaan in de cookie');
    }else{
        console.log(data.planned);
        if(data.planned == 1){
            $('.filteredItem[guildid='+GuildId+'] .addToList').addClass('inlist');
        }else{
            $('.filteredItem[guildid='+GuildId+'] .addToList').removeClass('inlist');
        }
        console.log($('.filteredItem[guildid='+2+'] .addToList'));
    }
});

Jquery选择器仅适用于浏览器

1 个答案: