Question

我知道我很近，但需要在这里提示！我试图根据特定父级是否在类名中添加标签，从内部包含其他元素的某些类元素中选择数据。

<div class=something_1>
<div class=other>
<div class=other>
<div class=other_TEMPLATE_1>
<div class=other>
<div class=selected_thing>I DO NOT WANT THIS</div>
</div>
</div>
</div>
</div>
</div>

<div class=something_2>
<div class=other>
<div class=other>
<div class=other_TEMPLATE_2>
<div class=other>
<div class=selected_thing>I DO NOT WANT THIS</div>
</div>
</div>
</div>
</div>
</div>

<div class=something_3>
<div class=other>
<div class=other>
<div class=other>
<div class=other>
<div class=selected_thing>I WANT THIS</div>
</div>
</div>
</div>
</div>
</div>

<script>

$("[class^=something]").each(function(){

selectedThing = $(this).children().children().children(":not[class^=other_TEMPLATE]").children().children(".selected_thing").html();

console.log(selectedThing);

});

</script>

正如您所看到的，部分挑战是某些类名称可能略有改变，但我想忽略TEMPLATE出现的那些。

Answer 1

你可以用这个：

$('.selected_thing').each(function(){
    if($(this).parents('[class*="TEMPLATE"]').length==0){
    //do something
        $(this).css('color','red')
    }
})

<强> jsFiddle example

Answer 2

$("[class^=something]").each(function(){

selectedThing = $(this).find('div').not("[class^=other_TEMPLATE]").children().children(".selected_thing").html();

console.log(selectedThing);

});

Answer 3

var selectedThings = [];

$("[class^=something]").each(function(){

selected = $(this).find('div')
.not("[class^=other_TEMPLATE]")
.children().children(".selected_thing")
.html();

if(selected !== undefined)
{
    selectedThings.push(selected);
}

});

console.log(selectedThings);

编辑：放慢速度:( @nevermind赢了......

Answer 4

$('.selected_thing').not('[class*="TEMPLATE"] .selected_thing');

Accompanying JSfiddle

Jquery选择没有类的孩子

4 个答案: