or child in bsObj.find("table",{"id":"giftList"}).tr.next_sibling:print(child)改成for child in bsObj.find("table",{"id":"giftList"}).tr.next_siblings:if :continueelse:print(child)break我想只抓取<tr>标签的下一个兄弟标签,但是输出是空行
桃花长相依
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