1 var layIds = $("[lay-id]"); 2 $.each(layIds, function (index, value, array) { 3 var id = $(value).attr("lay-id"); 4 element.tabDelete('tab', id); 5 });
wuhailong 2017-08-17 19:00 原文
1 var layIds = $("[lay-id]"); 2 $.each(layIds, function (index, value, array) { 3 var id = $(value).attr("lay-id"); 4 element.tabDelete('tab', id); 5 });