Hello, tell me how to do that after the ajax request, the content that comes in does not replace the existing one, but is added. I will give an example:

there was this html code: <div> 1 </ div> <div> 2 </ div> <div> 3 </ div>

after the ajax request: <div> 1 </ div> <div> 2 </ div> <div> I was just loading under ajax </ div> <div> 3 </ div>

  • thanks after what you need)) - Lgunchik

1 answer 1

for this you can use .eq () and .after ()

$('div').eq(1).after('<div>Я только что под грузился с помощью ajax</div>');