ajax响应成功返回数据后,如何将html元素提取到变量中?这是包含以下返回数据的ajax调用 {
$.each(data, function(key, valuedon't know how to extract the strong element from the p element into a v
当单击(.refresh)按钮时,如何将$data_name传递给ajax,您将通过、()在php脚本的body字段中获取它?我想传递用于从数据库中选择数据的变量。2Scripts:ajax written in header field,
php for() written in body field(2) SQL