解决:JavaScript 在函数中使用Ajax获取的值作为函数的返回值,结果无法获取到返回值 原因:ajax默认使用异步方式,要将异步改为同步方式 案例:通过区域ID,获取该区域下所有的学校 var School = new Object(); //保存区下面所有学校 School.GetAllInArea = function (areaId) { var returnData; $.ajax({ url: "/Area/Home/Schools/" + areaId, type:
Generalized Intersection over Union: A Metric and A Loss for Bounding Box Regression 2019-05-20 19:34:55 Paper: https://arxiv.org/pdf/1902.09630.pdf Project page: https://giou.stanford.edu/ Code: https://github.com/generalized-iou 1. Background and M