电影天堂电影网提供影视作品从海底出击第三季高清全集在线观看的影视全集网，欧美剧《从海底出击第三季》全集作品的导演是Hans Steinbichler  丹尼斯·甘塞尔   ，由汤姆·弗拉席亚  弗兰茨·丁达  里克·奥肯  雷·史蒂文森  恩斯特·施托兹  路易斯·沃尔夫勒姆  艾丽莎·施洛特  安娜·辛特  乔安娜·里贝罗  弗里兰·坦科斯  皮尔·基维特  Alessandro Schuster  Konstantin Gries  亚特乔姆·吉尔茨  Aniol Kirberg  Yuri Völsch  乔·哈特利  弗丽茨·哈勃兰特  约翰·冯·比罗  特里斯坦·皮特  弗朗兹·哈特维希  主演，从海底出击第三季在豆瓣的评分为3.0，本片由小编于2026-01-30 22:19更新，希望大家喜欢，可以把《从海底出击第三季》推荐给你朋友，本作品的地址为 https://www.83h8.com/83/4765.html

《从海底出击第三季》简介：　　Season 3 follows the fortunes of a young U-boat crew as they engage in the Battle of the Atlantic, are hunted down by an obsessed Royal Navy Commander (Ray Stevenson) and are sent on a dangerous mission to the Southern Hemisphere. Together they form strong personal alliances under the command of Robert Ehrenberg (Franz Dinda).　　And yet Ehrenberg's thoughts remain in Kiel, where he has developed feelings for Greta (Elisa Schlott). The young mother is married to a man who is seriously injured and suffering from war neurosis. Also in Kiel, Wilhelm Hoffmann's daughter Hannie (Luise Wolfram) is trapped in a loveless marriage to Commander Albrecht Lessing (Florian Panzner), a bitter, emotionally distant man. Hannie suffers under the tight constraints of her status and the social expectations that go with it. When she meets the charming submarine commander Schulz (Pierre Kiwitt), she is playing with fire.　　Meanwhile in the exotic climes of neutral Lisbon where the exiles, spies and criminals of Europe rub shoulders with each other and their enemies, Hagen Forster (Tom Wlaschiha) discovers a lethal plot to steal a fortune in plundered war time gold. These discoveries and aspects of his own history deriving from dark experiences on the Eastern front, lead him to question his own actions and moral path. They also lead him to a shadowy figure caught up in a conspiracy that might change the course of the War - for the better. That man, very much changed by his own shattering experiences, is Klaus Hoffmann (Rick Okon). These two historically based strands come together in an unexpected, thrilling finale.