微信菜单view跳转获取code问题
想要跳转的地址REDIRECT如:http://a.com/b/?c.d
构建view地址(服务号):https://open.weixin.qq.com/connect/oauth2/authorize?appid=APPID&redirect_uri=REDIRECT&response_type=code&scope=snsapi_base&state=1#wechat_redirect
问题是,通过微信可以打开这个跳转的页面,但通过复制页面地址还是显示构建的view地址,说明并没有跳转回传code,而我想要获取的地址是:htt://a.com/b/?c.d&code=xxxxxxx&state=1
如果传这种地址就可以获取到code:htt://a.com/b,即不带参数不带斜线的情况,有人遇到过这个问题么?
回复内容:
想要跳转的地址REDIRECT如:http://a.com/b/?c.d
构建view地址(服务号):https://open.weixin.qq.com/connect/oauth2/authorize?appid=APPID&redirect_uri=REDIRECT&response_type=code&scope=snsapi_base&state=1#wechat_redirect
问题是,通过微信可以打开这个跳转的页面,但通过复制页面地址还是显示构建的view地址,说明并没有跳转回传code,而我想要获取的地址是:htt://a.com/b/?c.d&code=xxxxxxx&state=1
如果传这种地址就可以获取到code:htt://a.com/b,即不带参数不带斜线的情况,有人遇到过这个问题么?
授权的URL参数有没有对?把代码贴出来看看。

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