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如何自定义apiRequest

Open kongdemin opened this issue 10 months ago • 1 comments

文档写的有点不清楚,是要返回对象吗 还是要怎么的

kongdemin avatar Apr 15 '24 07:04 kongdemin

success({ appkey:"", token:"" }); fail("错误消息") 自己处理完,回调一下就可以了

参考微信小程序: https://github.com/xiangyuecn/Recorder/blob/b4fa90a66adffdd9138d5e4705e61ff825e13189/app-support-sample/miniProgram-wx/pages/test_asr/test_asr.js#L255-L286

xiangyuecn avatar Apr 16 '24 04:04 xiangyuecn