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WXApi.send(request) 返回false, 是什么问题?需要错误提示

AppID
wx22b32fd1b4dcc489

[wx] Universal Link check - step: WXULCheckStep(rawValue: 5), result: WXCheckULStepResult[success: 1, errorInfo: Universal Link check passed. The application is launched by WeChat via Universal Link, suggestion: ]

[wx]  [WXApi.m]+[WXApi checkUniversalLinkCallback:step:success:errorInfo:suggestion:]WXLog:WXCheckULStepResult[success: 1, errorInfo: Universal Link check passed. The application is launched by WeChat via Universal Link, suggestion: ]

[wx] Universal Link check - step: WXULCheckStep(rawValue: 6), result: WXCheckULStepResult[success: 1, errorInfo: All Check Passed!, suggestion: ]

[wx]  [WXApi.m]+[WXApi checkUniversalLinkCallback:step:success:errorInfo:suggestion:]WXLog:WXCheckULStepResult[success: 1, errorInfo: All Check Passed!, suggestion: ]

我也遇到这个问题,求解

[wx] sendAuthReq callback: false

          var request = SendAuthReq()

          request.scope = "snsapi_userinfo"

          request.state = "123"

            WXApi.send(request) { rst in

                print("[wx] sendAuthReq callback: \(rst)")

            }

WXApi.registerApp("wx22b32fd1b4dcc489", universalLink: "https://ipunk.cn/recipe/")

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2 个回答

  • 社区技术运营专员--许涛
    社区技术运营专员--许涛
    04-07

    你好,+ (void)startLogByLevel:(WXLogLevel)level logBlock:(WXLogBolock)logBlock;

    接一下sdk的log回调查看错误日志

    04-07
    有用
    回复
  • 智能回答 智能回答 本次回答由AI生成
    04-05
    有用
    回复
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