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req.callbackAbility = kDemoEntryAbility 返回原app未响应

// 1. 初始化微信API
const wxApi = wxOpenSdk.WXAPIFactory.createWXAPI('');
const context = getContext(this) as common.UIAbilityContext;

// 2. 创建网页对象
const webpageObject = new wxOpenSdk.WXWebpageObject();
webpageObject.webpageUrl = "http://www.qq.com"

// 3. 创建媒体消息
const mediaMessage = new wxOpenSdk.WXMediaMessage();
mediaMessage.mediaObject = webpageObject
mediaMessage.title = `${this.doctorInfo.doctorname}\u0020\u0020${this.doctorInfo.doctorrank}`
mediaMessage.description = `${this.doctorInfo.hospitalname}\u0020\u0020${this.doctorInfo.deptname}`

const thumbData = await getContext(this).resourceManager.getMediaContent($r('app.media.default_avatar'))
const thumbPixel = image.createImageSource(thumbData.buffer).createPixelMapSync()
const thumbBuffer = await image.createImagePacker().packToData(thumbPixel, { format: "image/png", quality: 100 })
mediaMessage.thumbData = new Uint8Array(thumbBuffer)

const req = new wxOpenSdk.SendMessageToWXReq();
const callbackAbility = `${context.abilityInfo.bundleName}/${context.abilityInfo.name}`;
console.log('a:: callbackAbility',callbackAbility)
req.callbackAbility = callbackAbility;
req.scene = wxOpenSdk.SendMessageToWXReq.WXSceneSession;
req.message = mediaMessage;

wxApi.sendReq(getContext(this) as common.UIAbilityContext, req);

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1 个回答

  • 社区技术运营专员--许涛
    社区技术运营专员--许涛
    2天前

    你好,复现问题时麻烦在手机微信那里上传下日志: 我->设置->帮助与反馈右上角有个上报日志的入口,麻烦提供一下微信号,时间点,bundleid、identifier、appid和opensdk的调用时间点(可以社区私信提供)

    2天前
    有用
    回复
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