关于小程序打开app传参问题

最新推荐文章于 2024-08-15 10:50:48 发布

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最新推荐文章于 2024-08-15 10:50:48 发布

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分类专栏： iOS 文章标签：小程序打开app传参

本文链接：https://blog.youkuaiyun.com/evenstyle1/article/details/95450077

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小程序打开app分两种情况：

app直接打开微信小程序之后的返回app

WXLaunchMiniProgramReq *launchMiniProgramReq = [WXLaunchMiniProgramReq object];
launchMiniProgramReq.userName = userName;
launchMiniProgramReq.path =path;
launchMiniProgramReq.miniProgramType = miniProgramType;
[WXApi sendReq:launchMiniProgramReq];

这种情况下，在微信小程序点击打开app的时候走的函数是:

- (void)onResp:(BaseResp *)resp{
    if ([resp isKindOfClass:[WXLaunchMiniProgramResp class]]) {
        NSString *errMsg = [NSString stringWithFormat:@"errorcode:%@",((WXLaunchMiniProgramResp*)resp).extMsg];
        NSLog(@"errcode:%@",errMsg);
    }
}

app分享到微信聊天，然后在聊天里面打开小程序，再跳转回app

WXMiniProgramObject *object = [WXMiniProgramObject object];
object.webpageUrl = url
object.userName = username;
object.hdImageData = data
object.path = path

WXMediaMessage *message = [WXMediaMessage message];
message.title = title;
message.description = descriptionStr;
message.thumbData = nil;
message.mediaObject = object;

SendMessageToWXReq *req = [[SendMessageToWXReq alloc] init];
req.bText = NO;
req.message = message;
req.scene = WXSceneSession;  //目前只支持会话
[WXApi sendReq:req];

这种情况下打开app传参走的是

- (void)onReq:(BaseReq *)req{
    NSLog(@"req is %@",req);
    if ([req isKindOfClass:[LaunchFromWXReq class]]) {
        LaunchFromWXReq *wxReq = (LaunchFromWXReq*)req;
        WXMediaMessage *message =  wxReq.message;
        NSString *messageExt = message.messageExt;
        NSLog(@"ext message is %@",messageExt);
    }
}

通常我们走的是第二种方式，这里很坑，微信文档也没有写详细。