ios - Deep Link 不包含有效的必需参数 - Flutter with Firebase Dynamic Link

Question

我在我的颤振应用程序中使用 Firebase 动态链接。我允许用户在应用程序中生成可以通过短信/电子邮件/whatsApp 等共享的动态链接。此链接适用于 Android，但对于 iOS，我收到此异常。

提前致谢。

下面是我的 Xcode 日志。

[Firebase/Analytics][I-ACS023001] Deep Link does not contain valid required params. URL params: {
    "_cpb" = 1;
    "_cpt" = cpit;
    "_fpb" = "XXAHEJ4DGgVlbXXXXX==";
    "_iumchkactval" = 1;
    "_iumenbl" = 1;
    "_osl" = "https://app.XXXXXXX.com/PPZbgKsKpKvukDWZ8";
    "_plt" = 1400;
    "_uit" = 1400;
    amv = 1;
    apn = "com.xxx.xxxx";
    cid = 000000;
    ibi = "com.xxx.xxxx";
    imv = 1;
    isi = X4967XXXXX;
    link = "https://app.xxxxx.com/data?userid=Bzhm1TScavV2&refcode=3DWIN11329206";
}

以下是我的 Firebase 动态链接详细信息

Link name
Invite Friends

Deep link
https://app.xxxxx.com/data?userid=Bzhm1TScavV2&refcode=WIN11329206

Android app
com.xxx.xxxx

iOS app
com.xxx.xxxx

Long Dynamic Link
https://app.xxxxx.com/?link=https://app.xxxxx.com/data?userid%3DBzhm1TScavV2%26refcode%3DWIN11329206&apn=com.xxx.xxxx&isi=X4967XXXXX&ibi= com.xxx.xxxx


Short Dynamic Link
https://app.xxxxx.com/invitefriends 

标签： iosxcodefirebaseflutterfirebase-dynamic-links