不是“字符串”类型的子类型。从 Firebase 单个集合中检索 4 个字段

Question

我正在尝试从 firebase 集合中获取 4 个字段（同一集合中的所有字段，每个文档都有这 4 个字段）到一个列表中，在本地获取数据后需要以不同的方式对其进行过滤

我收到此错误类型“speakerNamesModel”不是“String”类型的子类型。我的错误是什么？请给出示例代码。

    class speakerNamesModel {
    String speakerlanguage;
    String speakername_en;
    String speakername_lo;
    String speach_audiourl;
    speakerNamesModel(
    this.speakerlanguage, this.speakername_en,              this.speakername_lo, this.speach_audiourl);

    speakerNamesModel.fromJson(Map<dynamic, dynamic> parsedJSON)
    : speakerlanguage = parsedJSON['speakerlanguage'],
    speakername_en = parsedJSON['speakername_en'],
    speakername_lo = parsedJSON['speakername_lo'];
    speach_audiourl = parsedJSON['speach_audiourl'];
    }

    Future<List<speakerNamesModel>> getspeakerNamesList() async {
    QuerySnapshot qShot =
    await FirebaseFirestore.instance.collection                 ('speakernames').get();
    return qShot.docs
    .map((doc) => speakerNamesModel(
    doc.data()["speachdate"],     //timestamp data
    doc.data()["speakerlanguage"],//string
    doc.data()["speakername_en"], //name in roman letters string
    doc.data()["speakername_lo"],//name in local language unicode 
    doc.data()["speach_audiourl"],  //string
    ))
    .toList();
    }

    @override
    void initState() {
    getspeakerNamesList().then((List data) {
    setState(() {
    speakernames = filteredspeakernames = data;
    });
    });

Answer 1

仅用于跟踪，正如@subramanian-v 所提到的，通过如下声明它们的变量来解决问题：

List<SpeakerNamesModel> Speakernames;
List<SpeakerNamesModel> filteredSpeakernames;