flutter - Flutter typeAhead错误：queryrow不是字符串的子类型

Question

我尝试使用预先输入在文本字段中显示建议，但我收到错误“queryro 不是字符串的子类型”

Future<List<dynamic>> _getSuggestions(String query) {
return OrderProvider.db.rawQuery(
    "select TypeOfProduct from Items where TypeOfProduct = ?", [query]);

} **这是我用来从数据库中获取数据的代码**

TypeAheadFormField(
              textFieldConfiguration: TextFieldConfiguration(
                controller: this._typeController,
                autocorrect: true,
                autofocus: true,
                decoration: InputDecoration(hintText: "Product Name"),
              ),
              suggestionsCallback: (suggestion) async {
                return _getSuggestions(suggestion);
              },
              itemBuilder: (context, suggestion) {
                return ListTile(
                  title: Text(suggestion),
                );
              },
              onSuggestionSelected: (suggestion) {
                this._typeController.text = suggestion;
              },
            ),

标签： fluttersqflite

Answer 1

OrderProvider.db.rawQuery(
    "select TypeOfProduct from Items where TypeOfProduct = ?", [query]);

这将返回一个List<Map>（如果您使用的是 sqfLite），suggestsCallback 会将该列表传递给 itemBuilder，但 itemBuilder 正在使用这些建议来构建一个 Text Widget（并且 Text 需要一些 type String，而不是 type Map）。_getSuggestions(String query)因此您可以在方法中使用 return 来返回 List之前减少 Map或执行类似的操作

itemBuilder: (context, suggestion) {//suggestion is of type Map because thats what you returned
   return ListTile(
     title: Text(suggestion['product']), //or whatever the key name is
   );
},