模型提供程序中的颤振搜索
模型提供程序中的颤振搜索
提问于 2021-04-28 17:35:47
几天后它不起作用，我合并了模型上的数据，最初的例子:三星有很多数据，我已经将它组合成一个数组，我想问一下如何根据模型的品牌名称进行搜索？
@override
Widget build(BuildContext context) {
remoteModelId.clear();
isLoading = true;
final products = Provider.of<List<Brands>>(context);
return Scaffold(
...
body: products != null
? ListView.separated(
itemCount: products.length,
itemBuilder: (context, index) {
var lol = [];
var idModel = [];
var sublist = [].join();
var countList =[];
//var allList =[];
//var subLol = lol.indexOf(lol);
for(var ok in products){
lol.add(ok.brandName);
idModel.add(ok.ids);
countList.add(lol);
if(lol.contains(lol)){
sublist.compareTo(lol[index]);
break;
}
}
distinctIds = lol.toSet().toList();
hasilakhir = Set.of(distinctIds).toList();
newDataList = List.from(distinctIds);
templist.add(hasilakhir);
final myMap = Map();
lol.forEach((element) {
if(!myMap.containsKey(element)){
myMap[element] = 1;
return false;
}else{
myMap[element] += 1;
return false;
}
});
//newDataList = newDataList.map((brand)=>brand.toLowerCase()).toList();
return ListTile(
title: Text(hasilakhir[index]),谢谢
回答 1
Stack Overflow用户
发布于 2021-04-29 17:13:22
我不确定您需要什么，但以下是一些示例代码：
// Get products with a specific brandName
print(
'Number of Samsung=${products.where((p) => p.brandName == 'Samsung').length}');
// Count products by brandName - like your code
final map = Map();
products.forEach((product) {
if (!map.containsKey(product.brandName)) {
map[product.brandName] = 1;
} else {
map[product.brandName] += 1;
}
});
print('map=$map');
// Group products by brandName, with the count and list of ids
final map2 = Map();
products.forEach((product) {
if (!map2.containsKey(product.brandName)) {
map2[product.brandName] = {
'count': 1,
'ids': [product.ids]
};
} else {
var current = map2[product.brandName];
current['count']++;
current['ids'].add(product.ids);
}
});
print('map2=$map2');我的测试数据的输出是：
Number of Samsung=3
map={Samsung: 3, Apple: 3}
map2={Samsung: {count: 3, ids: [S1, S2, S3]}, Apple: {count: 3, ids: [iPhone1, iPhone2, iPhone3]}}我不知道它是否有用，但我已经在下面注释了您的一些代码：
// Get list of distinct brandNames
var distinctIds = lol.toSet().toList();
// Get list of distinct brandNames - same as distinctIds
var hasilakhir = Set.of(distinctIds).toList();
// Copy list distinctIds
var newDataList = List.from(distinctIds);
List templist = [];
// Make a list with one element which is the list of brandNames
templist.add(hasilakhir);页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/67297605
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