为什么我不能将数字转换为双数？

Question

weight是一个字段(Firestore中的数字),设置为100.

int weight = json['weight'];
double weight = json['weight'];

int weight工作正常,100按预期返回,但double weight崩溃(Object.noSuchMethod异常)而不是返回100.0,这是我的预期.

但是,以下工作:

num weight = json['weight'];
num.toDouble();

Answer 1

100从Firestore 解析(实际上不支持"数字类型",但转换它)时,它将通过标准解析为int.
Dart不会自动"巧妙"地施放这些类型.事实上,你不能投int一个double,这是你面临的问题.如果可能,您的代码将正常工作.

解析

相反,你可以自己解析它:

double weight = json['weight'].toDouble();

铸件

还有什么用,将JSON解析为a num,然后将其分配给a double,将其转换num为double.

double weight = json['weight'] as num;

这看起来有点奇怪,实际上Dart分析工具(例如内置在VS Code和IntelliJ的Dart插件中)会将其标记为"不必要的演员",但事实并非如此.

double a = 100; // this will not compile

double b = 100 as num; // this will compile, but is still marked as an "unnecessary cast"

double b = 100 as num编译因为num是超阶级的double和飞镖投射超级亚类型,即使没有明确的强制转换.
一个显式转换将是follwing:

double a = 100 as double; // does not compile because int is not the super class of double

double b = (100 as num) as double; // compiles, you can also omit the double cast

这是关于"Dart中的类型和铸造" 的精彩读物.

说明

你怎么了以下几点:

double weight;

weight = 100; // cannot compile because 100 is considered an int
// is the same as
weight = 100 as double; // which cannot work as I explained above
// Dart adds those casts automatically

Answer 2

您可以在一行中完成：

 double weight = (json['weight'] as num).toDouble();

Answer 3

您可以解析数据，如下所示：

这里的文档是一个Map<String,dynamic>

double opening = double.tryParse(document['opening'].toString());