获取 ObservableHQ 单选按钮的标签
Get the label of an ObservableHQ radio button
有没有办法获取单选按钮的标签?
比如我有
viewof classification = radio({
options: [
{ label: "Underweight", value: "Underweight" },
{ label: "Healthy Weight", value: "Normal" },
{ label: "Overweight", value: "Overweight" },
{ label: "Obese", value: "Obese" }
],
value: "Underweight"
})
如果我想访问“健康体重”标签而不是“正常”的 classification
值,是否可行?
如果您可以控制收音机的定义(即此收音机不是从您不拥有的笔记本中导入的),您可以对其进行一些重组以使其更容易。
例如在一个单元格中:
options = new Map([
["Underweight", "Underweight"],
["Normal", "Healthy Weight"],
["Overweight", "Overweight"],
["Obese", "Obese"]
]);
和无线电小区:
viewof classification = radio({
options: [...options.entries()].map(([value, label]) => ({label, value})),
value: "Underweight"
})
并且在单元格中您需要引用标签而不是值:
label = options.get(classification)
有没有办法获取单选按钮的标签?
比如我有
viewof classification = radio({
options: [
{ label: "Underweight", value: "Underweight" },
{ label: "Healthy Weight", value: "Normal" },
{ label: "Overweight", value: "Overweight" },
{ label: "Obese", value: "Obese" }
],
value: "Underweight"
})
如果我想访问“健康体重”标签而不是“正常”的 classification
值,是否可行?
如果您可以控制收音机的定义(即此收音机不是从您不拥有的笔记本中导入的),您可以对其进行一些重组以使其更容易。
例如在一个单元格中:
options = new Map([
["Underweight", "Underweight"],
["Normal", "Healthy Weight"],
["Overweight", "Overweight"],
["Obese", "Obese"]
]);
和无线电小区:
viewof classification = radio({
options: [...options.entries()].map(([value, label]) => ({label, value})),
value: "Underweight"
})
并且在单元格中您需要引用标签而不是值:
label = options.get(classification)