我想根据布尔属性将对象数组减少为字符串数组。我的对象数组如下所示：diet: [ { slug: "dairy-free", selected: true, }, { slug: "fish", selected: false, }, { slug: "meat", selected: false, }, { slug: "poultry", selected: true, }, { slug: "vegetarian", selected: true, }, ]如果所选值为 true，我的理想结果将是一个字符串数组。["dairy-free", "poultry", "vegetarian"]
4 回答
江户川乱折腾
TA贡献1851条经验 获得超5个赞
你只需要filter和map。
使用filter你将得到那些selected真正有价值的人。用于将这些对象的值map映射slug到数组中。
var diet = [ { slug: "dairy-free", selected: true, }, { slug: "fish", selected: false, }, { slug: "meat", selected: false, }, { slug: "poultry", selected: true, }, { slug: "vegetarian", selected: true, }, ]
var newArray = diet.filter(f => f.selected).map(m => m.slug)
console.log(newArray)
慕桂英4014372
TA贡献1871条经验 获得超13个赞
您可以使用方法来过滤选择值为 true 的filter记录。这将返回满足条件的对象数组
var diet = [ { slug: "dairy-free", selected: true, }, { slug: "fish", selected: false, }, { slug: "meat", selected: false, }, { slug: "poultry", selected: true, }, { slug: "vegetarian", selected: true, }, ]
console.log(diet.filter(e => e.selected))
如果只想返回slug属性，可以使用filter和map方法组合
var diet = [ { slug: "dairy-free", selected: true, }, { slug: "fish", selected: false, }, { slug: "meat", selected: false, }, { slug: "poultry", selected: true, }, { slug: "vegetarian", selected: true, }, ]
console.log(diet.filter(e => e.selected == true).map(e => e.slug))
桃花长相依
TA贡献1860条经验 获得超8个赞
Array像这样使用函数filter和的组合map：
let diet= [ { slug: "dairy-free", selected: true, }, { slug: "fish", selected: false, }, { slug: "meat", selected: false, }, { slug: "poultry", selected: true, }, { slug: "vegetarian", selected: true, }, ]
let selected=diet.filter(type=>type.selected).map(selectedType=>selectedType.slug)
console.log(selected)
不幸的是，这需要对数组进行两次迭代，一次过滤掉selected，一次只返回slug
reduce您也可以使用这样的方法获得相同的结果
let diet= [ { slug: "dairy-free", selected: true, }, { slug: "fish", selected: false, }, { slug: "meat", selected: false, }, { slug: "poultry", selected: true, }, { slug: "vegetarian", selected: true, }, ]
let reduced = diet.reduce((filtered, option)=> {
if (option.selected) {
filtered.push(option.slug);
}
return filtered;
}, []);
console.log(reduced)
reduce将是推荐的方法，因为您只在数组上迭代一次
德玛西亚99
TA贡献1770条经验 获得超3个赞
另一种方式可能是
final newList = diet.fold<List>([], (previousValue, e) {
if (e["selected"]) {
previousValue.add(e["slug"]);
}
return previousValue;
});
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