Question

我的尝试：

var objects = [
  {"min":"0","max":"50","name":"name1"},
  {"min":"50","max":"100","name":"name2"},
  {"min":"100","max":"150","name":"name3"}
];

var value = 40;

objects.forEach(function(item, i){
    if(Number(item.min) >= value &&  value <= Number(item.max) ){
    console.log("name is ", item.name );
  }
});

将控制台设为：

name is  name2
name is  name3

但我期望name is name2通过方案获取范围值的正确方法是什么？

Answer 1

移动value变量，以便检查它是否大于或等于min，否则你要检查min是否大于或等于value ： -

＆＃13;

let objects = [
    {"min":"0","max":"50","name":"name1"},
    {"min":"50","max":"100","name":"name2"},
    {"min":"100","max":"150","name":"name3"}
];

let value = 40;

objects.forEach(function(item, i){
    if (value >= Number(item.min) && value <= Number(item.max)) {
        console.log("name is ", item.name );
    }
});

＆＃13;

Answer 2

使用Array#find方法：

var objects = [
  {"min":"0","max":"50","name":"name1"},
  {"min":"50","max":"100","name":"name2"},
  {"min":"100","max":"150","name":"name3"}
];

var value = 90;

let res = objects.find(obj => Number(obj.min) <= value && value <= Number(obj.max));
console.log(res ? res.name : 'Not found');

Answer 3

我们走了，你只需要交换变量：

if(Number(item.min) >= value &&  value <= Number(item.max) ){

if(value >= Number(item.min) &&  value <= Number(item.max) ){

var objects = [
  {"min":"0","max":"50","name":"name1"},
  {"min":"50","max":"100","name":"name2"},
  {"min":"100","max":"150","name":"name3"}
];

var value = 40;

objects.forEach(function(item, i){
    if(value >= Number(item.min) &&  value <= Number(item.max) ){
    console.log("name is ", item.name );
  }
});

Answer 4

如果范围增加50，则根本不需要迭代。我们将value除以50并向下舍入，为我们提供索引。

const index = n && Math.floor(n/50) - !(n%50);

x1到x50的任何范围都在较低的范围内，因此会处理150之类的值。 0值包含在第一个范围内。

var objects = [
  {"min":"0","max":"50","name":"name1"},
  {"min":"50","max":"100","name":"name2"},
  {"min":"100","max":"150","name":"name3"}
];

test(0);
test(10);
test(50);
test(51);
test(100);
test(149);
test(150);
test(151);

function test(n) {
  const index = n && Math.floor(n/50) - !(n%50);
  const res = objects[index];
  console.log("value:", n, ", item:", res ? res.name : 'Not found');
}

如何按最近的范围值查找对象名称？

4 个答案: