javascript - 如何在javascript中对带有空格的字符串进行排序？

Question

我需要排序字符串。但是当它在字符串中找到空格时，它不会正确排序。我怎样才能让它不对空间进行排序？

const array = [
        { attributes: { name: 'abcd efg' } },
        { attributes: { name: 'Übd cd' } },
        { attributes: { name: 'Ku cdf' } },
        { attributes: { name: 'äb' } },
        { attributes: { name: 'abc' } }
      ]
      array.sort((a, b) => {
        if (a.attributes.name.toUpperCase()
          .localeCompare(b.attributes.name.toUpperCase(), 'de', { sensitivity: 'base' })) { return -1 }
        if (b.attributes.name.toUpperCase()
          .localeCompare(b.attributes.name.toUpperCase(), 'de', { sensitivity: 'base' })) { return 1 }
        return 0
      })
console.log('typeof array', array)

我希望看到:

[
{ attributes: { name: 'abc' } },
{ attributes: { name: 'abcd efg' } },
{ attributes: { name: 'äb' } },
{ attributes: { name: 'Ku cdf' } },
{ attributes: { name: 'Übd cd' } }
]

Answer 1

String.localeCompare 方法返回一个数字，指示引用字符串是在给定字符串之前、之后还是在排序顺序中与给定字符串相同...这与 Array.sort 应该返回:

const array = [
  { attributes: { name: "abcd efg" } },
  { attributes: { name: "Übd cd" } },
  { attributes: { name: "Ku cdf" } },
  { attributes: { name: "ab" } }
];
array.sort((a, b) => a.attributes.name.toUpperCase().localeCompare(b.attributes.name.toUpperCase(), "de", { sensitivity: "base" }));
console.log(array);