I need sort strings. But it doesn't sort correctly when it finds spaces in string. How can I make it not to sort spaces?
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)I expect to see:
[
{ attributes: { name: 'abc' } },
{ attributes: { name: 'abcd efg' } },
{ attributes: { name: 'äb' } },
{ attributes: { name: 'Ku cdf' } },
{ attributes: { name: 'Übd cd' } }
]
The String.localeCompare method returns a number indicating whether a reference string comes before, or after, or is the same as the given string in sort order... which is the same as what Array.sort is supposed to return:
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);The way localCompare works is, if the first string is smaller i.e. comes before the second string, it will return a negative number.
And if the first string is greater i.e. it comes after the second string, it will return a positive number.
This line:
if (a.attributes.name.toUpperCase()
.localeCompare(b.attributes.name.toUpperCase(), 'de', { sensitivity: 'base' })) { return -1 }
will be true even if the first string is greater or the second string is.
The problem is that if (-1) or if(any_negative_value) is considered true. Even if localeCompare() returns a negative value, your first if statement will always execute. The second if statement will never be executed. Therefore, no matter if the a.attributes.name is lexicographically greater or b.attributes.name is greater, the first if statement will always be executed.
You do not need the if statements. The sort function just needs the number returned by localCompare().
Hence, you can simply return the value of localeCompare() and it will sort the attributes correctly.
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('typeof array', array)Try this one....
var hasLeading = s => /^\S+\s\S+\s\S+$/.test(s);
var array = [
{ attributes: { name: 'abcd efg' } },
{ attributes: { name: 'Übd cd' } },
{ attributes: { name: 'Ku cdf' } },
{ attributes: { name: 'ab' } }
];
array.sort((a, b) => hasLeading(b.attributes.name.toUpperCase()) - hasLeading(a.attributes.name.toUpperCase()) || a.attributes.name.toUpperCase() > b.attributes.name.toUpperCase() || -(a.attributes.name.toUpperCase() < b.attributes.name.toUpperCase())
);
console.log(array);