JavaScript two-dimensional array sorting problem

Question

[['10','8'],['11','4'],['18','7'],['7','6'],['8','7'],['9','5']]

JavaScript How to make an array like this
[['7','6'],['8','7'],['9','5'],['10',' 8'],['11','4'],['18','7']]
In this order

大神

Answer 1

array = [['10','8'],['11','4'],['18','7'],['7','6'],['8',' 7'],['9','5']]
function sortNumber(a,b){return a[0]-b[0]}
let newArray = array.sort(sortNumber)

newArray should be the result you want

Answer 2

var arr = [['10','8'],['11','4'],['18','7'],['7','6'],['8','7'],['9','5']]
arr.sort(function(a, b) {
    return a[0]-b[0]
})

Answer 3

The result of the arrangement of your two-dimensional array seems to be compared with the first subscripted element in each element in the array (array). Then it would be bad if it were converted into a one-dimensional array and sorted. ? The one-dimensional array is sorted, and then the original two-dimensional array is sorted using index to correspond to the previous one-dimensional array.
Logical implementation idea:
var twoArray = [['10','8'],['11','4'],['18','7'],['7','6'], ['8','7'],['9','5']];

    var oneArray = [];
    twoArray.map(function (item ,index) {
        oneArray.push({value: item[0], index: index});
    });
    console.log(oneArray);
    function sortNumber (a, b) {
        return a.value - b.value;
    }
    console.log(oneArray.sort(sortNumber));
    // 此时的oneArray已排好序
    var newTwoArray = [];
    oneArray.map(function (item) {
        newTwoArray.push(twoArray[item.index]);
    });
    console.log(newTwoArray);// 即你要的排序

Answer 4

var groupNum = 3;

var arr1 = [['10','8'],['11','4'],['18','7'],['7','6'],['8','7'],['9','5']];

var arr2 = [];

for (var i = arr1.length; i > 0 ; i -= groupNum) {
  arr2.push(arr1.slice(i - groupNum, i));
}

console.log(arr2) // [['7','6'],['8','7'],['9','5']，['10','8'],['11','4'],['18','7']]