对于使用ES2015 (ES6)的用户

你现在可以使用扩展语法来连接数组:

const arr1 = [0, 1, 2],
      arr2 = [3, 4, 5];

const result1 = [...arr1, ...arr2]; // -> [0, 1, 2, 3, 4, 5]

// or...

const result2 = [...arr2, ...arr1]; // -> [3, 4, 5, 0, 1, 2]

你可以用这个-

let array2d = [[1, 2, 3], [5, 4], [7, 8]];

let array1d = array2d.reduce((merged, block) => {
                          merged.push(...block); 
                          return merged; 
                     }, []);

console.log(array1d); // [1, 2, 3, 5, 4, 7, 8]

或者从上面的一个答案中我喜欢这个-

let array2d = [[1, 2, 3], [5, 4], [7, 8]];
console.log([].concat(...array2d)); // [1, 2, 3, 5, 4, 7, 8]

或者我发现的这个

let array2d = [[1, 2, 3], [5, 4], [7, 8]];
console.log(array2d.join().split(',').map(Number); // [1, 2, 3, 5, 4, 7, 8]

[].concat.apply([], [array1, array2, ...])

效率证明:http://jsperf.com/multi-array-concat/7

Tim Supinie mentions in the comments that this may cause the interpreter to exceed the call stack size. This is perhaps dependent on the js engine, but I've also gotten "Maximum call stack size exceeded" on Chrome at least. Test case: [].concat.apply([], Array(300000).fill().map(_=>[1,2,3])). (I've also gotten the same error using the currently accepted answer, so one is anticipating such use cases or building a library for others, special testing may be necessary no matter which solution you choose.)

最快的10倍是迭代数组，就像它们是一个数组一样，而不实际连接它们(如果可以的话)。

我很惊讶concat比push稍微快一点，除非测试不公平。

const arr1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i']; const arr2 = ['j', 'k', 'l', 'i', 'm', 'n', 'o', 'p', 'q', 'r', 's']; const arr3 = ['t', 'u', 'v', 'w']; const arr4 = ['x', 'y', 'z']; let start; // Not joining but iterating over all arrays - fastest // at about 0.06ms start = performance.now() const joined = [arr1, arr2, arr3, arr4]; for (let j = 0; j < 1000; j++) { let i = 0; while (joined.length) { // console.log(joined[0][i]); if (i < joined[0].length - 1) i++; else { joined.shift() i = 0; } } } console.log(performance.now() - start); // Concating (0.51ms). start = performance.now() for (let j = 0; j < 1000; j++) { const a = [].concat(arr1, arr2, arr3, arr4); } console.log(performance.now() - start); // Pushing on to an array (mutating). Slowest (0.77ms) start = performance.now() const joined2 = [arr1, arr2, arr3, arr4]; for (let j = 0; j < 1000; j++) { const arr = []; for (let i = 0; i < joined2.length; i++) { Array.prototype.push.apply(arr, joined2[i]) } } console.log(performance.now() - start);

如果你抽象它，你可以让不加入的迭代更干净，它仍然是原来的两倍:

const arr1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i']; const arr2 = ['j', 'k', 'l', 'i', 'm', 'n', 'o', 'p', 'q', 'r', 's']; const arr3 = ['t', 'u', 'v', 'w']; const arr4 = ['x', 'y', 'z']; function iterateArrays(arrays, onEach) { let i = 0; while (joined.length) { onEach(joined[0][i]); if (i < joined[0].length - 1) i++; else { joined.shift(); i = 0; } } } // About 0.23ms. let start = performance.now() const joined = [arr1, arr2, arr3, arr4]; for (let j = 0; j < 1000; j++) { iterateArrays(joined, item => { //console.log(item); }); } console.log(performance.now() - start);

如果N个数组是从数据库中获得的，并且没有硬编码，i将使用ES6这样做

let get_fruits = [...get_fruits , ...DBContent.fruit];

其中'n'是数组的个数，可能是数组的数组…

Var答案= _。Reduce (n, function(a, b){返回a.concat(b)})