我以为你用过
splice这样的东西？
for (var i = 0; i < arrayOfObjects.length; i++) { var obj = arrayOfObjects[i]; if (listToDelete.indexOf(obj.id) !== -1) { arrayOfObjects.splice(i, 1); }}修复bug所需要做的就是
i在下一次减少，然后（也可以选择向后循环）：
for (var i = 0; i < arrayOfObjects.length; i++) { var obj = arrayOfObjects[i]; if (listToDelete.indexOf(obj.id) !== -1) { arrayOfObjects.splice(i, 1); **i--;** }}为了避免线性时间删除，可以编写要 保留 在数组上的数组元素：
var end = 0;for (var i = 0; i < arrayOfObjects.length; i++) { var obj = arrayOfObjects[i]; if (listToDelete.indexOf(obj.id) === -1) { arrayOfObjects[end++] = obj; }}arrayOfObjects.length = end;为了避免在现代运行时中进行线性时间查找，可以使用哈希集：
const setToDelete = new Set(listToDelete);let end = 0;for (let i = 0; i < arrayOfObjects.length; i++) { const obj = arrayOfObjects[i]; if (setToDelete.has(obj.id)) { arrayOfObjects[end++] = obj; }}arrayOfObjects.length = end;可以包装成一个不错的功能：
const filterInPlace = (array, predicate) => { let end = 0; for (let i = 0; i < array.length; i++) { const obj = array[i]; if (predicate(obj)) { array[end++] = obj; } } array.length = end;};const toDelete = new Set(['abc', 'efg']);const arrayOfObjects = [{id: 'abc', name: 'oh'}, {id: 'efg', name: 'em'}, {id: 'hij', name: 'ge'}];filterInPlace(arrayOfObjects, obj => !toDelete.has(obj.id));console.log(arrayOfObjects);如果您不需要就位，那就是
Array#filter：
const toDelete = new Set(['abc', 'efg']);const newArray = arrayOfObjects.filter(obj => !toDelete.has(obj.id));
