Question

我们知道jQuery.extend(true, obj1, obj2)方法可以将对象的属性从obj2深度复制到obj1。在数组的情况下，它根据索引复制属性。但我需要根据一些属性进行复制（例如我的情况下为id），如下例所示：

obj1 = [{id:"id1", name:"name1"},{id:"id2", name:"name2"}]
obj2 = [{id:"id3", name:"name3"}{id:"id1", name:"name1_modified"}]

jQuery.extend将返回：

[{id:"id3", name:"name3"}{id:"id1", name:"name1_modified"}]

但我需要输出：

[{id:"id1", name:"name1_modified"},{id:"id2", name:"name2"}{id:"id3", name:"name3"}]

是否有任何方法/库可以完成此任务？

Answer 1

我在处理项目时遇到了同样的问题。所以我jquery扩展以基于其属性完成数组合并。如果要基于属性合并数组，请将属性名称作为合并函数中的最后一个参数传递。我创建了一个jsfiddle，在浏览器控制台中查看结果。

function merge() {
    var options, name, src, copy, copyIsArray, clone, targetKey, target = arguments[0] || {}, i = 1, length = arguments.length, deep = false;
    var currentId = typeof arguments[length - 1] == 'string' ? arguments[length - 1] : null;
    if (currentId) {
        length = length - 1;
    }
    // Handle a deep copy situation
    if (typeof target === "boolean") {
        deep = target;
        target = arguments[1] || {};
        // skip the boolean and the target
        i = 2;
    }

    // Handle case when target is a string or something (possible in deep
    // copy)
    if (typeof target !== "object" && !jQuery.isFunction(target)) {
        target = {};
    }

    // extend jQuery itself if only one argument is passed
    if (length === i) {
        target = this;
        --i;
    }

    for (; i < length; i++) {
        // Only deal with non-null/undefined values
        if ((options = arguments[i]) != null) {
            // Extend the base object
            for (name in options) {
                if (!options.hasOwnProperty(name)) {
                    continue;
                }
                copy = options[name];
                var mm = undefined, src = undefined;
                if (currentId && jQuery.isArray(options) && jQuery.isArray(target)) {
                    for (mm = 0; mm < target.length; mm++) {
                        if (currentId && (isSameString(target[mm][currentId], copy[currentId]))) {
                            src = target[mm];
                            break;
                        }
                    }
                }
                else {
                    src = target[name];
                }

                // Prevent never-ending loop
                if (target === copy) {
                    continue;
                }
                targetKey = mm !== undefined ? mm : name;
                // Recurse if we're merging plain objects or arrays
                if (deep && copy && (jQuery.isPlainObject(copy) || (copyIsArray = jQuery.isArray(copy)))) {
                    if (copyIsArray) {
                        copyIsArray = false;
                        clone = src && jQuery.isArray(src) ? src : [];

                    }
                    else {
                        clone = src && jQuery.isPlainObject(src) ? src : {};
                    }

                    // Never move original objects, clone them
                    if (currentId) {
                        target[targetKey] = merge(deep, clone, copy, currentId);
                    }
                    else {
                        target[targetKey] = merge(deep, clone, copy);
                    }

                    // Don't bring in undefined values
                }
                else if (copy !== undefined) {
                    target[targetKey] = copy;
                }
            }
        }
    }

    // Return the modified object
    return target;
};
 function isSameString (a , b){
        return a && b && String(a).toLowerCase() === String(b).toLowerCase();
    }
obj1 = [ {
    id : "id1",
    name : "name1"
}, {
    id : "id2",
    name : "name2"
} ]
obj2 = [ {
    id : "id3",
    name : "name3"
}, {
    id : "id1",
    name : "name1_modified"
} ];
console.log(merge(true, obj1, obj2, "id"));

如何基于属性（而不是索引）合并2个数组

1 个答案: