[英]How can I merge an array of objects?
假设我有一系列文章,每篇文章可能有也可能不超过1个图像对象,现在由于mysql无法将对象组合在一起,您必须自己做。 所以结果是你接近重复的文章对象......唯一的区别是图像对象。 通过接近重复,我的意思是返回结果的唯一区别是图像对象。
我试图找到一种方法来循环遍历articles
数组并消除重复对象并创建一个images
数组来保存每个article
图像对象。
例如,给出以下文章数组:
{ articles: [
{article: {articleId: 10, articleText: 'blah'}, image: {articleId: 10, imageUrl: 'http://differentUrlTooBelow'} },
{article: {articleId: 10, articleText: 'blah'}, image: {articleId: 10, imageUrl: 'http://AnotherUrlDifferentToAbove'} },
{article: {articleId: 11, articleText: 'ajsdnfa'}, image: {articleId: 11, imageUrl: 'http://DifferentUrlTooBelow'} },
{article: {articleId: 11, articleText: 'asfdjnasjdf'}, image: {articleId: 11, imageUrl: 'http://AnotherUrlDifferentToAbove'} },
]}
我们可以使用 lodash reduce 函数,但它似乎只返回最后article (articleId: 11)
...它似乎也没有找到它的所有图像。 它只会在每篇文章中创建一个图像数组。 请注意, res是文章对象的数组(如上所示)
var z = _.reduce(res, function(result, value, key) {
if (key == 0) return;
if(value.article['articleId'] == res[key-1].article['articleId']) {
result = value;
result.images = [];
result.images.push(result.image);
}
return result;
}, {});
最终结果将如下所示:
{ articles: [
{article: {articleId: 10, articleText: 'blah'},
images: [
{articleId: 10, imageUrl: 'http://differentUrlTooBelow'},
{articleId: 10, imageUrl: 'http://AnotherUrlDifferentToAbove'}
]},
{article: {articleId: 11, articleText: 'blah'},
images: [
{articleId: 11, imageUrl: 'http://differentUrlTooBelow'},
{articleId: 11, imageUrl: 'http://AnotherUrlDifferentToAbove'}
]},
]}
您可以使用_.groupBy()
来匹配article.articleId
,然后_.map()
使用第一项关联每篇文章,然后通过_.map()
获取所有image
项来设置images
属性。
var data = { articles: [ { article: {articleId: 10, articleText: 'blah'}, image: { articleId: 10, imageUrl: 'http://differentUrlTooBelow' } }, { article: {articleId: 10, articleText: 'blah'}, image: { articleId: 10, imageUrl: 'http://AnotherUrlDifferentToAbove' } }, { article: {articleId: 11, articleText: 'ajsdnfa'}, image: { articleId: 11, imageUrl: 'http://DifferentUrlTooBelow' } }, { article: {articleId: 11, articleText: 'asfdjnasjdf'}, image: { articleId: 11, imageUrl: 'http://AnotherUrlDifferentToAbove' } } ] }; var result = { articles: _(data.articles) .groupBy('article.articleId') .map(function(items) { return _(items[0]) .omit('image') .set('images', _.map(items, 'image')) .value(); }) .value() }; document.write('<pre>' + JSON.stringify(result, 0, 4) + '</pre>');
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