我看不到这个问题的答案 - 有没有简单的方法可以做到这一点?我想获取一些对象 x 的列表并将其变成一个没有任何 hacky 代码的 json 对象?我正在做一些丑陋的事情,但想将此列表放入一个名为“数据”的对象中。我可以以某种方式将其映射到对象“数据”吗?
private def renderArticleJson(articles: Iterable[GraphedArticle]): String = {
val listToConvert = for (article <- articles) yield {
JsObject(
"articleId" -> Json.toJson(article.getArticleId)
:: "articleUrl" -> Json.toJson(article.getArticleUrl)
:: "graphId" -> Json.toJson(article.asVertex().getId.toString)
:: "fullName" -> Json.toJson(article.getTitle)
:: "imageUrl" -> Json.toJson(article.getImageUrl)
:: Nil
)
}
}
按要求编辑:添加了我想退出的内容(由于第一个答案的帮助,现在解决了)
{
"data": [
{
"articleId": null,
"articleUrl": null,
"graphId": "#8:24",
"fullName": "hey",
"imageUrl": "hey"
},
{
"articleId": null,
"articleUrl": null,
"graphId": "#8:25",
"fullName": "hey",
"imageUrl": "hey"
},
{
"articleId": "b23c162d-b0af-4ce3-aebf-f33943492f95",
"articleUrl": null,
"graphId": "#8:26",
"fullName": "hey",
"imageUrl": "hey"
},
{
"articleId": "8afe310c-8337-4a8a-8406-5670249ba0a7",
"articleUrl": "hey",
"graphId": "#8:27",
"fullName": "hey",
"imageUrl": "hey"
}
]
}