2

假设我有以下地图：

(def m {"a" {"d" {}
             "e" {}}
        "b" {"f" {}
             "g" {"h" {}
                  "i" {}}}
        "c" {}})

我需要像这样渲染它：

(om.dom/div #js {} "a" 
  (om.dom/div #js {} "d")
  (om.dom/div #js {} "e"))

(om.dom/div #js {} "b" 
  (om.dom/div #js {} "f")
  (om.dom/div #js {} "g"
    (om.dom/div #js {} "h")
    (om.dom/div #js {} "i")))

(om.dom/div #js {} "c")

我该怎么做呢？我已经搞砸了clojure.walk，但无法让它先调用om.dom/div叶子，然后调用直接父母等。

我认为该解决方案可能涉及将递归函数映射到vals给定子图的 。它会将地图分开，直到它看到一片叶子，然后将om.dom/div调用冒泡到地图上。

到目前为止，我有这个功能：

(defn build-tree [data]
  (apply dom/div #js {}
         (->> (clojure.walk/postwalk
                #(cond (map? %) (vec %)
                       (vector? %) %
                       (string? %) %) data)
              (clojure.walk/postwalk
                #(if (and (vector? %) (string? (first %)))
                   (apply dom/div #js {} %) %)))))

结果是：

在检查器中使用这个：

dom/ul生成嵌套和元素的奖励积分dom/li..

4

4 回答 4

2

在解决此类问题时，有时在 REPL 中生成类似于所需调用树的树结构会很有帮助。一旦结果看起来没问题，将其转换为实际的调用树通常很简单。

例如，这是一个om.dom/div为您的示例生成调用树的函数；要在 ClojureScript 中使用，请按照评论中的说明进行调整：

(defn div [m]
  (for [[k v] m]
    ;; replace with (apply om.dom/div #js {} k (div v)) in CLJS
    (list* 'om.dom/div {} k (div v))))

示例调用：

(div {"a" {"d" {}
           "e" {}}
      "b" {"f" {}
           "g" {"h" {}
                "i" {}}}
      "c" {}})

上述输出：

;; in CLJS, you'll want to use this as the ... in
;; (apply create-container-component initial-args ...)
((om.dom/div {} "a"
   (om.dom/div {} "d")
   (om.dom/div {} "e"))
 (om.dom/div {} "b"
   (om.dom/div {} "f")
   (om.dom/div {} "g"
     (om.dom/div {} "h")
     (om.dom/div {} "i")))
 (om.dom/div {} "c"))
于 2015-03-06T17:30:13.990 回答
0

clojure.walk很棒，但有时会变得更难。我有这个：

 (defn recursive-component [el]
   (cond
    (empty? el) nil
    (vector? el) (apply dom/div nil 
                      (first el)
                      (recursive-component (second el)))
    (map? el) (map recursive-component el)))


 (defn main-component [data owner]
   (om/component
    (apply dom/div nil "Main"
           (recursive-component m))))
于 2015-03-06T11:45:41.430 回答
0

以下代码生成的 DOM 结构看起来与您正在寻找的非常接近，除了子节点（即使没有子节点）被包裹在 a 中div，这应该很容易修复：

(defn tree-component
  [tree]
  (apply dom/div nil
         (keep (fn [[k v]]
                 (when k
                   (dom/div nil k
                            (tree-component v))))
               tree)))

(defn app-view
  [data owner]
  (reify
    om/IRender
    (render [_]
      (tree-component
        {"a" {"d" {}
              "e" {}}
         "b" {"f" {}
              "g" {"h" {}
                   "i" {}}}
         "c" {}}))))
于 2015-03-06T11:25:44.687 回答
0

对于ul和li：

(defn recursive-component [el]
  (cond
   (and (vector? el) (empty? (second el)))
   (dom/li nil (first el))

   (vector? el) (apply dom/ul nil 
                     (first el)
                     (recursive-component (second el)))

   (map? el) (map recursive-component el)))

但它显示为li元素{:c {}}。

于 2015-03-06T11:54:21.507 回答