2

我有一个试剂表，并设置了表头的 col-span = 2。

 (defn my-table []
  [:table.table.table-striped.table-bordered
   [:thead
    [:tr
     [:th "Col1"]
     [:th "Col2"]
     (doall (for [id @ids]
          ^{:key id}
          [:th {:data-id (id :id)
                :col-span "2"}
           (str (get id :name))]))]]
   [:tbody
    (doall (for [i (range (count @table-items))]
         (let [table-item (cursor table-items [i])]
           ^{:key i}
           [item-row table-item])))]])

但是在每一行中，见item-row下文，我正在循环通过一个数据结构来获取该行中拆分列的数据并且遇到了问题。如果这样做，我可以让数据正确呈现

[:div [:td [:span (:id id)]]
             [:td [:span (:name id)]]]

我知道这是不正确/无效的 html。

 (defn item-row [item]
    (fn []
      [:tr
       [:td [:span (:key1 @item)]]
       [:td [:span (:key2 @item)]]
       (doall (for [id @ids]
            ^{:key id}
            [[:td [:span (:x id)]]
             [:td [:span (:y id)]]]))]))

有谁知道如何:td在 for 循环中渲染两个元素？我的想法是将这两个:td元素包装在一个向量中，并让试剂处理渲染拆分列。

 (doall (for [id @ids]
            ^{:key id}
            [[:td [:span (:x id)]]
             [:td [:span (:y id)]]])

我还收到以下错误： core.cljs:4793 Uncaught Error: No item [:td "."] in vector of length 2 at Object.cljs$core$vector_index_out_of_bounds [as vector_index_out_of_bounds]

这与[:td [:span (:y id)]]

4

1 回答 1

3

您可以利用您只是返回数据这一事实，因此一种方法是:tr使用前两列构建向量的初始内容，然后:td在 using 中添加其余元素into。这是一个我认为可以解决您的问题的示例：

cljs.user=> (def ids [{:x 1 :y 2} {:x 3 :y 4}])
#'cljs.user/ids
cljs.user=> (let [tr [:tr 
       #_=>           [:td "first col"]
       #_=>           [:td "second col"]]]
       #_=>   (reduce (fn [acc id]
       #_=>             (into acc
       #_=>                   [[:td [:span (:x id)]]
       #_=>                    [:td [:span (:y id)]]]))
       #_=>           tr
       #_=>           ids))
[:tr [:td "first col"] [:td "second col"] [:td [:span 1]] [:td [:span 2]] [:td [:span 3]] [:td [:span 4]]]
cljs.user=>
于 2018-03-22T21:58:26.170 回答