所以我有以下功能香草JS代码:
function get_menu(menu_id) {
wp.api.loadPromise.done(function() {
const menus = wp.api.collections.Posts.extend({
url: wpApiSettings.root + 'menus/v1/menus/' + menu_id,
});
const Menus = new menus();
Menus.fetch().then(
posts => {
let post_list = posts.items;
console.log(post_list);
});
});
}
get_menu(4);
遍历这些对象并在其中呈现 HTML 的最佳方法是什么?因此,假设我想遍历每个对象并获取 post_title
并输出 HTML <div> + post_title + </div>
。
感谢所有帮助!
更新:
需要循环渲染:
<div class="column is-one-third is-flex py-0">
<a href=" ***url*** " class="dropdown-item px-2 is-flex is-align-items-center">
<figure class="image is-32x32 is-flex">
<img src=" ***image*** + ***post_title*** '.svg'; ?>">
</figure>
<span class="pl-2"><?= ***post_title*** ?></span>
</a>
</div>
答案 0 :(得分:2)
您可以遍历数组并创建一个dom树
var invalid = "";
var rowWithError = "";
data.forEach((row, i) => {
if (row.join(",") == "Please Check Name, , , , , , , , ") {
var wrongName = ss.getRange('A' + (i + 1)).getValue();
invalid = wrongName;
rowWithError = row.join(",");
}
if (invalid.length != 0){
var ui = SpreadsheetApp.getUi();
ui.alert(
"The sheet had detected an invalid Name - '" + invalid + "', please check row " + rowWithError + " and try again.",)
}
});