Question

我定义的主要模式适用于以下性质的段落：

: Identifier
1. some text
2. ...
3. some more text

: New Identifier

: Another Identifier
some text

我想写一个名为defun的{{1}}，它会返回一个如下所示的列表：

get-paragraphs

如何在Emacs Lisp中删除这样的文本：

是否有一个函数可以迭代它们（然后根据自己的喜好将它们切掉）？我应该使用正则表达式吗？有更简单的方法吗？

Answer 1

您应该遍历缓冲区并收集文本（未经测试）：

(defun get-paragraphs ()
  (save-excursion
    (goto-char (point-min))
    (let ((ret '()))
      (while (search-forward-regexp "^: " nil t)
        (let ((header (buffer-substring-no-properties (point) (line-end-position)))
              (body '()))
          (forward-line)
          (while (not (looking-at "^$"))
            (push (buffer-substring-no-properties (point) (line-end-position)) body)
            (forward-line))
          (push (cons header (list (reverse body))) ret)))
      (nreverse ret))))

Answer 2

这里，请使用此Lisp代码：

(defun chopchop ()
  (mapcar 
   (lambda (x)
     (destructuring-bind (head &rest tail)
         (split-string x "\n" t)
       (list head tail)))
   (split-string (buffer-substring-no-properties
                  (point-min)
                  (point-max)) "\n?: *" t)))

如何在Emacs Lisp中获取所有段落？

2 个答案: