Question

我有以下函数来打印指向* scratch *缓冲区的行，

(defun print-line ()
  (print (thing-at-point 'line) (get-buffer "*scratch*")))

但它甚至打印出像这样的已知信息

#(" OFFICE
" 0 2 (fontified t org ...

如何放弃已完成信息的打印。

Answer 1

扩展Daimrod提到的buffer-substring-no-properties ......

M-x apropos RET no-properties RET

buffer-substring-no-properties
  Function: Return the characters of part of the buffer, without the
            text properties.
field-string-no-properties
  Function: Return the contents of the field around POS, without text
            properties.
insert-buffer-substring-no-properties
  Function: Insert before point a substring of BUFFER, without text
            properties.
match-string-no-properties
  Function: Return string of text matched by last search, without text
            properties.
minibuffer-contents-no-properties
  Function: Return the user input in a minibuffer as a string, without
            text-properties.
substring-no-properties
  Function: Return a substring of STRING, without text properties.

您可以阅读手册中的文字属性：

M - ：（info“（elisp）Text Properties”） RET

Answer 2

在操作org-table中的字符串时，我需要eredis类似的东西。您可以在显示字符串时使用`set-text-properties'删除它们。

(defun strip-text-properties(txt)
  (set-text-properties 0 (length txt) nil txt)
      txt)

(defun print-line ()
 (print (strip-text-properties 
         (thing-at-point 'line))
    (get-buffer "*scratch*")))

Answer 3

我尝试过一些东西，但这很奇怪，我真的不明白文字属性是如何运作的。

例如：

(type-of (thing-at-point 'line)) => string

正如你所说，如果有人试图打印它，那么也会打印属性，但是如果有人试图插入它：

(insert (format "%s" (thing-at-point 'line)))

仅打印字符串，而不是属性。

所以在我看来，这些属性只是绑定到字符串，但你可以照常操作字符串：

(lenght (thing-at-point 'line))
(substring (thing-at-point 'line) 0 2)

但是，如果您想要的只是行，而只有行，您可以使用buffer-substring-no-properties：

(defun print-line ()     
  (print (buffer-substring-no-properties (point-at-bol) (point-at-eol))))

仅打印文本丢弃文本属性

3 个答案: