问题描述
有没有办法删除所有字符,直到 Sublime 中的第一个空格。我知道您可以使用 ctrl+delete
来执行此操作,但它会在非单词字符("、:、&、* 等)处停止。当您尝试从末尾删除 aaa aaa 2+a
时,它会将删除 2+a
直到 +
符号,但它会删除 aaa
直到空格。我需要更改它,以便它会删除 2+a
直到第一个空格。解决方案可以是任何东西;更改设置、插件。
解决方法
我找到了解决方案。这是通过这个插件:
https://packagecontrol.io/packages/KeyboardNavigation
它的关键是:
{ "keys": ["ctrl+backspace"],"command": "delete_to_beg_of_contig_boundary","args": {"forward": false} }
它从右到左删除任何字符,直到第一个空格。
,我编写了一个 Sublime Text 插件,可以根据需要删除文本。它几乎与 ST 的 delete_word
命令相同,但仅在空格/非空格处中断。
当插件被调用时,该插件会删除从光标到下一组或上一组字符的文本,该分组被定义为空白字符或非空白字符。因此,如果连续运行多次,它将在删除光标前后的空白和非空白字符组之间交替。 run()
方法的 forwards 参数(即命令的 arg)控制删除方向。
将插件保存在配置 Packages
文件夹层次结构中的某个位置。例如
.../sublime-text-3/Packages/User/DeleteToWhitespace.py
将键绑定添加到您的用户 .sublime-keymap
文件。例如
//
// These key bindings override the ST 'delete_word' keys but use whatever keys you want.
// You could use `super+delete` and `super+backspace` and keep ST's delete keys intact.
//
{ "keys": ["ctrl+delete"],"command": "delete_to_whitespace","args": {"forwards": true} },{ "keys": ["ctrl+backspace"],"args": {"forwards": false} },
下面是 DeleteToWhitespace.py
插件。它已上传到 this GitHub Gist - 这直接链接到 the raw source code。
#
# Name: Delete To Whitespace
# Requires: Plugin for Sublime Text v3
# Command: delete_to_whitespace
# Args: forwards: bool (delete backwards if false)
# License: MIT License
#
import sublime,sublime_plugin,re
class DeleteToWhitespaceCommand(sublime_plugin.TextCommand):
"""
A Sublime Text plugin that deletes text from the cursor to the next or
previous group of characters,the grouping being defined as either
whitespace or non-whitespace characters. Thus if run several times in
succession it will alternate between deleting groups of whitespace and
non-whitespace ahead or behind the cursor. The forwards parameter of the
run() method (i.e. the command's arg) controls the deletion direction.
"""
def run(self,edit,forwards=True):
self.edit = edit
self.forwards = forwards
if forwards:
self.delete_forwards()
else:
self.delete_backwards()
def delete_forwards(self):
whitespace_regex = "^\s+"
non_whitespace_regex = "^\S+"
for sel in self.view.sel():
if sel.size() > 0:
self.view.erase(self.edit,sel)
continue
# ∴ sel.a == sel.b == sel.begin() == sel.end()
# view.full_line() includes the trailing newline (if any).
cursor = sel.a
line = self.view.full_line(cursor)
cursor_to_eol = sublime.Region(cursor,line.end())
cursor_to_eol_str = self.view.substr(cursor_to_eol)
match = re.search(whitespace_regex,cursor_to_eol_str)
if match:
self.erase_matching_characters(cursor,match)
continue
match = re.search(non_whitespace_regex,match)
continue
def delete_backwards(self):
whitespace_regex = "\s+$"
non_whitespace_regex = "\S+$"
for sel in self.view.sel():
if sel.size() > 0:
self.view.erase(self.edit,sel)
continue
# ∴ sel.a == sel.b == sel.begin() == sel.end()
# view.line() excludes the trailing newline (if any).
cursor = sel.a
line = self.view.line(cursor)
cursor_to_bol = sublime.Region(cursor,line.begin())
cursor_to_bol_str = self.view.substr(cursor_to_bol)
# Delete the newline of the 'previous' line.
if cursor_to_bol.size() == 0 and cursor > 0:
erase_region = sublime.Region(cursor,cursor - 1)
self.view.erase(self.edit,erase_region)
continue
match = re.search(whitespace_regex,cursor_to_bol_str)
if match:
self.erase_matching_characters(cursor,match)
continue
def erase_matching_characters(self,cursor,match):
match_len = match.end() - match.start()
if self.forwards:
erase_region = sublime.Region(cursor,cursor + match_len)
else:
erase_region = sublime.Region(cursor,cursor - match_len)
self.view.erase(self.edit,erase_region)