vim：删除光标处的所有空白字符
是否有办法删除与光标同一行上的每个字符，一直到光标？例如，我可能有一行如下所示的代码：
foo = [cursor] Bar.new
如果我的光标位于上面的占位符处，是否可以删除每个空白字符（不使用正则表达式？），以便将 Bar.new 放置在光标处？
is there anyway to delete every character on the same line as a cursor, all the way up to the cursor? for instance, I might have a line of code that looks like the following:
foo = [cursor] Bar.new
If my cursor is at the place holder above, is it possible to delete every whitespace character (without using regex?) so that Bar.new is placed at the cursor?
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d w
d w
根据你的例子，尼尔的答案是正确的。但是，根据您最初的问题，
你会输入 d 0
实际上，你在标题中问第三个问题......删除光标之前的所有空白字符。我不知道如果没有正则表达式该怎么做。 d g e 将删除所有向后引导的空白字符，直到出现非空白字符，但它也会删除第一个非空白字符。
Based on your example, Neall's answer is correct. However, based on your initial question,
you would type d 0
Actually, you're asking a third question in your title... delete all whitespace characters up to the cursor. That one I'm not sure how to do without regex. d g e would remove all the whitespace characters leading backwards until a non-whitespace, but it also deletes the first non-whitespace character.
不完全是你想要的，但也许 d i w 会有所帮助 - 在上面的例子中，它会删除所有
=和Bar之间的空白。也许 c i w space 会给你你正在寻找的结果？Not quite what you want, but perhaps d i w would help - in the example above, it would delete all the whitespace between the
=and theBar. Perhaps c i w space would give you the result you are looking for?d t B
将删除“B”之前的任何字符，但不包括“B”
d t B
Will delete any character up to, but not including the 'B'