VI 删除除模式之外的所有内容 [英] VI delete everything except a pattern

问题描述

我有一个巨大的 JSON 输出，我只需要删除每行中除了一个小字符串之外的所有内容.

I have a huge JSON output and I just need to delete everything except a small string in each line.

字符串的格式为

"title": "someServerName"

someServerName"(引号内的部分)可能变化很大.

the "someServerName" (the section within quotes) can vary wildly.

我最近的一次是这样的:

The closest I've come is this:

:%s/\("title":\s"*"\)

但这只是设法删除

"title": "

我唯一想要在每一行中留下的是

The only thing I want left in each line is

"title": "someServerName"

编辑以回答发布的问题:

EDIT to answer the posted question:

我将要处理的文本的格式类似于

The Text I'm going to be working with will have a format similar to

{"_links": {"self": {"href": "/api/v2/servers/32", "title": "someServerName"},tons_of_other_json_crap_goes_here

我想要的最后是:

"title": "someServerName"

推荐答案

应该是 .* 而不是 * 来匹配一组任意字符.这可以完成工作:

It should be .* rather than * to match a group of any characters. This does the job:

%s/^.*\("title":\s".*"\).*$/\1/

各部分说明:

• %s/ 在每个匹配的行上替换.
• ^.* 忽略从行首开始的任何字符.
• \("title":\s".*"\) 捕获标题和服务器名称.".*" 将匹配引号之间的任何字符.
• .*$ 忽略该行的其余部分.
• /\1/ 替换的结果将是第一个捕获的组.该组由括号 \(...\) 捕获.

• %s/ Substitute on each matching line.
• ^.* Ignore any characters starting from beginning of line.
• \("title":\s".*"\) Capture the title and server name. ".*" will match any characters between quotes.
• .*$ Ignore the rest of the line.
• /\1/ The result of the substitution will be the first captured group. The group was captured by parentheses \(...\).

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