Linux Sed：如何替换在特定模式位于文件中后找到的字符串
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Sed: How to replace a string found after a specific pattern is located in a file
提问by Isopycnal Oscillation
If I have the following list in a file:
如果我在文件中有以下列表：
integer, parameter :: ni = 1024
integer, parameter :: nj = 256
integer, parameter :: nk = 16
and want to search based on the string 'ni =', and then replace the string that follows (in this case '1024') with a new string like '512' for example (I would like to preserve the space). How can I use sed for this? Note that I would like to just essentially wipe anything that comes after the equal sign, this is because sometimes the string will not be a simple integer, it might be something like '1.D0'. And in some cases there may be comments ahead. So I just want to wipe out whatever is in front of the equal sign and replace with the new value.
并且想要根据字符串“ni =”进行搜索，然后将后面的字符串（在本例中为“1024”）替换为一个新字符串，例如“512”（我想保留空格）。我该如何使用 sed 呢？请注意，我只想擦除等号后面的任何内容，这是因为有时字符串不是简单的整数，它可能类似于“1.D0”。在某些情况下，可能会提前发表评论。所以我只想清除等号前面的任何内容并替换为新值。
The result would be:
结果将是：
integer, parameter :: ni = 512
integer, parameter :: nj = 256
integer, parameter :: nk = 16
采纳答案by hwnd
GNU sed supports extended regular expressions if you give it the -rflag.
如果你给它这个-r标志，GNU sed 支持扩展的正则表达式。
sed -re 's/(:: ni =)[^=]*$/ 512/' file
Better yet, match for multiple whitespace.
更好的是，匹配多个空格。
sed -re 's/(::\s+ni\s+=)[^=]*$/ 512/' file
The \1is a reference to what's matched in parentheses ( ), so we replace with \1and a new value.
该\1是什么括号匹配的参考( )，所以我们将其替换\1和新的价值。
回答by Jonathan Leffler
sed -e 's/:: ni = [0-9][0-9]*$/:: ni = 512/'
This looks for plausible context around the match specified to minimize the chance of finding nisomewhere in another string.
这会在指定的匹配项周围寻找合理的上下文，以最大限度地减少ni在另一个字符串中找到某处的机会。
回答by user2719058
If I understand you correctly, something like this should do it:
如果我理解正确的话，应该这样做：
sed 's/\(ni = \).*/\1REPLACEMENT/'
sed 's/\(ni = \).*/\1REPLACEMENT/'
