"sed"命令删除与第一个单词的确切字符串匹配的行 [英] "sed" command to remove a line that match an exact string on first word

问题描述

我正在使用一个小的bash脚本.我需要删除仅与该行的第一个单词中的确切单词匹配的整行.

I'm working in a small bash script. I need to remove an entire line that match an exact word in only the first word of the line.

这就是文本文件的外观:

That's how the text file looks:

John is a dumb
Maria is awesome
Toni and Maria are funny

我要删除第二行.

现在，我可以匹配精确的单词"maria"，但它也删除了第三行:

Right now, I can match the exact word "maria" but it removes the third line too:

sed -i "/\b\(maria\)\b/d" file.txt

如何仅指定第一个单词?

How to specify just the first word?

谢谢！

推荐答案

当前，您正在查找被单词边界(由 \ b 表示)包围的玛丽亚".注意，我还删除了不必要的括号并添加了/I 标志，这将使搜索不区分大小写.您的原件不会匹配"Maria".

Currently you are looking for "maria" surrounded by a word boundary (represented by \b.) Instead look for "maria" preceded by the start of line (represented by ^.) Note I've also removed unnecessary parentheses and added the /I flag, which will make the search case-insensitive. Your original wouldn't have matched "Maria".

sed -i "/^maria\b/Id" file.txt

将不区分大小写的标志固定为I而不是i！

fixed case insensitive flag to I instead of i!

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