bash “sed”命令删除与第一个单词上的确切字符串匹配的行
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"sed" command to remove a line that match an exact string on first word
提问by Dani Polo
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.
我正在使用一个小的 bash 脚本。我需要删除仅在该行的第一个单词中与精确单词匹配的整行。
That's how the text file looks:
这就是文本文件的样子：
John is a dumb
Maria is awesome
Toni and Maria are funny
I want to remove the just the second line.
我想删除第二行。
Right now, I can match the exact word "maria" but it removes the third line too:
现在，我可以匹配确切的单词“maria”，但它也删除了第三行：
sed -i "/\b\(maria\)\b/d" file.txt
How to specify just the first word?
如何仅指定第一个单词？
Thanks!
谢谢！
回答by miken32
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 /Iflag, which will make the search case-insensitive. Your original wouldn't have matched "Maria".
目前您正在寻找“玛丽亚”，由一个单词边界包围（表示为\b。）相反寻找“玛丽亚”，由行的开始之前（以表示^）。注意我还删除不必要的括号内添加的/I标志，这将使搜索不区分大小写。您的原件不会与“玛丽亚”匹配。
sed -i "/^maria\b/Id" file.txt
Edit: fixed case insensitive flag to I instead of i!
编辑：将不区分大小写的标志固定为 I 而不是 i！
