bash awk 替换行中的字符串
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Awk replace string in line
提问by tlorin
NOTE: This question looks like a previously posted one, but as mentioned in the comments, it turned out to be a chameleon question.I accepted the answer, and I'm posting here the same problem but with slightly different "solution conditions".
注意：这个问题看起来像以前发布的一个，但正如评论中提到的，它原来是一个变色龙问题。我接受了答案，我在这里发布了同样的问题，但“解决方案条件”略有不同。
I have a file test.txtlike this (but containing many more lines)
我有一个test.txt这样的文件（但包含更多行）
/foo/bar/how /SOME_TEXT_HERE/hello
/foo/bar/are hello/SOME_OTHER_TEXT
/foo/bar/you hello
I want to get this output:
我想得到这个输出：
/foo/bar/how /SOME_TEXT_HERE/how
/foo/bar/are are/SOME_OTHER_TEXT
/foo/bar/you you
I have tried this:
我试过这个：
while read line
do
bla=$(echo $line | cut -f4 -d"/" | cut -f1 -d" ")
sed -i "s/hello/$bla/" test.txt
done <test.txt
But the output is:
但输出是：
/foo/bar/how /SOME_TEXT_HERE/how
/foo/bar/are how/SOME_OTHER_TEXT
/foo/bar/you how
NOTE:
注意：
I would like the solution to:
我想解决以下问题：
allow me to define a variable (here,
bla) that I will define manually and differently from one file to the other (so not using sthg like basic field position), but usingcutor other command as in my example)replace a specific string somewhere else in the line by this variable (so not a field position like
$2, but really something likes/hello/$bla)
允许我定义一个变量（在这里，
bla），我将手动定义一个文件到另一个文件（所以不要像基本字段位置那样使用 sthg），而是cut像我的例子一样使用或 其他命令）用这个变量替换行中其他地方的特定字符串（所以不是像 那样的字段位置
$2，而是像 那样s/hello/$bla）
I'm not sure of this is possible though (obviously I don't know how to do this by myself...), but thanks for your time trying! :)
我不确定这是否可行（显然我自己不知道如何做到这一点......），但感谢您的时间尝试！:)
回答by Claes Wikner
awk '{sub(/are hello/,"are are")sub(/you hello/,"you you")}1' file
/foo/bar/how /SOME_TEXT_HERE/hello
/foo/bar/are are/SOME_OTHER_TEXT
/foo/bar/you you
