bash 如何修改GREP输出？
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How to modify GREP output?
提问by knittl
I have a script which uses grep to output like this:
我有一个脚本，它使用 grep 输出如下：
12 Apples
10 Grapes
11 Mangoes
I want to the output as
我想输出为
Apples: 12 Grapes: 10 Mangoes: 11
How to modify the grep output?
如何修改grep输出？
回答by FatalError
You could pipe to awk like:
您可以通过管道传输到 awk，例如：
$ echo "
12 Apples
10 Grapes
11 Mangoes
" | awk 'BEGIN { OFS=": "; ORS=" "; } NF >= 2 { print , ; }'
Apples: 12 Grapes: 10 Mangoes: 11
replacing, of course, the echo cmd with the grepyou are doing.
当然，将 echo cmd 替换为grep您正在执行的操作。
回答by Dan Fego
I'd use awk:
我会用awk：
$ grep regex file | awk '{ printf "%s: %d ", , }; END { printf "\n" }'
Apples: 12 Grapes: 10 Mangoes: 11
This would use printfto print out your lines to the specified format, and then add a newline at the end, since that's what you probably want. If you're not familiar with awk, $1and $2represent positional parameters, which is to say, the first (in this case, space-separated) field, and the second.
这将用于printf将您的行打印为指定的格式，然后在末尾添加一个换行符，因为这可能是您想要的。如果您不熟悉awk,$1并$2表示位置参数，也就是说，第一个（在本例中为空格分隔）字段和第二个字段。
回答by knittl
You can use sed:
您可以使用 sed：
grep … | sed 's/^\([0-9]*\) \(.*\)$/: /'
回答by William Pursell
If you are willing to have no spaces around the colon in the output you can simply do:
如果您愿意在输出中的冒号周围没有空格，您可以简单地执行以下操作：
$ grep ... | tr ' 2' ': '; echo
This just changes the spaces to colons and the newlines to spaces. The echo is there to give a trailing newline.
这只是将空格更改为冒号，将换行符更改为空格。回声在那里给出一个尾随换行符。
回答by ДМИТРИЙ МАЛИКОВ
Using ghc:
使用ghc：
$> cat ./text
12 Apples
10 Grapes
11 Mangoes
$> cat ./text | ghc -e "getContents >>= putStrLn . concatMap ((\(num,name) -> (drop 1 name) ++ \": \" ++ num ++ \" \") . break (== ' ')) . lines"
Apples: 12 Grapes: 10 Mangoes: 11
回答by potong
This might work for you:
这可能对你有用：
echo -e "12 Apples\n10 Grapes\n11 Mangoes" |
sed ':a;$!{N;ba};s/\(\S*\) \(\S*\)\n*/: /g'
Apples: 12 Grapes: 10 Mangoes: 11
