将多行字符串转换成单逗号分隔 [英] Turning multi-line string into single comma-separated

问题描述

假设我有以下字符串:

something1:    +12.0   (some unnecessary trailing data (this must go))
something2:    +15.5   (some more unnecessary trailing data)
something4:    +9.0   (some other unnecessary data)
something1:    +13.5  (blah blah blah)

如何将其简单化

+12.0,+15.5,+9.0,+13.5

在bash中?

推荐答案

您可以使用awk和sed:

awk -vORS=, '{ print $2 }' file.txt | sed 's/,$/\n/'

或者如果您想使用管道:

Or if you want to use a pipe:

echo "data" | awk -vORS=, '{ print $2 }' | sed 's/,$/\n/'

要分解它:

• awk非常适合处理细分为字段的数据
• -vORS=,将输出记录分隔符"设置为,，这就是您想要的
• { print $2 }告诉awk为每个记录(行)打印第二个字段
• file.txt是您的文件名
• sed只是删除了结尾的,并将其转换为换行符(如果不需要换行符，可以执行s/,$//)

• awk is great at handling data broken down into fields
• -vORS=, sets the "output record separator" to ,, which is what you wanted
• { print $2 } tells awk to print the second field for every record (line)
• file.txt is your filename
• sed just gets rid of the trailing , and turns it into a newline (if you want no newline, you can do s/,$//)

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