bash 如何使用 sed/awk 替换逗号分隔字符串中的第 n 列/字段？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/9838160/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to replace the nth column/field in a comma-separated string using sed/awk?
提问by Peter Meier
assume I have a string
假设我有一个字符串
"1,2,3,4"
Now I want to replace, e.g. the 3rd field of the string by some different value.
现在我想用一些不同的值替换，例如字符串的第三个字段。
"1,2,NEW,4"
I managed to do this with the following command:
我设法使用以下命令做到了这一点：
echo "1,2,3,4" | awk -F, -v OFS=, '{="NEW"; print }'
Now the index for the column to be replaced should be passed as a variable. So in this case
现在，要替换的列的索引应作为变量传递。所以在这种情况下
index=3
How can I pass this to awk? Because this won't work:
我怎样才能把它传递给 awk？因为这行不通：
echo "1,2,3,4" | awk -F, -v OFS=, '{$index="NEW"; print }'
echo "1,2,3,4" | awk -F, -v OFS=, '{$($index)="NEW"; print }'
echo "1,2,3,4" | awk -F, -v OFS=, '{$$index="NEW"; print }'
Thanks for your help!
谢谢你的帮助！
采纳答案by Jens
Have the shell interpolate the index in the awk program:
让 shell 在 awk 程序中插入索引：
echo "1,2,3,4" | awk -F, -v OFS=, '{$'$index'="NEW"; print }'
Note how the originally single quoted awk program is split in three parts, a single quoted beginning '{$', the interpolated index value, followed by the single quoted remainder of the program.
请注意最初的单引号 awk 程序是如何分成三个部分的，一个单引号开头的 '{$'，插入的索引值，然后是程序的单引号其余部分。
回答by potong
This might work for you:
这可能对你有用：
index=3
echo "1,2,3,4" | awk -F, -v OFS=, -v INDEX=$index '{$INDEX="NEW"; print }'
or:
或者：
index=3
echo "1,2,3,4" | sed 's/[^,]*/NEW/'$index
回答by Lev Levitsky
Here's a seductive way to break the awkwardness:
这是sed打破awk守望的有效方法：
$ echo "1,2,3,4" | sed 's/,/\n/g' | sed -e $index's/.*/NEW/'
This is easily extendable to multiple indexes just by adding another -e $newindex's/.*/NEWNEW/'
只需添加另一个索引即可轻松扩展到多个索引 -e $newindex's/.*/NEWNEW/'
回答by baz
# This should be faster than awk or sed.
str="1,2,3,4"
IFS=','
read -a f <<< "$str"
f[2]='NEW'
printf "${f[*]}"
回答by JimR
With plain awk (I.E. Not gawk etc) I believe you'll have to use split( string, array, [fieldsep] );change the array entry of choice and then join them back together with sprintfor similar in a loop.
使用普通的 awk（IE 不是 gawk 等），我相信您必须使用split( string, array, [fieldsep] );更改选择的数组条目，然后将它们sprintf以循环方式或类似方式重新连接在一起。
gawk allows you to have a variable as a field name, $index in your example. See here.
gawk 允许您将变量作为字段名称，在您的示例中为 $index 。看到这里。
gawk is usually the default awk on Linux, so change your invocation to gawk "script" and see if it works.
gawk 通常是 Linux 上的默认 awk，因此将您的调用更改为 gawk "script" 并查看它是否有效。
