Linux 使用awk打印从第n到最后的所有列
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原文地址: http://stackoverflow.com/questions/2961635/
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Using awk to print all columns from the nth to the last
提问by Andy
This line worked until I had whitespace in the second field.
这条线一直有效，直到我在第二个字段中有空格。
svn status | grep '\!' | gawk '{print ;}' > removedProjs
is there a way to have awk print everything in $2 or greater? ($3, $4.. until we don't have anymore columns?)
有没有办法让 awk 以 2 美元或更高的价格打印所有内容？（3 美元，4 美元 .. 直到我们不再有列？）
I suppose I should add that I'm doing this in a Windows environment with Cygwin.
我想我应该补充一点，我是在带有 Cygwin 的 Windows 环境中执行此操作的。
采纳答案by zed_0xff
will print all but very first column:
将打印除第一列之外的所有内容：
awk '{=""; print awk '{==""; print awk '{out=""; for(i=2;i<=NF;i++){out=out" "$i}; print out}'
}' somefile
}' somefile
will print all but two first columns:
将打印除第一列之外的所有列：
awk '{for(i=2;i<=NF;i++){printf "%s ", $i}; printf "\n"}'
回答by VeeArr
You could use a for-loop to loop through printing fields $2 through $NF (built-in variable that represents the number of fields on the line).
您可以使用 for 循环来循环打印字段 $2 到 $NF（表示行上字段数的内置变量）。
Edit: Since "print" appends a newline, you'll want to buffer the results:
编辑：由于“打印”附加换行符，您需要缓冲结果：
awk '{print substr(awk '{out=; for(i=3;i<=NF;i++){out=out" "$i}; print out}'
,length()+1);}' < file
Alternatively, use printf:
或者，使用 printf：
svn status | grep '\!' | cut -d\ -f2-
回答by whaley
Would this work?
这行得通吗？
echo "1 2 3 4 5 6" | awk '{ $NF = ""; print ls -lthr | awk '{out=; for(i=7;i<=NF;i++){out=out" "$i}; print out}'
}'
It leaves some whitespace in front though.
不过，它在前面留下了一些空白。
回答by Wim
ls -lthr | awk '{ORS=" "; for(i=6;i<=NF;i++) print $i;print "\n"}'
My answer is based on the one of VeeArr, but I noticed it started with a white space before it would print the second column (and the rest). As I only have 1 reputation point, I can't comment on it, so here it goes as a new answer:
我的答案基于VeeArr 之一，但我注意到它在打印第二列（和其余列）之前以空格开头。由于我只有 1 个声望点，我无法对此发表评论，因此这里作为一个新答案：
start with "out" as the second column and then add all the other columns (if they exist). This goes well as long as there is a second column.
以“out”作为第二列开始，然后添加所有其他列（如果存在）。只要有第二列，这就会顺利进行。
回答by Joshua Goldberg
There's a duplicate question with a simpler answerusing cut:
有一个重复的问题，使用 cut有一个更简单的答案：
ls -l | awk '{sub(/[^ ]+ /, ""); print awk -F" " '{ for (i=4; i<=NF; i++) print $i }'
}'
-dspecifies the delimeter (space), -fspecifies the list of columns (all starting with the 2nd)
-d指定分隔符（空格），-f指定列的列表（都从第 2 个开始）
回答by Kaushal Jha
@m=`ls -ltr dir | grep ^d | awk '{print $6,$7,$8,$9}'`;
foreach $i (@m)
{
print "$i\n";
}
this one uses awk to print all except the last field
这个使用 awk 打印除最后一个字段之外的所有内容
回答by Manuel Parra
This is what I preferred from all the recommendations:
这是我从所有建议中更喜欢的：
Printing from the 6th to last column.
从第 6 列到最后一列打印。
##代码##or
或者
##代码##回答by savvadia
Printing out columns starting from #2 (the output will have no trailing space in the beginning):
打印从 #2 开始的列（输出开头没有尾随空格）：
##代码##回答by koullislp
I personally tried all the answers mentioned above, but most of them were a bit complex or just not right. The easiest way to do it from my point of view is:
我个人尝试了上面提到的所有答案，但大多数答案都有些复杂或不正确。从我的角度来看，最简单的方法是：
##代码##Where -F" " defines the delimiter for awk to use. In my case is the whitespace, which is also the default delimiter for awk. This means that -F" " can be ignored.
Where NF defines the total number of fields/columns. Therefore the loop will begin from the 4th field up to the last field/column.
Where $N retrieves the value of the Nth field. Therefore print $i will print the current field/column based based on the loop count.
其中 -F" " 定义了 awk 使用的分隔符。在我的例子中是空格，它也是 awk 的默认分隔符。这意味着 -F" " 可以被忽略。
其中 NF 定义字段/列的总数。因此循环将从第 4 个字段开始直到最后一个字段/列。
其中 $N 检索第 N 个字段的值。因此 print $i 将根据循环计数打印当前字段/列。
回答by pkm
Perl:
珀尔：
##代码##