All fields after $4

[columb]

This should be simple but I can't find it in the O'Reilly book.
What I want to do is the equivalent of

cut -d ' ' -f 7-

OK, the -d ' ' bit is easy (and the default) but how do I get every field after the 7th?

Thanks

Ceci n'est pas une signature
Columb Healy

 

[columb]

Er, excuse the inconsistancy between the heading and the 'cut' statement. It's the principle I'm after.

Thanks

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Columb Healy

 

[Annihilannic]

What you posted seems to work fine for me on Linux and Solaris, what operating system exactly?

[tt]$ echo one two three four five six | cut -d ' ' -f 3-
three four five six
$[/tt]

Annihilannic.

 

[columb]

Sorry, I need the awk equivalent. What I have a the moment is something along the lines of

ls -l | grep wantedfile | cut -d ' ' -f 6-

What I want is

ls -l | awk '/wantedfile/ {print ????}'

Apart from wishing to increase my awk knowlege whenever I see a grep with something else it always say to me that awk ought to be able to do it better.

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Columb Healy

 

[p5wizard]

You want the awk equivalent? Off-hand, something like

awk '{for (i=1;i<=4;i++) $i=""; print}'

I would think for 5th field and beyond.

I believe cut is rather pecky about the field separator - tab not space is the default, and multiple field separators mean that there's an empty field in between. (not so for awk).


HTH,

p5wizard

 

[Annihilannic]

Unfortunately it's not exactly brief:

[tt]ls -l | awk '/wantedfile/ {for (i=6;i<=NF;i++) {t=t $i OFS};print t}'[/tt]

Annihilannic.

 

[columb]

Thanks p5wizard and Annihilannic. I was hoping to do it without the loop but you've confirmed that I can't.

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Columb Healy

 

[p5wizard]

a dirty way without a loop if it's just a couple of fields:

awk '{$1=$2=$3=$4="";print}'

but the field separators get left behind.

HTH,

p5wizard

 

[vgersh99]

shift left by 4 fields

nawk '{$1=$2=$3=$4="";sub("^" FS "*","")}1' shift.txt

vlad
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