shell script writing

ust · 01-23-2005, 04:57 AM

I am new to shell writing , could suggest how to write the below script,

I have the below file , the content as below
#vi file.txt

10 11 12 13
20 21 22 23
30 31 32 33

then run the script , the new output as below,

10 13 11 12
20 23 21 22
30 33 31 32

could suggest how to write the script ? thx

frob23 · 01-23-2005, 05:28 AM

Not really a shell script... one command to do that.

Code:

awk 'BEGIN{FS=" "} {print $1, $4, $2, $3}' file.txt

ust · 01-23-2005, 08:09 PM

thx reply , it work,

but how to add space to each column , eg. if I want to have 3 space line between $1 and $4 , how to set it ? thx

homey · 01-23-2005, 08:29 PM

Just put the spaces in quotes
awk '{print $1" "$4" "$2" "$3}' file.txt

ust · 01-24-2005, 12:45 AM

thx reply , but still have a problem if add the space to each column

the output is not tidy now, because the space seems start from the last character of previous word ,as the file.txt not exactly as the above exapmle

my file.txt should be as below ,

10 133 11 12
200 23333 211 22
30000 33 3111 32

if I add 4 spaces to each column , the out will be as below.

10 133 11 12
200 23333 211 22
30000 33 3111 32,

could suggest how to make the result is more tidy as below? thx

homey · 01-24-2005, 09:17 AM

Quote:

the output is not tidy now, because the space seems start from the last character of previous word

I think tabs would be better in this case. If some nunbers are quite large, you may want more than one tab..... "\t\t"

Code:

awk '{print $1"\t"$4"\t"$2"\t"$3}' file.txt

LasseW · 01-24-2005, 01:06 PM

You can also use c style format strings with the awk printf command:

awk '{printf "%6d%6d%6d%6d\n",$1,$2,$3,$4}'