UNIX for Dummies Questions & Answers

This is what I need to do

I have a file that has a field with values like this

1111 2222
3333 4444
55555 666
333333333
444444444

I need for my command to out put only those fields that do not have spaces in them. So my output for the above file would be

333333333
444444444


how do I go about this?

Code:

nawk 'NF==1' myFile

Thanks man....it worked...

Wat is nawk??? I tried awk 'NF==1' myFile and tht worked too!

an 'N' in 'nawk' stands for 'New'.

Thanks vgersh appreciate your help....I also came up with the following during one of my searches

"After the authors got over their initial consternation, they decided to accept the fact, and enhance Awk to make it a better general-purpose programming tool. The new version of Awk was released in 1985. Since the old Awk implementation is still the standard in UN*X systems, the new version is often, if not always, known as Nawk ("New Awk") to distinguish it from the old one."

from http://www.vectorsite.net/tsawk3.html

Hi

can you also tell me how i can acheive the below


i have a code in which i assign a value to a str variable.
Say line contains,
line=Field|Location ES|comp|dur|10

str=`echo $line | awk '{split($0, date,"|"); print date[2]"|"date[3]"|"date[4] }'`

i want to grep for str value in a file in the script.

for j in `cat uniqfile | egrep "${str}"`

but this is giving me the output as
Field|location
ES|comp|dur|10

Whereas i want the output as

Field|Location ES|comp|dur|10


how can i suppress the space here?

could you please help. Thanks, Mike

This is because the default input separator is a space. You could get away with reassigning IFS, though I would highly recommend you use a tool like awk to do this.

Code:

line=Field|Location ES|comp|dur|10

str=`echo $line | awk '{split($0, date,"|"); print date[2]"|"date[3]"|"date[4] }'`

export IFS="
"

for j in `cat uniqfile | egrep "${str}"`
do echo $j
done

export IFS=" "

for j in `cat uniqfile | egrep "${str}"`
do echo $j
done

awk ' NF > 0 ' uniqfile