Print lines where there's no indent on the first field

Shell Programming and Scripting
1

Hi All,

I need a code to print those lines where there's NO indents on the 1st field
Example shown below.
I tried to use the below codes but i am not able to see the expected result.
Can any expert give any advise ?

My Code

cat filename| awk '$1 ~ /^[0-9]+$/ {print $0}'

Input

1199  ccc  zzz
1200  bbb  nnn
      1200  mmm  kkk
      1200  ppp  yyy
      1200  uuu  ttt
1201  vvv  ppp
1202  ccc  qqq

Output

1199  ccc  zzz
1200  bbb  nnn
1201  vvv  ppp
1202  ccc  qqq
2 
grep -v '^[[:blank:]]' filename
3

Hi cfajohnson,

I tried your code but it does not get the expected result.

4

Try this:

grep '^   *'  filename

So it was a NOT??? Didn't noticed that. Post below is correct

5 
sed '/^ /d' file
6

Which character do you use to indent? space or tab..you need to use that char after the ^ in the reg exp.

grep  -v '^ ' filename
7

So what did you get? In what way was it not what you wanted?

8

Hi cfajohnson,

Perhaps i misinterpret your code because i copied the whole line exactly.
Now i notice that the word "blank " in your code is suppose to have empty spaces in it.
Anyway, thanks alot for your advice.
The below code works perfectly. :b:

grep -v '^   *' filename