awk -repeated pattern match

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1

Hi.
How can I write this differently:

awk '$3 ~ /0001/{print}'

Is there a way to write 0001 differently. I am looking for the pattern 01, with 3 or more 0 and 3 or more 1 in a pattern.
Thanks.

2 
 
/0000*1111*/
3 
awk '$3 ~ /0{3,}1{3,}/' file
4

/0000*1111*/ is nice but what if I want to match 0 twenty times?
I am not going to type 000000000....

awk '$3 ~ /0{3,}1{3,}/' file is not working :frowning:

5

Could you explain what doesn't work?

Which OS?

If Solaris, try nawk.

6

In sed, it's a bit different, \ and second count limited to 99; maybe in awk, too:

 
/0\{3,99\}1\{3,99\}/
7

I am using a Mac OS X.
Not working means that I am not getting anything back from the command even though I know that I should. It looks like nothing matches the awk pattern, but I am positive that the pattern I am looking for exists.

8

Please post the actual line that you want to match

9

For example:
0000000111111112222222
00111111111111111111
2222222444444401111111

The awk should only return the first line, not the last two lines.

10

I'm also using OS X.

$ system_profiler SPSoftwareDataType
Software:

    System Software Overview:

      System Version: Mac OS X 10.6.5 (10H574)
....

Interestingly, your sample data has only one ($1) field?

$ awk '$1 ~ /0{3,}1{3,}/' file
0000000111111112222222
11

Thanks!

12 
 awk '/000+/{print}' file
13

That'd be /000+111+/ if I recall the user requirement correctly, except he wants to specify the min length not in unary!