4

I want to delete the lines containing empty fields in the last row $7

File:

1 1479870 5022248660 1 40001 189445122 740020
1 1911574 3015889020 1 33001 162049034 633004
1 1569783 5029193930 1 22001 133687297 522216
1 2069616 1025856960 2 25001 185608704 
1 1741598 5021128160 1 44001 164870942 644027
1 1052941 5020319300 1 10001 156161802 610007
1 1686734 5020347480 1 13001 131405824 513304
1 1872263 5029089700 1 23001 185092353 723017

Desired output :

1 1479870 5022248660 1 40001 189445122 740020
1 1911574 3015889020 1 33001 162049034 633004
1 1569783 5029193930 1 22001 133687297 522216
1 1741598 5021128160 1 44001 164870942 644027
1 1052941 5020319300 1 10001 156161802 610007
1 1686734 5020347480 1 13001 131405824 513304
1 1872263 5029089700 1 23001 185092353 723017
  • awk 'NF >= 7' <file, note that it does not guarantee that the last field is empty, example the line were also deleted when one of other fields is empty. – cuonglm Feb 27 '17 at 10:06
  • How are your fields separated? Any amount of blank characters or one space character ? Is 1 2  4 5 6 7 (where there are two spaces in between 2 and 4) a line with a missing 3rd field or one with a missing 7th field? – Stéphane Chazelas Feb 27 '17 at 10:12
7

Try with this one :

awk  '$7!=""' file > final_output
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  • @Nydenn If helped mark it as answered. – Rakesh.N Feb 27 '17 at 10:12
0

If the fields are separated by one space character as opposed to any sequence of blank characters, that if

1 2  4 5 6 7

Is a line that has a 7th field but no 3rd field, you'd do:

awk -F '[ ]' '$7 != ""' < file > final_output

[ ] is a regular expression that matches the space character. We don't use -F ' ' as a single space as the field separator has a special meaning (gives the default field splitting behaviour where any sequence of one or more blank characters act as one separator and leading and trailing blanks are ignored).

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