UNIX for Dummies Questions & Answers

Hi,


To load a big file in a table,I have a make sure that all rows in the file has same number of the columns .

So in my file if I am getting any rows which have columns not equal to 6 , I need to delete it . Delimiter is space and columns are optionally enclosed by "".

This can be identified like this :
COLUMN_1 COLUMN2 COLUMN_3 COLUMN4 COLUMN_5 COLUMN6

Correct Records :
123 1 3 test "abc def test" 200
234 1 3 "test" "abc def test" 200
456 1 3 "test" "abc def test" "abc 200"

Records to be deleted from the file
678 1 3 test "abc def test"
789 1 3 "test" "abc def test" 200 1234


Row 678 .. has number of columns as 5 instead of 6 and row 789 has number of columns at 7 instead of 6. So need to delete these records from the file.


Can count the number of columns using this , but not sure to proceed further.
perl -nle 's/".*?"//g;print s/,//g+1'


Any pointed further will be appreciated.

Assuming that the caret character will never occur in your input, try:

Code:

awk -F\" '{t=$0;for(i=2;i<=NF;i+=2) gsub(/[ \t]+/,"^",$i)}split($0,a,OFS)==6{print t}' file