regex - Unix pattern datetime match -


i want edit line:

1987,4,12,31,4,1987-12-31 00:00:00.0000000,ua,19977,ua,,631,12197,1219701,31703,hpn,white plains, ny,ny,36,new york,22,13930,1393001,30977,ord,chicago\, il,il,17,illinois,41,756,802,483.2,6,6,0,0,0700-0759,,,,,914,938,600.8,24,24,1,1,0900-0959,0,,0,138,156,,1,738,3,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,us1njbg0005,us1ilck0027,,,,,,,,,,,,,1987-12-31 08:09:12.0000000,519494350

and want output :

1987,4,12,31,4,1987-12-31 00:00:00.000,ua,19977,ua,,631,12197,1219701,31703,hpn,white plains, ny,ny,36,new york,22,13930,1393001,30977,ord,chicago\, il,il,17,illinois,41,756,802,483.2,6,6,0,0,0700-0759,,,,,914,938,600.8,24,24,1,1,0900-0959,0,,0,138,156,,1,738,3,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,us1njbg0005,us1ilck0027,,,,,,,,,,,,,1987-12-31 08:09:12.000,519494350

i want find each pattern of: ****-**-** **:**:**.0000000

and erase last 4 digits ( 0000 ) ****-**-** **:**:**.000.

if helpful date format in 6th columns , n-1 columns.

to value of 6th column , erase last 4 digits can use:

awk -f, '{print substr($6, 0, length($6)-4) }'

similarly, n-1 column can reached by:

awk -f, '{print substr( $(nf-1), 0, length($(nf-1))-4) }'

edit:

to replace values in columns, still print use:

awk 'begin{ fs=","; ofs=","}  { $6=substr($6, 0, length($6)-4);    $(nf-1)=substr( $(nf-1), 0,length($(nf-1))-4);    print $0}'

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