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Recent content by beardown
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awk print records with column is not blank
how to find/print all records that are not blank in column 2 it must be some variation of this # delete ALL blank lines from a file (same as "grep '.' ") awk NF awk '/./' -
awk compare column with date format
file1 25480820~a~b~c~d~e~f~g~h~09/10/09 26701789~a~b~c~d~e~f~g~h~12/06/10 26701789~a~b~c~d~e~f~g~h~03/28/13 22907330~a~b~c~d~e~f~g~h~07/29/11 file2 25480820~a~b~c~d~e~f~g~h~09/10/09 25480820~a~b~c~d~e~f~g~h~11/12/10 26701789~a~b~c~d~e~f~g~h~12/06/10 26701789~a~b~c~d~e~f~g~h~01/11/11... -
awk compare column with date format
I have this code to compare columns 1 and 10 between file1 and file 2 and give me all records that match column 1 but dont match column 10 However column 10 is date format mm/dd/yy and awk cant read it and compare ...i tried CODE awk < file1 -F~ '{print $10}' CODE and it gave blank screen...
