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04-17-2008
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awk print the next line on the current line
Ok I have a file with hundreds of lines, four columns, space delimited, TESTB.TXT for example
TESTB.TXT --- AA ZZ 12 34 BB YY 56 78 CC XX 91 23 DD VV 45 67 --- I want a new file that has 7 columns, the first four are identical, and the next 3 are the last three of the next line...so for example TESTB2.TXT --- AA ZZ 12 34 YY 56 78 BB YY 56 78 XX 91 23 CC XX 91 23 VV 45 67 DD VV 45 67 --- I hope I typed that right and I hope the pattern is obvious. Obviously the last line of the file wont have a next line, so its last3 columns are blank. no problem there. or if you know of a similar script just point me to it and I'll hack out the details. 
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04-17-2008
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$[/tmp] > more test.txt
AA ZZ 12 34 BB YY 56 78 CC XX 91 23 DD VV 45 67 $[/tmp] > echo "" > test1.txt $[/tmp] > more test1.txt $[/tmp] > cat test1.txt test.txt > test2.txt $[/tmp] > more test2.txt AA ZZ 12 34 BB YY 56 78 CC XX 91 23 DD VV 45 67 $[/tmp] > cat test.txt | awk '{print $2 " " $3 " " $4}' > test3.txt $[/tmp] > more test3.txt ZZ 12 34 YY 56 78 XX 91 23 VV 45 67 $[/tmp] > paste test2.txt test3.txt > test4.txt $[/tmp] > more test4.txt ZZ 12 34 AA ZZ 12 34 YY 56 78 BB YY 56 78 XX 91 23 CC XX 91 23 VV 45 67 DD VV 45 67
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04-17-2008
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With awk:
Code:
awk '
NR==1{printf("%s", $0);next}
{print FS $2,$3,$4;printf("%s", $0)}
END{if(!(NR%2)){print ""}}
' file
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