Need to delete all lines where any line meets a condition

Shell Programming and Scripting
1

Here is an example of a file...

foo1,good
foo1,good
foo2,error
foo2,good

Note that both rows for foo1 have good in the 2nd field, but one of the foo2 rows has error...

I need something in ksh/awk/perl that will delete ALL foo2 lines if ANY of them have error in the 2nd field...so:

foo1,good
foo1,good
foo2,error
foo2,good

becomes:

foo1,good
foo1,good

Please help...I know it should be simple, but I can't figure it out.

Thanks!

2

try this...

grep -v $(awk -F, '$2 ~ /error/{print $1}' file1) file1
3 
{ grep -q "foo2,error" file1 && grep -v "foo2," file1 ; } || cat file1
4

awk version:

awk -F, '$2=="error"{F[$1]} FNR!=NR && !($1 in F)' infile infile

(the input file is specified twice)

5

Sorry for the late replies...I've been trying to figure this out as much as possible myself and I'm just plain stuck...

This gives me
awk: syntax error near line 1
awk: bailing out near line 1

My flavor of Unix doesn't support grep -q

This simply displays my file as is and doesn't make any changes.

6

it can't write to file. you need to redirect it.

grep .... file > temp 
mv temp file

also try with nawk

7

try:

awk -F, '
{
 a[NR]=$0;
 if ($0 ~ /foo2.*error/) e=$1;
}
END {
 for (i=1; i<=NR; i++) {
   if (!e) {
     print a[1];
   } else {
     if (a !~ e) print a;
   }
 }
}
' file1
8

On Solaris use /usr/xpg4/bin/awk or nawk rather than awk