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sed command to grep multiple pattern present in single line and delete that line

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Top Forums Shell Programming and Scripting sed command to grep multiple pattern present in single line and delete that line
# 8  
Old 05-14-2013
Then this should do it:
Code:
awk '!(/7800/ &&  /REJECT/ || /3306/ && /DROP/ || /5070/ && /REJECT|DROP/)' fileName >newfile

No output to console, only to a new file named newfile
# 9  
Old 05-14-2013
okay... but as i said before i dont want to redirect to new file... if i redirect to same file the content will be lost.

i just want a command which after executing it, it should remove the lines from that file. Now for the above process i need to execute two command one for redirecting to new file and again copy it to old file or rename the file.
# 10  
Old 05-14-2013
Code:
awk '!(/7800/ &&  /REJECT/ || /3306/ && /DROP/ || /5070/ && /REJECT|DROP/)' fileName >newfile
rm fileName 
mv newfile fileName

This User Gave Thanks to Jotne For This Post:
vivek d r
# 11  
Old 05-14-2013
Code:
sed -i '/7800.*REJECT/d; /3306.*DROP/d; /5070.*REJECT/d; /5070.*DROP/d;' fileName

This User Gave Thanks to MadeInGermany For This Post:
vivek d r
# 12  
Old 05-14-2013
Thanks Jotne and MadeinGermany... :-)
# 13  
Old 05-14-2013
Code:
/5070.*REJECT/d; /5070.*DROP/d;

This regex /5070.*(REJECT|DROP)/ seems to work on awk, but not sed, any idea.

Edit
Shorten my awk
Code:
awk '!(/7800.*REJECT/ || /3306.*DROP/ || /5070.*(REJECT|DROP)/)' fileName

PS Using regex this way 7800 needs to come before REJECT. It will not work the other way around.
Last edited by Jotne; 05-14-2013 at 08:17 AM..
# 14  
Old 05-14-2013
GNU sed takes /5070.*\(REJECT\|DROP\)/
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