How to remove all words starting from a matching word in a line

Shell Programming and Scripting
1

Hi Guys,

I have a file like this:

 
wwwe 1 ioie ewew yyy uuu 88
erehrlk 4 ihoiwhe lkjhassad lkhsad yyy mmm 45
jhash lhasdhs lkhsdkjsn ouiyrshroi oihoihswodnw oiyhewe yyy ggg 77

I want to remove everything after "yyy" and including "yyy" from each line in the file.
So I want:

wwwe 1 ioie ewew 
erehrlk 4 ihoiwhe lkjhassad lkhsad 
jhash lhasdhs lkhsdkjsn ouiyrshroi oihoihswodnw oiyhewe

How do I do this?

Thanks.

2

Hi.

You can do it with sed:

$ cat file1
wwwe 1 ioie ewew yyy uuu 88
erehrlk 4 ihoiwhe lkjhassad lkhsad yyy mmm 45
jhash lhasdhs lkhsdkjsn ouiyrshroi oihoihswodnw oiyhewe yyy ggg 77

$ sed "s/yyy.*//" file1
wwwe 1 ioie ewew 
erehrlk 4 ihoiwhe lkjhassad lkhsad 
jhash lhasdhs lkhsdkjsn ouiyrshroi oihoihswodnw oiyhewe 

# I suppose more accurate would be
sed "s/\wyyy.*//" file1
# or
sed "s/  *yyy.*//" file1
3

You can use this to get everything before the pattern yyy.
Here I used yyy pattern as field seperator.

 
cat file|awk -F 'yyy' '{print $1}'

Thanks