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Shell Programming and Scripting

Parsing problem

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# 1  
Old 04-21-2011
Parsing problem
Hello,
I have a similar problem so I continue this thread.
I have:
Code:
my_script_to_format_nicely_bdf.sh | grep "RawData" |tr -s ' '|cut -d' ' -f 4|tr -d '%'

So it supposed to return the percentage used of RawData FS:
80
(Want to use it in a alert script)
However I also have a RawData2 FS so it will output both:
80
60

So in the grep part how can I declare that there is no character after RawData like RawData2?

Thanks in advance

Moderator's Comments:
Mod Comment Even though it is similar, as many things can be similar, it is better to open up a new post instead of "hi-jacking" an existing one Smilie Here you go.
Last edited by zaxxon; 04-21-2011 at 08:43 AM.. Reason: splitting post to separate thread
# 2  
Old 04-21-2011
Easiest would be to post the unfiltered output of your script.
# 3  
Old 04-21-2011
Sorry for hijacking. I thought it will be better to continue as it is a similar case.

What do you mean unfiltered output?

---------- Post updated at 02:51 PM ---------- Previous update was at 02:48 PM ----------

./bdf_rep.sh | grep "RawData" |tr -s ' '|tr -d '%'

138048 122619 15429.5 89 /RawData
193632 116719 76399.9 60 /RawData2

So i need only to display the first line how do I filter this out with grep ?
# 4  
Old 04-21-2011
To grep FS 80% or more :

Code:
df -k | grep [890][0-9]%

# 5  
Old 04-21-2011
Does not work cause the output of df is messy ( due to long FS names)
# 6  
Old 04-21-2011
Try this

Code:
./bdf_rep.sh | grep "RawData$"

regards,
Ahamed
# 7  
Old 04-21-2011
try :

df -k | cat | grep [890][0-9][%]

---------- Post updated at 02:17 PM ---------- Previous update was at 02:17 PM ----------

could you post an example of output of
./bdf_rep.sh
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