I think that is the question as I understood it. I made the assumption that it should match words like "zffz": if that is not acceptable then you can just pipe the result through "grep -v 'z.*z' ".

All you have to do is translate this:
into a regular expression. The translation is very direct.

hi

Yes, the words found should only contain a single z so zffz is not acceptable.

\(either the beginning of a line OR a non-vowel, non-z character \) z \(either the end of a line OR a non-vowel, non-z character \)

=>is this it?
\(^[^aeiouzAEIOUZ]\)z\([^aeiouzAEIOUZ]$\)

If this is it then i dun tink i can get hte answer for both grep or egrep.
Sorry to say this bu ti dun quite understand ur pseudocode for the above, I just touch on LInux at the beginning of this month aka newbie woth poor english.

Mayb if you could furthur elaborate? or point out my mistakes?

That's almost exactly it. You need to use '\|' when you mean OR, and you might want to change the z in the middle to [zZ]. Alternatively, you could leave everything lowercase and use the -i option to grep.

I've written this for grep, not egrep. You probably have to add or remove some backslashes in order to switch between them.

hi

ok so with what you said, this shld be it

'\(^\|[^aeiouzAEIOUZ]\)z\([^aeiouzAEIOUZ]\|$\)'

and the full command is

grep -i '\(^\|[^aeiouzAEIOUZ]\)z\([^aeiouzAEIOUZ]\|$\)' /usr/share/dict/words

and yes i tink i got it.

ok u r good.

sorry for the previous tries.

(anyway i din know we can use () inside the expression. I hve nvr come across the use of this ().)

=>ok mayb tis is how i see ur command.

beginning wif z, no vowels infront of z, no vowels after z, ending wif z

is that rite?...just use it like a mathematical quesiotn? multiply z into the brackets and split them up?.m i correct?

If im correct, can u help me with another prob.

im trying to get lower case words in the same path as above but i cant. this aim is part of another question but i would wan to tackle this part 1st.

i try
grep [:lower:] /usr/share/dict/words
but they still show upper case letters.
patterns that i tried include
[[:lower:]], '[:lower:]', '[[:lower:]]', "[:lower:]", "[[:lower:]]", '[^[:upper:]]', "[^[:upper:]]"

but to no avail. How do i do that? its such a simple command and yet i failed to achieve that. I m clueless. i try egrep oso, bt to no avail.

anyway the whole question requires these conditions

all words containing 'ae', beginning with lowercase letter, not beginning or ending with 'ae', assume 'ae' appears only once in each line and do this in 1 regular expression.

i m able to get not beginning or ending with ae with this
grep '[^ae]\|[ae$]' /usr/share/dict/words

containing wif 'ae'
grep 'ae' /usr/share/dict/words

but not the small caps words. Can you help?

For the first command, don't forget to pipe the result through something like "grep -vi 'z.*z' " to get rid of words with two nonadjacent z's.

For lowercase words, you're on the right track but you forgot that grep finds *any* occurence of what you search for. So for example grep '[[:lower:]]' would find any word with at least one lowercase letter. You could solve this in one of two ways:

• Use the -v switch to grep to find words not containing an uppercase letter
• Search for the beginning of a line followed by some lowercase letters followed by the end of a line

Hi

thanks for the tip. But in the 1st command w/o using the grep command, there are no words that contain 2 non adjacent z. Thanks.

Oh yes thanks for reminding about the -v option and the using of the [:upper:].

Can I also check with you, if this command works for you?
grep 'z{2}' /usr/share/dict/words

or is this command?
grep z\{2\} /usr/share/dict/words

or this command?
grep 'z\{2\}' /usr/share/dict/words

which command will display words with 2 z in it? I tried all 3, and only the last command works. May I know why isit so?

the second command (the one without 's) doesn't work because the shell interprets the \s as escaping the {s before ever calling grep, so the second command is equivalent to the first. In the third, the 's keep the shell from interpreting the stuff inside them, so grep gets all of what you typed (excepting the 's of course). The \s turn on grep's special meaning for {.

If you use egrep, the first two will work but the third won't, since { has special meaning by default in egrep, and \ turns it off.

Hi,

thanks. some how I was told the other way round. No wonder i cant figure out whose the rite 1. Since u clarify it for me, im grateful. thanks.

ok here's another prob that i encounter...mayb u can guide me like wat u did initially,as in tell me in pseudocode or something like tat. thanks

I have a text file, which contains 4 columns of words and numbers separated with a tab

1st col= ID (numeric)
2nd col= Name (alphabets)
3rd col= Hours worked (Numeric)
4th col= Hourly Pay (Numeric)

I would need to get the ID and hours worked for employees whose hourly pay is more than 20.
I tried
cut -f1,3,4 | grep '.*[2-9].*'
cut -f1,3,4 | grep '.*[^0-1].*'

Any advice?

Are you familiar with awk? If not, see its man page. This becomes *really* easy if you do use awk.

Hi,

ya i do know abt awk a little. but i was told to skip awk...i hve no idea why...seems so easy with awk. sigh. anyway can i ask you if new hampshire is in UK or US?

Hi,

do you know about ed editor?.

i tried

a testing
a "testing"
a 'testing'

in order to add a few lines but to no avail. any advice?

try

grep -Ev '(za)|(ze)|(zi)|(zo)|(zu)' FILENAME

?

tat is another idea of doing it...let me try n get back to u..thanks

you can try with this... this should work..

grep "[z]\{1\\}" file.txt | grep -v "z[aeiouAEIOU]" | grep -v "[aeiouAEIOU]z"

where file.txt is the name of your file