grep in ubuntu for text occurrence - grep

I currently have a text file which contains the following lines of text in it. I need to use grep to display the number of lines the word test appears in the file for different scenario. Below are some things I have tried, but the output is wrong. I am pretty new to using grep, so hope someone could assist to advice where I have done wrong.
test.txt:
this is some testing for my own test purpose.
testing testing testing 1 2 3 test
test-case 1 is for delete purpose
tester’s is coming to test out the new devices
Tester test 3 test case
Scenario 1:
Report on how many lines in the text file test contain the case-sensitive word “test”. It should not include count of word that is a substring of another word. E.g testing, tester’s, test-case
Ans: grep “\btest\b” test | wc -l
Scenario 2:
Report on how many times the case-sensitive word “test” appears in the text file test (within other words is ok)
Ans: grep –o “test” test | wc -w
Scenario 3:
Report on how many times the word “test” appears in the last 2 lines of the text file test
Ans: tail –n 2 test | grep –o “\btest\b” | wc -w

You can try the following :
Scenario 1:
$ grep -i -c "[^a-z]test[^a-z]" test.txt
Scenario 2:
$ grep –o “test” test.txt | wc -l
Scenario 3:
$ tail -2 test.txt | grep -o "[^a-z]test[^a-z]" | wc -l

Related

print the rest of input along with matching line

I am new to linux and I am experimenting with basic terminal commands. I found out that I can list all users using compgen -u but what if I only want to display the bottom line outputs ?
Ok lets say the output of compgen -u goes like this:
extra
extra
extra
extra
extra
extra
extra
extra
extra
John
William
Kate
Harold
I can only use grep to find a single text (ex. compgen -u | grep John). But what if I want to use grep to display John as well as all the remaining entries after it ?
sed or awk solution would be easier, but if you can only use grep, then the option --after-context (or -A) might do:
grep -A 5 John file
The drawback is that you need to know the number of lines to display after the matching (or use an arbitrary big number for the rest of the file).
compgen -u | grep -A$(compgen -u| wc -l) John
Explanation:
From man grep
-A NUM, --after-context=NUM
Print NUM lines of trailing context after matching lines. Places a line containing a group separator (described under --group-separator) between
contiguous groups of matches.
grep -A -- print number of rows after pattern
$() -- Execute unix command
compgen -u| wc -l --> Get total number of rows of output of command.
You can use the following one-liner :
n=$( compgen -u | grep -n John | head -1 | cut -d ":" -f 1 ) && compgen -u | tail -n +$n
This finds out the line number for first occurrence of John, and prints everything starting that line.

Cutting a length of specific string with grep

Let's say we have a string "test123" in a text file.
How do we cut out "test12" only or let's say there is other garbage behind "test123" such as test123x19853 and we want to cut out "test123x"?
I tried with grep -a "test123.\{1,4\}" testasd.txt and so on, but just can't get it right.
I also looked for example, but never found what I'm looking for.
expr:
kent$ x="test123x19853"
kent$ echo $(expr "$x" : '\(test.\{1,4\}\)')
test123x
What you need is -o which print out matched things only:
$ echo "test123x19853"|grep -o "test.\{1,4\}"
test123x
$ echo "test123x19853"|grep -oP "test.{1,4}"
test123x
-o, --only-matching show only the part of a line matching PATTERN
If you are ok with awkthen try following(not this will look for continuous occurrences of alphabets and then continuous occurrences of digits, didn't limit it to 4 or 5).
echo "test123x19853" | awk 'match($0,/[a-zA-Z]+[0-9]+/){print substr($0,RSTART,RLENGTH)}'
In case you want to look for only 1 to 4 digits after 1st continuous occurrence of alphabets then try following(my awk is old version so using --re-interval you could remove it in case you have latest version of ittoo).
echo "test123x19853" | awk --re-interval 'match($0,/[a-zA-Z]+[0-9]{1,4}/){print substr($0,RSTART,RLENGTH)}'

Grep: Capture just number

I am trying to use grep to just capture a number in a string but I am having difficulty.
echo "There are <strong>54</strong> cities | grep -o "([0-9]+)"
How am I suppose to just have it return "54"? I have tried the above grep command and it doesn't work.
echo "You have <strong>54</strong>" | grep -o '[0-9]' seems to sort of work but it prints
5
4
instead of 54
Don't parse HTML with regex, use a proper parser :
$ echo "There are <strong>54</strong> cities " |
xmllint --html --xpath '//strong/text()' -
OUTPUT:
54
Check RegEx match open tags except XHTML self-contained tags
You need to use the "E" option for extended regex support (or use egrep). On my Mac OSX:
$ echo "There are <strong>54</strong> cities" | grep -Eo "[0-9]+"
54
You also need to think if there are going to be more than one occurrence of numbers in the line. What should be the behavior then?
EDIT 1: since you have now specified the requirement to be a number between <strong> tags, I would recommend using sed. On my platform, grep does not have the "P" option for perl style regexes. On my other box, the version of grep specifies that this is an experimental feature so I would go with sed in this case.
$ echo "There are <strong>54</strong> 12 cities" | sed -rn 's/^.*<strong>\s*([0-9]+)\s*<\/strong>.*$/\1/p'
54
Here "r" is for extended regex.
EDIT 2: If you have the "PCRE" option in your version of grep, you could also utilize the following with positive lookbehinds and lookaheads.
$ echo "There are <strong>54 </strong> 12 cities" | grep -o -P "(?<=<strong>)\s*([0-9]+)\s*(?=<\/strong>)"
54
RegEx Demo

grep using a list to find matches in a file, and print only the first occurrence for each string in the list

I have a file, for example, "queries.txt" that has hard return separated strings. I want to use this list to find matches in a second file, "biglist.txt".
"biglist.txt" may have multiple matches for each string in "queries.txt". I want to return only the first hit for each query and write this to another file.
grep -m 1 -wf queries.txt biglist.txt > output
only gives me one line in output. I should have output that is the same number of lines as queries.txt.
Any suggestions for this? Many thanks! I searched for past questions but did not find one that was exactly the same sort of case after a few minutes of reading.
If you want to "reset the counter" after each file, you could do
cat queries.txt | xargs -I{} grep -m 1 -w {} biglist.txt > output
This uses xargs to call grep once for each line in the input… should do the trick for you.
Explanation:
cat queries.txt - produce one "search word" per line
xargs -I{} - take the input one line at a time, and insert it at {}
grep -m 1 -w - find only one match of a whole word
{} - this is where xargs inserts the search term (once per call)
biglist.txt - the file to be searched
> output - the file where the result is to be written
An alternate method without xargs (which one should indeed learn):
(this method assumes there are no spaces in the lines in queries.txt)
cat queries.txt | while read target; do grep -m 1 $target biglist.txt; done > outr
I might not fully understand your question, but it sounds like something like this might work.
cat queries.txt | while read word; do grep "$word" biglist.txt | tee -a output.txt; done

basic grep

I have a large file where each line contains a substring such as ABC123. If I execute
grep ABC file.txt
or
grep ABC1 file.txt
I get those lines back as expected, but if I execute
grep ABC12 file.txt
grep fails to find the corresponding lines.
This seems pretty trivial functionality, but I'm not a heavy user of grep so perhaps I'm missing some gotcha.
Use something like
od -x -a < filename
to dump out the file contents in hex. That'll immediately show you if what you have in your file is what you expect. Which I suspect it isn't :-)
Note: od has lots of useful options to help you here. Too many to list, in fact.
Is there a chance your file contains some hidden character, such as 0x00 ?
This doesn't make sense. Are you sure the file contains "ABC123"?
You can verify this by running following command in a shell
echo "ABC123" | grep ABC12
If the lines contain ABC123, then "grep ABC12" should get them. Do you perhaps mean that you want to match several different strings, such as ABC1, ABC2 and ABC3? In that case you can try this:
grep -E 'ABC1|ABC2|ABC3'
I'm not sure what the problem is.. grep works exactly as it should.. For example, the contents of my test file:
$ cat file.txt
ABC
ABC1
ABC12
ABC123
..and grep'ing for ABC, ABC1, ABC12, ABC123:
$ grep ABC file.txt
ABC
ABC1
ABC12
ABC123
$ grep ABC1 file.txt
ABC1
ABC12
ABC123
$ grep ABC12 file.txt
ABC12
ABC123
$ grep ABC123 file.txt
ABC123
grep is basically a filter, any line containing the first argument (ABC, or ABC1 etc) will be displayed. If it doesn't contain the entire string, it will not be displayed

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