Commands from history without command number - grep

I am trying to create a function to add to my bash_profile file in order to be able to extract a list of commands from the history grepping a given word and removing the numbers of the commands. The thing is that I can do it directly like this
$history | grep tar | cut -c 8-
but when I put it into a function in bash_profile with this
function hrep() {
history | grep $1 | cut -c 8-
}
It does not take into account the cut command. It prints out the command grepped but it doesn't trim the command numbers.
Do you know what am I missing?
Thanks a lot in advance,
Juanpe.

I'd rather use a regex than telling cut to remove a specific numbers of characters. Your history could go far more than a thousand lines.
hrep() {
history | grep $1 | sed 's/^[ ]*[0-9]*[ ]*//g'
}

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.

Only one type of list may be specified?

So I have a bash script which is below, and whenever I try to execute it, it says only one type of list may be specified, any clue as to whats wrong with this code? I interpret this as we use grep to find the word (-word) of the first character ($1) typed by the user in the file femalenames.txt and then using the cut command we print the 2nd field of characters 16-20? sort of confused there.
#!/bin/bash
grep -w $1 femalenames.txt | cut -f2 -c16-20
Just pipe it ( | ) again & it'll work.
Command guide here
#!/bin/bash
grep -w $1 femalenames.txt | cut -f2 | cut -c16-20

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 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

Is there a way in grep to find out how many lines matched the grep result?

Suppose I write a grep query to find out the occurrence of a method call on an object like this:
// might not be accurate, but irrelevant
grep -nr "[[:alnum:]]\.[[:alnum:]](.*)" .
This would give many results. How to find out how many such results are obtained?
What about using | wc -l to count the number of result lines?
What about
man grep | grep "count"
It outputs
-c, --count
Suppress normal output; instead print a count of matching lines for each input file. [...]
Previous answers are OK, I just want to put it into command line instructions in order to have copy-paste versions (from explicit to simplest) for the future:
grep --count "PATTERN" FILE
Is exactly the same as:
grep -c "PATTERN" FILE
And it is equivalent to:
grep "PATTERN" FILE | wc -l
As a bonus, below i give you a version where a file with a list of patterns is used.
grep -count --file=PATTERNFILE FILE
or simply
grep -cf PATTERNFILE FILE

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