Hello How can I grep only match before : mark?
If I run grep test1 file, it shows all three lines.
test1:x:29688:test1,test2
test2:x:22611:test1
test3:x:25163:test1,test3
But I would like to get an output test1:x:29688:test1,test2
I would appreciate any advice.
If the desired lines always start with test1 then you can do:
grep '^test1' file
If it's always followed by : but not the other (potential) matches then you can include it as part of the pattern:
grep 'test1:' file
As your data is in row, columns delimited by a character, you may consider awk:
awk -F: '$1 == "test1"' file
I think that you just need to add “:” after “test1”, see an example:
grep “test1:” file
Related
I have some files, and I want grep to return the lines, where I have at least one string Position:"Engineer" AND at least one string which does have Position not equal to "Engineer"
So in the below file should return only first line:
Position:"Engineer" Name:"Jes" Position:"Accountant" Name:"Criss"
Position:"Engineer" Name:"Eva" Position:"Engineer" Name:"Adam"
I could write something like
grep 'Position:"Engineer"' filename | grep 'Position:"Accountant"'
And this works fine (I get only first line), but the thing is I don't know what are all of the possible values in Position, so the grep needs to be generic something like
grep 'Position:"Engineer"' filename | grep -v 'Position:"Engineer"'
But this doesn't return anything (as both grep contradict each other)
Do you have any idea how this can be done?
This line works :
grep "^Position:\"Engineer\"" filename | grep -v " Position:\"Engineer\""
The first expresion with "$" catch only the Position at the begining of line, the second expression with " " space remove the second "Postion" expression.
You can avoid the pipe and additional subshell by using awk if that is allowed, e.g.
awk '
$1~/Engineer/ {if ($3~/Engineer/) next; print}
$3~/Engineer/ {if ($1~/Engineer/) next; print}
' file
Above just checks if the first field contains Engineer and if so checks if field 3 also contains Engineer, and if so skips the record, if not prints it. The second rule, just swaps the order of the tests. The result of the tests is that Engineer can only appear in one of the fields (either first or third, but not both)
Example Use/Output
With your sample input in file, you would have:
$ awk '
$1~/Engineer/ {if ($3~/Engineer/) next; print}
$3~/Engineer/ {if ($1~/Engineer/) next; print}
' file
Position:"Engineer" Name:"Jes" Position:"Accountant" Name:"Criss"
Use negative lookahead to exclude a pattern after match.
grep 'Position:"Engineer"' | grep -P 'Position:"(?!Engineer)'
With two greps in a pipe:
grep -F 'Position:"Engineer"' file | grep -Ev '(Position:"[^"]*").*\1'
or, perhaps more robustly
grep -F 'Position:"Engineer"' file | grep -v 'Position:"Engineer".*Position:"Engineer"'
In general case, if you want to print the lines with unique Position fields,
grep -Ev '(Position:"[^"]*").*\1' file
should do the job, assuming all the lines have the format specified. This will work also when there are more than two Position fields in the line.
I have a file and I want to know how many times does a word is inside that file.(NOTE: A row can have the same word)
You can use this command. Hope this wil help you.
grep -o yourWord file | wc -l
Use the grep -c option to count the number of occurences of a search pattern.
grep -c searchString file
awk solution:
awk '{s+=gsub(/word/,"&")}END{print s}' file
test:
kent$ cat f
word word word
word
word word word
kent$ awk '{s+=gsub(/word/,"&")}END{print s}' f
7
you may want to add word boundary if you want to match an exact word.
Yes, i know you want a grep solution, but my favorite perl with the rolex operator can't missing here... ;)
perl -0777 -nlE 'say $n=()=m/\bYourWord\b/g' filename
# ^^^^^^^^
if yoy want match the YourWord surrounded with another letters like abcYourWordXYZ, use
perl -0777 -nlE 'say $n=()=m/YourWord/g' filename
grep (GNU grep) 2.14
Hello,
I have a log file that I want to filter on a selected word. However, it tends to filter on many for example.
tail -f gateway-* | grep "P_SIP:N_iptB1T1"
This will also find words like this:
"P_SIP:N_iptB1T10"
"P_SIP:N_iptB1T11"
"P_SIP:N_iptB1T12"
etc
However, I don't want to display anything after the 1. grep is picking up 11, 12, 13, etc.
Many thanks for any suggestions,
You can restrict the word to end at 1:
tail -f gateway-* | grep "P_SIP:N_iptB1T1\>"
This will work assuming that you have a matching case which is only "P_SIP:N_iptB1T1".
But if you want to extract from P_SIP:N_iptB1T1x, and display only once, then you need to restrict to show only first match.
grep -o "P_SIP:N_iptB1T1"
-o, --only-matching show only the part of a line matching PATTERN
More info
At least two approaches can be tried:
grep -w pattern matches for full words. Seems to work for this case too, even though the pattern has punctuation.
grep pattern -m 1 to restrict the output to first match. (Also doable with grep xxx | head -1)
If the lines contains the quotes as in your example, just use the -E option in grep and match the closing quote with \". For example:
grep -E "P_SIP:N_iptB1T1\"" file
If these quotes aren't in the text file, and there's blank spaces or endlines after the word, you can match these too:
# The word is followed by one or more blanks
grep -E "P_SIP:N_iptB1T1\s+" file
# Match lines ending with the interesting word
grep -E "P_SIP:N_iptB1T1$" file
I want to grep for the string THREAD: 2. It has a space in between. Not able to figure out how.
I tried with grep "THREAD:[ \2]", but its not working
Please let me know.
Try grep "THREAD: 2" <filename>? You just want a literal '2', right?
If you are using GNU grep you could try using the alias egrep or grep -e with 'THREAD: 2$'
You might have to use '^.*THREAD: 2$'
grep reports back the entire line that has matched your pattern. If you wish to look at lines that contains THREAD: 2 then the following should work -
grep "THREAD: 2" filename
However, if you wish to fetch lines that could contain THREAD: and any number then you can use a character class. So in that case the answer would be -
grep "THREAD: [0-9]" filename
You can add + after the character class which means one or more numbers so that you can match numbers like 1,2,3 or 11,12,13 etc.
If you only want to fetch THREAD: 2 from your line then you will have to use an option of grep which is -o. It means show me only my pattern from the file not the entire line.
grep -o "THREAD: 2" filename
You can look up man page for grep and play around with all the options.
I have lines in a file which look like the following
....... DisplayName="john" ..........
where .... represents variable number of other fields.
Using the following grep command, I am able to extract all the lines which have a valid 'DisplayName' field:
grep DisplayName="[0-9A-Za-z[:space:]]*" e:\test
However, I wish to extract just the name (ie "john") from each line instead of the whole line returned by grep. I tried piping the output into the cut command but it does not accept string delimiters.
This works for me:
awk -F "=" '/DisplayName/ {print $2}'
which returns "john". To remove the quotes for john use:
awk -F "=" '/DisplayName/ {gsub("\"","");print $2}'
Specifically:
sed 's/.*DisplayName="\(.*\)".*/\1/'
Should do, sed semantics is s/subsitutethis/forthis/ where "/" is delimiter. The escaped parentheses in combination with escaped 1 are used to keep the part of the pattern designated by parentheses. This expression keeps everything inside the parentheses after displayname and throws away the rest.
This can also work without first using grep, if you use:
sed -n 's/.*DisplayName="\(.*\)".*/\1/p'
The -n option and p flag tells sed to print just the changed lines.
More in: http://www.grymoire.com/Unix/Sed.html