I have a set of strings and want to write out all lines of a file that start with either of these.
Tried this that I found on the internet but that writes out the whole file... May I ask for a help? Thanks!
grep -Ev "^(58|11518|11909|11910|11911|11912|11913|11914|11915|11916|11917|11918|11919|11920|11921|11922|11923|11924|11925|11926|12055|12056|12060|12102|12103|12104|12105|12106|12107|12108|12109|12110|12111|12112|12113|12114|12115|12116|12117|12118|12119|12120|12121|12122|12123|12124|13813)" dead_end1_model.inp > newfile.txt
Your current invocation of grep only prints the inverse -v of the matches. I.e. any line not starting with any of those strings gets matched.
By your description you want to match the lines that do start with any of the strings, so just remove the -v from the invocation.
Related
I am a windows user having basic idea about LINUX and i encountered this command:
cat countryInfo.txt | grep -v "^#" >countryInfo-n.txt
After some research i found that cat is for concatenation and grep is for regular exp search (don't know if i am right) but what will the above command result in (since both are combined together) ?
Thanks in Advance.
EDIT: I am asking this as i dont have linux installed. Else, i could test it.
Short answer: it removes all lines starting with a # and stores the result in countryInfo-n.txt.
Long explanation:
cat countryInfo.txt reads the file countryInfo.txt and streams its content to standard output.
| connects the output of the left command with the input of the right command (so the right command can read what the left command prints).
grep -v "^#" returns all lines that do not (-v) match the regex ^# (which means: line starts with #).
Finally, >countryInfo-n.txt stores the output of grep into the specified file.
It will remove all lines starting with # and put the output in countryInfo-n.txt
This command would result in removing lines starting with # from the file countryInfo.txt and place the output in the file countryInfo-n.txt.
This command could also have been written as
grep -v "^#" countryInfo.txt > countryInfo-n.txt
See Useless Use of Cat.
I was just wondering what command i need to put into the terminal to read a text file, eliminate all lines that do not contain a certain keyword, and then print those lines onto a new file. for example, the keyword is "system". I want to be able to print all lines that contain system onto a new separate file. Thanks
grep is your friend.
For example, you can do:
grep system <filename> > systemlines.out
man grep and you can get additional useful info as well (ex: line numbers, 1+ lines prior, 1+lines after, negation - ie: all lines that do not contain grep, etc...)
If you are running Windows, you can either install cygwin or you can find a win32 binary for grep as well.
grep '\<system\>'
Will search for lines that contain the word system, and not system as a substring.
below grep command will solve ur problem
grep -i yourword filename1 > filename2
with -i for case insensitiveness
without -i for case sensitiveness
to learn how grep works on ur server ,refer to man page on ur server by the following command
man grep
grep "system" filename > new-filename
You might want to make it a bit cleverer to not include lines with words like "dysystemic", but it's a good place to start.
I'm trying to parse through an application.log that has many lines that follow the same syntax below.
"Error","jrpp-237","10/13/11","02:55:04",,"File not found: /indexUsa~.cfm The specific sequence of files included or processed is: c:\websites\pj7fe4\indexUsa~.cfm '' "
I need to use some type of command to pull out what is listed between c:\websites\ and the next \
e.g. in this case it would be pj7fe4
I thought that the following command would work..
bin/sed -n '/c:\\websites\\/,/\\/p' upload/test.log
Unfortunately from reading further I now understand that this will return the entire line containing c:\websites through the \ and I need to know the in between, not the whole line.
To be more difficult I need to match all of the directory sub paths, not just one particular line as this is for multiple sites.
You're using range patterns incorrectly. You can't use it to limit the command (print in this case) to a part of the line, only to a range of lines. You also don't escape the backspaces.
Try this: sed 's/.*c:\\websites\\\([0-9a-zA-Z]*\)\\.*/\1/'
There's a good sed tutorial here: Sed - An Introduction and Tutorial by Bruce Barnett
grep way:
grep -Po "(?<=c:\\\websites\\\)[^\\\]+(?=\\\)" yourFile
test:
kent$ echo '"Error","jrpp-237","10/13/11","02:55:04",,"File not found: /indexUsa~.cfm The specific sequence of files included or processed is: c:\websites\pj7fe4\indexUsa~.cfm '' "'|grep -Po "(?<=c:\\\websites\\\)[^\\\]+(?=\\\)"
pj7fe4
For example suppose I have the following piece of data
ABC,3,4
,,ExtraInfo
,,MoreInfo
XYZ,6,7
,,XyzInfo
,,MoreXyz
ABC,1,2
,,ABCInfo
,,MoreABC
It's trivial to get grep to extract the ABC lines. However if I want to also grab the following lines to produce this output
ABC,3,4
,,ExtraInfo
,,MoreInfo
ABC,1,2
,,ABCInfo
,,MoreABC
Can this be done using grep and standard shell scripting?
Edit: Just to clarify there could be a variable number of lines in between. The logic would be to keep printing while the first column of the CSV is empty.
grep -A 2 {Your regex} will output the two lines following the found strings.
Update:
Since you specified that it could be any number of lines, this would not be possible as grep focuses on matching on a single line see the following questions:
How can I search for a multiline pattern in a file?
Regex (grep) for multi-line search needed
Why can't i match the pattern in this case?
Selecting text spanning multiple lines using grep and regular expressions
You can use this, although a bit hackity due to the grep at the end of the pipeline to mute out anything that does not start with 'A' or ',':
$ sed -n '/^ABC/,/^[^,]/p' yourfile.txt| grep -v '^[^A,]'
Edit: A less hackity way is to use awk:
$ awk '/^ABC/ { want = 1 } !/^ABC/ && !/^,/ { want = 0 } { if (want) print }' f.txt
You can understand what it does if you read out loud the pattern and the thing in the braces.
The manpage has explanations for the options, of which you want to look at -A under Context Line Control.
I am trying to extract no.s from a file, so I created a script, but grep is giving error:grep: line too long. Can anyone tell me where am I wrong. command is:
echo $(cat filename|grep '\<[0-9]*\>')
Thanks in advance
grep is line-oriented; it will print matching lines to output. Probably you have a huge line in your file, and the resulting line cannot be converted into a string value by shell, as $(...) requires.
First of all, try just cat filename | grep '\<[0-9]*\>' > results and see what is in the results file. Maybe it's enough.
But if you have multiple numbers in a line and you want to extract them all, use -o: grep -o '\<[0-9]*\>'. This will print only matching parts, every match on a new line, even if original matches are on the same line. If you need line numbers, too, add -n.