How to search a unix log file for a specific error code and display only the variable that matches that search?I have tried using grep and egrep commands.
I am not sure you have tried this
grep TextYouWantToExclude YourLogfile.abc | grep -v ERROR
And did grep or egrep not work for you?
I use vim for this. The command :g/<search> comes in very handy because it filters out all lines that do not match <search>. <search> can be a regular expression just as one would use with grep.
grep -i error_code log_file
This will display all the instances of error code found in log file.
Related
Related question: How do I find all files containing specific text on Linux?
I have been using the command mentioned in the answer of above question to search for string occurences in all files:
grep -rnw '/path/to/somewhere/' -e "pattern"
However lately I encountered a problem, shown in the following picture:
Looks like this command only recognizes strings that stand out as a word or something. How should I modify the command to improve my search result?
explainshell helpfully explains your command, and gives an excerpt from man grep:
-w, --word-regexp
Select only those lines containing matches that form whole words.
So just remove -w since that explicitly does what you don't want:
grep -rn '/path/to/somewhere/' -e "pattern"
I have a file in the location /home/someuser/sometext.txt . I want to count the number of lines in which a particular string occurs. What's the way to do that from Linux command line?
grep with -c switch is what you need:
grep -c "pattern" /home/someuser/sometext.txt
Alternate solution using awk:
awk '/regex/{c++}END{print c+0}' /home/someuser/sometext.txt
You're looking for the grep command. Here's a basic tutorial. It's extremely useful for string searching in files. It also has support for regular expressions.
It looks like you'll do something like this:
grep -c "mystring" /home/someuser/sometext.txt
The -c argument is short for --count and tells grep to print out the number of lines that contain the string.
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 exctract error lines from my log file:
I used this :
more report.txt | grep -E (?i)(error)
I'm getting this error msg :
bash: syntax error near unexpected token `('
What am I doing wrong? I'm trying to extract all lines containing "Error" ignoring the case sensitivity so it can be error, Error, ERROR etc.
The problem with your line is that the parens are picked up by the shell instead of grep, you need to quote them:
grep -E '(?i)(error)' report.txt
For this particular task the other answers are of course correct, you don't even need the parens.
You can do:
grep -i error report.txt
There is really no need to more the file and then pipe it to grep. You can pass the filename as an argument to grep.
To make the search case insensitive, you use the -i option of grep.
And there is really no need to go for -E option as the pattern you are using error is not an extended regex.
The cause of the error you are seeing is that your pattern (?i)(error) is interpreted by the shell and since the shell did not except to see ( in this context it throws an error, something similar to when you do ls (*).
To fix this you quote your pattern. But that will not solve your problem of finding error in the file because the pattern (?i)(error) looks for the string 'ierror' !!
you can use
grep -i error report.txt
this will achieve the same result
cat report.txt | grep -i error
and if you want to paginate the results:
cat report.txt | grep -i error | more