I want to change all instances of the variable Status to status in my code.
However there are some lines where Status is on the same line as strings like Current_Status and Check_Status_After_Write, etc.
I want to replace the variable name only, and not change definitions of other variables or matching comment strings if possible.
I tried to use:
grep -nwrs Status ./Status.txt | xargs sed -i 's/i\<Status\>/status/g'
This returns:
sed: no input files
I tried simplifying it to use:
grep -rl Status ./Status.txt | xargs sed -i 's/i\<Status\>/status/g'
But this fails to work as I wanted.
Am I over complicating this? Can anyone offer a solution?
As an aside, I've had a few failed attempts to do this and am a bit paranoid about the string replacement, is there a way to ask for verification before the replace happens without writing a script?
can you not use the pattern
^Status$
to match only the word status, with nothing before or after it?
Related
I am sorry for posting this but this is driving me crazy. I am very new to bash scripting and am really struggling. I have files with the following format 8_S58_L001.sorted.bam and I would like to take the first digit (8 in this case) from many files and generate a csv file. This will give me the order in which samples were processed by a downstream function.
The script is as follows and it works, however I get an error (-bash: syntax error near unexpected token `done') everytime I run it and am struggling to understand why. So far I have spent 2 days trying to get to the bottom of it and have searched extensively through various forums.
do
test=$(ls -LR | grep .bam$| sed 's/_.*//'| awk '{print}' ORS=',' | sed 's/*$//')
echo $test>../SampleOrder/fileOrder2.csv
done
If I just run
test=$(ls -LR | grep .bam$| sed 's/_.*//'| awk '{print}' ORS=',' | sed 's/*$//')
echo $test>../SampleOrder/fileOrder2.csv
Then I get the desired output and no errors but if it is incorporated within an do statement I get the above error. I am hoping to incorporate this into a larger script so I want to deal with this error first.
I should say that this is being run on a linux based cluster.
Can someone with more experience tell me where I am going wrong.
Thanks in advance
Sam
bash doesn't have a do statement, and done is a reserved word when it is the first word in a command.
So in
do
something
something
done
do is a syntax error. do is only useful in the context of for and while loops, where it serves to separate the condition from the body of the loop.
Since you're reporting a syntax error on the done as opposed to the do, my guess is that you've let Windows line-endings creep into your file. Bash doesn't regard the \r (CR) character as special, so if your file actually contains do\r, then that will be considered to be the name of an external command.
You should be aware that grep .bam$ doesn't do what you are expecting it to do. The dot is a grep wildcard which matches any single character, so the pattern .bam$ will match any string of 4 or more characters that ends in "bam". If you are trying to match all strings that end in ".bam", you should escape the dot and write grep "\.bam$"
But as a previous commenter correctly noted, you should be using shell wildcards (ls *.bam) instead of grep (ls | grep .bam$)
Currently I have a file that has a unique line with the pattern
alphanumeric_ChangeMe_moreAlphaNumeric
Actual looks like this:
127.0.0.1 local.com localhost HostType_test_HostNumber2
I'm trying to replace the string test with a variable determined by another command, run as another user using the following code.
site=$(su admin -c get_local_site | less | sed 's/Local Site Name: //')
sed -i -e "s/RecoverPoint_[[:alnum:]]*_RPA/RecoverPoint_$site_RPA/" fakehostfile
I've tested the individual codes and they echo the correct values, but when I try to use the $site variable in the second it fails to replace the section.
I can't seem to find the correct syntax to replace just what's between the underscores with a string (containing only alphanumerics) that's stored in a variable
I've already been looking on here, and found some similar problems, however the solutions don't seem to work, since part of the replacement string is a variable. I've tried to concatenate the string as three separate variables, but it replaces things strange (Maybe due to the underscores?)
What am I missing here??
Questions with similar problems that didn't work:
sed variable replacement does not seem to work
Sed replacement not working when using variables
As others recommended, no less command required, and if you need show the line of Local Site Name only, use -n option in Sed.
Second, put varies in braces, it should fix your problem.
site=$(su admin -c get_local_site | sed -n 's/Local Site Name: //p')
sed -i "s/RecoverPoint_[[:alnum:]]*_RPA/RecoverPoint_${site}_RPA/" fakehostfile
I've read lots of posts to understand how to correctly escape white spaces and special characters inside strings using sed, but still i can't make it, here's what i'm trying to achieve.
I have a file containing the some strings like this one:
JAVA_OPTS="$JAVA_OPTS -Dorg.apache.catalina.jsessionid=some_value"
and i'm trying to replace 'some_value' using the following:
sed -i "s/^\(JAVA_OPTS=\"\$JAVA_OPTS[ \t]*-Dorg\.apache\.catalina\.jsessionid*=\s*\).*\$/\1$DORG_APACHE_CATALINA_JSESSIONID/" $JBOSS_CONFIGURATION/jboss.configuration
$JBOSS_CONFIGURATION is a variable containing an absolute Linux path.
jboss.configuration is a file i'm pointing as the target for replace
operations.
$DORG_APACHE_CATALINA_JSESSIONID contains the value i want instead
of 'some_value'.
Please note that the pattern:
JAVA_OPTS="$JAVA_OPTS -D
Is always present, and org.apache.catalina.jsessionid is an example of a variable value i'm trying to replace with this script.
What's missing/wrong ? i tried also escaping whitespaces using \s without success,
and echoing the whole gives me the following:
echo "s/^\(JAVA_OPTS=\"\$JAVA_OPTS[ \t]*-Dorg\.apache\.catalina\.jsessionid*=\s*\).*\$/\1$DORG_APACHE_CATALINA_JSESSIONID/"
s/^\(JAVA_OPTS="$JAVA_OPTS[ \t]*-Dorg\.apache\.catalina\.jsessionid*=\s*\).*$/\1/
is echo interpreting the search pattern as sed does ?
any info/help/alternative ways of doing it are highly welcome,
thank you all
echo 'JAVA_OPTS="$JAVA_OPTS -Dorg.apache.catalina.jsessionid=some_value"' | (export DORG_APACHE_CATALINA_JSESSIONID=FOO/BAR/FOOBAR; sed "s/^\(JAVA_OPTS=\"\$JAVA_OPTS[ \t]*-Dorg\.apache\.catalina\.jsessionid*=\s*\).*\$/\1${DORG_APACHE_CATALINA_JSESSIONID////\/}\"/")
Note the bash expansion (in order to escape any / that may trip up sed) and the extra \" after $DORG_APACHE_CATALINA_JSESSIONID in order to properly close the double quote. Other than that your sed expression works for me and the above command outputs the follwoing result:
JAVA_OPTS="$JAVA_OPTS -Dorg.apache.catalina.jsessionid=FOO/BAR/FOOBAR"
You can use sed like this:
sed -r '/\$JAVA_OPTS -D/{s/^(.+=).*$/\1'"$DORG_APACHE_CATALINA_JSESSIONID"'/;}' $JBOSS_CONFIGURATION/jboss.configuration
You can specify a pattern that'll match the desired string rather than trying to specify it exactly.
The following should work for you:
sed -i 's#^\(JAVA_OPTS.*Dorg.apache.catalina.jsessionid\)=\([^"]*\)"#\1='"$DORG_APACHE_CATALINA_JSESSIONID"'"#' $JBOSS_CONFIGURATION/jboss.configuration
sed 's/=\w.*$/='"$DORG_APACHE_CATALINA_JSESSIONID"'/' $JBOSS_CONFIGURATION/jboss.configuration
I am trying to remove 2 lines from all my Javascript files on my Linux shared hosting. I wanted to do this without writing a script as I know this should be possible with sed. My current attempt looks like this:
find . -name "*.js" | xargs sed -i ";var
O0l='=sTKpUG"
The second line is actually longer than this but is malicious code so I have not included it here. As you guessed my server has been hacked so I need to clean up all these JavaScript files.
I forgot to mention that the output I am getting at the moment is:
sed: -e expression #1, char 4: expected newer version of sed
The 2 lines are just as follows consecutively:
;var
O0l='=sTKpUG
except that the second line is longer, but the rest of the second line should not influence the command.
He meant removing two adjacent lines.
you can do something like this, remember to backup your files.
find . -name "*.js" | xargs sed -i -e "/^;var/N;/^;var\nO0l='=sTKpUG/d"
Since sed processes input file line by line, it does not store the newline '\n' character in its buffer, so we need to tell it by using flag /N to append the next line, with newline character.
/^;var/N;
Then we do our pattern searching and deleting.
/^;var\nO0l='=sTKpUG/d
It really isn't clear yet what the two lines look like, and it isn't clear if they are adjacent to each other in the JavaScript, so we'll assume not. However, the answer is likely to be:
find . -name "*.js" |
xargs sed -i -e '/^distinctive-pattern1$/d' -e '/^alternative-pattern-2a$/d'
There are other ways of writing the sed script using a single command string; I prefer to use separate arguments for separate operations (it makes the script clearer).
Clearly, if you need to keep some of the information on one of the lines, you can use a search pattern adjusted as appropriate, and then do a substitute s/short-pattern// instead of d to remove the short section that must be removed. Similarly with the long line if that's relevant.
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.