For example I have:
Hello1 :
Hello2 :
Hello3 :
How Could I delete all of these with a shell script. The number reaches up all the way to 1000.
sed -i '/^Hello[[:digit:]]\+\>/d' file.txt
Or, if you want to output to a different file:
sed '/^Hello[[:digit:]]\+\>/d' file.txt > newfile.txt
If you wish to delete all the lines that contain only Hello(number) : use below :
Sample Input in file
Hell
Hello1 :
No hello stuff here
Unjulating stuff
Hello2 :
Some sentence
Hello99 :
Script
sed -Ei '/^Hello[[:digit:]]+ :$/d' file
Sample Output in the modified file
Hell
No hello stuff here
Unjulating stuff
Some sentence
What happens above
Using the ^ in the pattern we check for the beginning of the line.
We check the pattern Hello(number) : using Hello[[:digit:]]+ :$. Note that I used -E to enable sed extended regular expressions so I need not escape the + ie (\+). Here [[:digit:]] is a class which contains
all the decimal digits and + is used to check if the pattern before it matches at least one time.
Check the end of the line using $
For a matched pattern, delete it line using the d option
I have also used the sed inplace edit option -i so that the changes
are directly saved to the file.
If you wish to change the a line the begins with Hello(number) : then use the below script
sed -Ei '/^Hello[[:digit:]]+ :/d' file
You might have notices that I just removed the $, so our pattern matches any line that starts with Hello(number) :
Hope this helps.
Related
I am looking for a script to search a line if it contains two strings and then find if the first charater of that line contains certain character and remove it.
Eg. if line contains two strings as "abc" and "xyz" it should look for first character of the line and if it contains # , it should remove it and vice-versa.
It tried to run below command in crontab and got the result
crontab -l | grep az2-er32-cxv-iz| grep aze
Output
#5,10 * * * * /opt/apps/scripts/dsm-rync -q -del -s az2-er32-cxv-iz /opt/apps/sdl/scripts/aze-dsm-rync.app.config
since , its difficult to update the crontab entry directly , i copied it to tmpfile.
crontab -l > tmpfile and tried to run sed 's/^#//' tmpfile but it is removing all # instead of the line matching with two strings
You may use gnu awk to do this easily:
awk -i inplace '/az2-er32-cxv-iz/ && /aze/{sub(/^#/, "")} 1' crontab
This will remove # from first position if line has az2-er32-cxv-iz and aze in it.
As mentioned by Shloim, I won't give you the code itself, but I'm giving you a piece of pseudo-code to start:
In order to know if a line contains at least two words, you might search for a space within that line. (grep " " <filename>)
In order to know if a line starts with a certain character, you might search for the character, followed by that one certain character. (grep "<beginning_of_line>#")
Replacing a character in a line can be done, using the sed command.
I am trying to replace a specific line in a txt file with my shell script, for example;
cat aa.txt:
auditd=0
bladeServerSlot=0
When I run my script I would like to change "bladeServerSlot" to 12 as following;
cat aa.txt:
auditd=0
bladeServerSlot=12
Could you please help me?
Using sed and backreferencing:
sed -r '/bladeServerSlot/ s/(^.*)(=.*)/\1=12/g' inputfile
Using awk , this will search for the line which contains bladeServerSlot and replace the second column of that line.
awk 'BEGIN{FS=OFS="="}/bladeServerSlot/{$2=12}1' inputfile
perl -pe 's/bladeServerSlot=\K\d+/12/' aa.txt > output.txt
The \K is a particular form of the positive lookbehind, which discards all previous matches. So we need to replace only what follows. The s/ is applied by default to $_, which contains the current line. The -p prints $_ for every line, so all other lines are copied. We redirect output to a file.
Is it really necessary to replace the line in your example? As bladeServerSlot is a variable you could reset the value.
bladeServerSlot=`any command`
Or you could just let this variable be filled by a Parameter provided to this script.
bladeServerSlot=$1
With $1being the first parameter of your script. I think this would be the cleaner way do solve your issue than to do fancy regex here. The sed/perl solutions will work, but they are not very clear to other people reading your script.
I have a file "test.txt" with the lines below and also lot bunch of extra stuff after the "version"
soainfra_metrics{metric_group="sca_composite",partition="test",is_active="true",state="on",is_default="true",composite="test123"} map:stats version:1.0
soainfra_metrics{metric_group="sca_composite",partition="gello",is_active="true",state="on",is_default="true",composite="test234"} map:stats version:1.8
soainfra_metrics{metric_group="sca_composite",partition="bolo",is_active="true",state="on",is_default="true",composite="3415"} map:stats version:3.1
soainfra_metrics{metric_group="sca_composite",partition="solo",is_active="true",state="on",is_default="true",composite="hji"} map:stats version:1.1
I tried:
egrep -r 'partition|is_active|state|is_default|composite' test.txt
It's displaying every line, but I need only specific mentioned fields like this below,ignoring rest of the data/stuff or lines
in a nut shell, i want to display only these fields from a line not the rest
partition="test",is_active="true",state="on",is_default="true",composite="test123"
partition="gello",is_active="true",state="on",is_default="true",composite="test234"
partition="bolo",is_active="true",state="on",is_default="true",composite="3415"
partition="solo",is_active="true",state="on",is_default="true",composite="hji"
If your version of grep supports Perl-style regular expressions, then I'd use this:
grep -oP '.*?,\K[^}]+' file
It removes everything up to the first comma (\K kills any previous output) and prints everything up to the }.
Alternatively, using awk:
awk -F'}' '{ sub(/[^,]+,/, ""); print $1 }' file
This sets the field separator to } so the part you're interested in is the first field. It then uses sub to remove the part up to the first comma.
For completeness, you could also use sed:
sed 's/[^,]*,\([^}]*\).*/\1/' file
This captures the part after the first , up to the } and replaces the content of the line with it.
After the grep to pick out the lines you want, use sed to edit the lines:
sed 's/.*\(partition[^}]*\)} map.*/\1/'
This means: "whenever you see anything .*, followed by partition and
any number of non-}, then } map and anything else, grab the part
from partition up to but not including the brace \(...\) as group 1.
The replacement text is just group 1 \1.
Use a pipe | to connect the output of egrep to the input of sed:
egrep ... | sed ...
As far as i understood your file might have more lines you don't want to see, so i would use:
sed -n 's/.*\(partition.*\)}.*/\1/p' file
we use -n p to show only lines where we made substitution. The substitution part just gets the part of the line you need substituting the whole line with the pattern.
This might work for you (GNU sed):
sed -r 's/(partition|is_active|state|is_default|composite)="[^"]*"/\n&\n/g;s/[^\n]*\n([^\n]*)\n[^\n]*/\1,/g;s/,$//' file
Treat the problem as if it were a "decomposed club sandwich". Identify the fillings, remove the bread and tidy up.
I wish to replace
x.y.z.zz=/a/b/c/d/
with
x.y.z.zz=/a/b/e/d/
i know x.y.z.zz in advance.I also know the line number in advance.
I have tried this
sed "11s/.*/x.y.z.zz=\/a\/b\/e\/d\/" filename
but this is giving error. Is there a better way to directly search and replace the string ?
sed replaces by using the sed 's/pattern/replacement/' syntax. In your case, you were missing the last /. So by saying this it will work:
sed '11s/.*/x.y.z.zz=\/a\/b\/e\/d\//' file
^
However, it may be cleaner to use another delimiter, so that the syntax is more clear. What about #? (It can also be ~, _, etc.):
sed '11s#.*#x.y.z.zz=/a/b/e/d/#' file
Test
$ cat a
a
x.y.z.zz=/a/b/c/d/
b
c
Let's replace line 2:
$ sed '2s#.*#x.y.z.zz=/a/b/e/d/#' a
a
x.y.z.zz=/a/b/e/d/
b
c
You can just replace c with e if you know your input will always have "x.y.z.zz=/a/b/c/d". e.g. just executing sed s/c/e/
will just replace c with e in the line. Also, you don't need to change the complete line always. You can just change a character or a word in the text.Additionally, if a line contains more than one occurrence of character/word, this command will only change the first one e.g. if input string is x.y.z.zz=/a/b/c/d/c, executing sed s/c/e/ will have output x.y.z.zz=/a/b/e/d/c
If all the occurrences need to be changed g (global) needs to be added in sed command e.g. sed s/c/e/g will give output x.y.z.zz=/a/b/e/d/eIf sed needs to be executed only for a particular line, line number shall be mentioned in the sed command itself, as done in the question. This is the link (http://www.grymoire.com/Unix/Sed.html), I always refer when in question with sed
In looking for a command to delete a line (or lines) from a text file that contain a certain string.
For example
I have a text file as follows
Sat 21-12-2014,10.21,78%
Sat 21-12-2014,11.21,60%
Sun 22-12-2014,09.09,21%
I want to delete all lines that have "21-12-2014" in them.
I'm not able to find a solution that works.
According to #twalberg there is more three alternate solution for this question, which I'm explaining is as follows for future reader of this question for more versatile solutions:
With grep command
grep -v 21-12-2014 filename.txt
explanations:
-v is used to find non-matching lines
With awk command
awk '! /21-12-2014/' filename.txt
explanations:
! is denoting it will print all other lines that contain match of the string. It is not operator signify ignorance.
With sed command
sed -e '/21-12-2014/d' < filename.txt
explanations:
-e is signify scripted regex to be executed
d is denoting delete any match
< is redirecting the input file content to command
Try doing this :
sed -i.bak '/21-12-2014/d' *
A bit of explanations :
sed : the main command line, put the mouse pointer on sed
-i.bak : replace the file in place and make a backup in a .bak file
// is the regex
d means: delete