I want to replace exact word by sed command with variable. My file looks like this:
//module xyz
module xyz
Suppose I have the following shell variables defined:
var1='module xyz'
var2='module abc'
I want to change xyz to abc in uncommented line only(module xyz)
So after executing command output should be
//module xyz
module abc
I do not want to change commented line (//module xyz)
currently I am using sed command as,
sed -i "s|$var1|$var2|g" file_name
But this command doesn't work. It also replace commented line. How can I only replace the line that isn't commented?
Assuming that you know the pattern is at the start of the line, you can use this:
sed "s|^$var1|$var2|" file_name
That is, add an anchor ^, so that the match has to be at the start of the line.
I removed the -i switch so you can test it and also the g modifier, which isn't necessary as you only want to do one substitution per line.
It's worth mentioning that using shell variables in sed is actually quite tricky to do in a reliable way, so you should take this into account.
Your shell variable assignment should be quoted if there is space. Like:
var1="foo bar blah"
You can add pattern, "the lines don't start with // " to your sed command, so that do your substitution only for those lines
This line should work for your example:
sed -i "\#^//#\!s/$var1/$var2/g" file
the ! needs to be escaped, because we used double quote
since your pattern (comment) has slash (/), I used other char as regex separator
This command will only do substitution on lines not starting with //. If there are leading spaces, you may want to adjust the pattern ^//
You need to identify a pattern so that lines containing that pattern should not be processed.
Assuming that // will exist only in commented lines you can use
sed -i '/\/\// !s/$var1/$var2/g' file_name
/\/\// will enable sed to identify lines which contain the pattern //, and !s will enable you to skip those lines.
I am trying to delete a line with the pattern matches and replacing the entire line with the another line using sed command.
File contents:Sample.txt
Testfile=xxxx
Testfile3=uuuu
Testfile4=oooo
Testfile5=iiii
Testfile2=ikeii
I am using sed command to delete a line contains Testfile3=* and replace by Testfile3=linechanged
sed -i 's/Testfile3=\*/Testfile3=linechanged/' Sample.txt.
But it just appends the replaceable string in the line as shown below
Testfile3=linechanged=uuuu.
I am expecting the output to be
Testfile3=linechanged.
What i am doing wrong?
The star is not matched right:
sed -i 's/Testfile3=.*/Testfile3=linechanged/' Sample.txt
# ^^
.* matches any character (.) for any length (*), so it will match everything till the end of the line.
You can use captured group to keep what will be preserved and use the desired replacement for the rest:
sed -i 's/^\(Testfile3=\).*/\1linechanged/' file.txt
In your case, escaping the Regex token * like \* will match * literally e.g. Testfile3=* would be matched then.
This might work for you (GNU sed):
sed '/Testfile3/cTestfile3=linechanged' file
This matches the line containing Testfile3 and changes it to the required string.
My question is probably rather simple. I'm trying to replace sequences of strings that are at the beginning of lines in a file. For example, I would like to replace any instance of the pattern "GN" with "N" or "WR" with "R", but only if they are the first 2 characters of that line. For example, if I had a file with the following content:
WRONG
RIGHT
GNOME
I would like to transform this file to give
RONG
RIGHT
NOME
I know i can use the following to replace any instance of the above example;
sed -i 's/GN/N/g' file.txt
sed -i 's/WR/R/g' file.txt
The issue is that I want this to happen only if the above patterns are the first 2 characters in any given line. Possibly an IF statement, although i'm not sure what the condition would look like. Any pointers in the right direction would be much appreciated, thanks.
just add the circumflex, remove g suffix (unnecessary, since you want at most one replacement), you can also combine them in one script.
sed -i 's/^GN/N/;s/^WR/R/' file.txt
Use the start-of-string regexp anchor ^:
sed -i 's/^GN/N/' file.txt
sed -i 's/^WR/R/' file.txt
Since sed is line-oriented, start-of-string == start-of-line.
I want to delete first sign in a file (without creating new file). That is the line (and this line isn't the first one or last one):
#$config['rrdcached'] = "unix:/var/run/rrdcached.sock";
I'm trying to do thuis with sed command but it doesn't work. That is my command:
sed -i "s/#$config\['rrdcached'\].*$/$config\['config'\]/g" text.txt
Any suggestions?
Just replace the first match of # with following command:
sed -i '1 s/#//' test.txt
The $ characters are causing two problems.
First, the shell is treating $config as a variable reference, and replacing it with the value. You need to escape the $ to prevent that.
Second, $ has special meaning in regular expressions, so you need to escape it at that level as well. So you need to escape the backslash and the $.
sed -i "s/^#\\\$config\['rrdcached'\].*\$/\$config['config']/" text.txt
There's no need for the g modifier since you only want to replace the first match on the line. And you should use the ^ anchor so it only matches this at the beginning of the line.
It's also not necessary to escape special regexp characters in the replacement string.
This command works, but i forgot that there is something after '='. At now, everything after that is deleted.
I wrote this:
sed -i "s/^#\\\$config\['rrdcached'\] = "unix:/var/run/rrdcached.sock";.*\$/\\\$config\['config'\]/ = "unix:/var/run/rrdcached.sock";" text.txt
I want to replace the first six , for each line in a text file using sed or something similar in linux.
There are more than six , on each line, but only the first six should be replaced by |.
Sed doesn't really support the notion of "the first n occurrences", only "the n-th occurrence"; GNU sed has one for "replace all matches from the n-th on", which is not what you want in this case. To get the first six commas replaced, you have to call the s command six times:
sed 's/,/|/;s/,/|/;s/,/|/;s/,/|/;s/,/|/;s/,/|/' infile
If, however, you know that there are no | in the file and you have GNU sed, you can do this:
sed 's/,/|/g;s/|/,/7g' infile
This replaces all commas with pipes, then turns the pipes from the 7th on back to commas.
If you do have pipes beforehand, you can turn them into something that you know isn't in the string first:
sed 's/|/~~/g;s/,/|/g;s/|/,/7g;s/~~/|/g' infile
This makes all | into ~~ first, then all , into |, then the | from the 7th on back into ,, and finally the ~~ back into |.
Testing on this input file:
,,,,,,X,,,,,,
,,,|,,,|,,,|,,,|
the first and third command result in
||||||X,,,,,,
||||||||,,,|,,,|
The second one would fail on the second line because there are already pipe characters.
This might work for you (GNU sed):
sed 'y/,/\n/;s/\n/,/7g;y/\n/|/' file
Translate all ,'s to \n's, then replace from the seventh \n to the end of line by ,'s, then replace the remaining \n's by |'s.
Use the following pattern in sed: sed 's/old/new/<number>'
Where <number> is the number of times you want this pattern applied.
You can replace <number> with g to apply the pattern to all occurrences.
You can try this sed,
sed -r ':loop; s/^([^,]*),/\1|/g; /^([^|]*\|){6}/t; b loop' file
(OR)
sed ':loop; s/^\([^,]*\),/\1|/g; /^\([^|]*|\)\{6\}/t; b loop' file
Test:
$ cat file
a,b,c,d,e,f,g,h,i,j,k
$ sed -r ':loop; s/^([^,]*),/\1|/g; /^([^|]*\|){6}/t; b loop' file
a|b|c|d|e|f|g,h,i,j,k
Note: This will work only if you do not have any pipe(|) before that.