I want to substitute a String from a file which is:
# - "server1"
My first attempt was something like this:
sed -i 's/#\ -\ "\server1"\.*/ChangedWord/g' file
But I get an error if I try it like this.
So there is to be another way to handle whitespaces, I guess I have to use \s or [[:space:]]. But for some how I am not able to make it work.
I think you are complicating the expression too much. This should be enough:
sed 's/^#[[:space:]]*-[[:space:]]*"server1".*/ChangedWord/' file
It looks for those lines starting with # followed by 0 to n spaces, then "server1" and then anything. In such case, it replaces the line with ChangedWord.
Note I am using [[:space:]] to match the spaces, since it is a more compatible way (thanks Tom Fenech in comments).
Note also there is no need to use g in the sed expression, because the pattern can occur just once per line.
Test
$ cat a
hello
# - "server1"
hello# - "server1"
$ sed 's/^#[[:space:]]*-[[:space:]]*"server1".*/ChangedWord/' a
hello
ChangedWord
hello# - "server1"
The actual fault was the missing escaping from the double quotes:
ssh -i file root#IP sed 's/^#[[:space:]]*-[[:space:]]*\"server1\".*/ChangedWord/' file
That did it for me. Thanks for all your support
rghome is right, you don't need those backslashes in front of spaces as the expression is wrapped in quotes. In fact, they're causing the error: sed is telling you that \<Space> is not a valid option. Just remove them and it should work as expected:
sed -i 's/# - "server1"/ChangedWord/' file
Related
I want to replace this line #discovery.seed_hosts: ["host1","host2"] with discovery.seed_hosts: ["${extraNode1}","${extraNode2}","${masterIP}"]. Need to remove the # and replace the host1 and host2 as per the given argument also need to add another value (3rd value) into the array as well.
sudo sed -i "/#discovery.seed_hosts: ["host1","host2"]/s/#discovery.seed_hosts: ["host1","host2"]/discovery.seed_hosts: ["${extraNode1}","${extraNode2}","${masterIP}"]/" check.yml
I tried the above command to do this but it is giving error because of the ["host1","host2"] in the command.
sed: -e expression #1, char 49: Invalid range end - Error received
You need to use the s (substitute) command, and escape ., in addition to [. Like this:
sudo sed -i 's/#\(discovery\.seed_hosts: \["\)host1","host2"]/\1${extraNode1}","${extraNode2}","${masterIP}"]/' check.yml
If you don't scape the ., the sed command will match any character where the . is, like #discovery7seed_hosts: ["host1","host2"].
The sed command is pretty straight forward. I just added parentheses around the part of the match that I wanted to reuse in the substitution which creates a group. The \1 is replace with "group 1", the contents of what's in between the parentheses, which must be escaped too.
EDIT: The ", double quotes, don't need to be escaped because the sed command is in single quotes: 's/.../.../'. Also, the ], closing square bracket, doesn't need to be escaped as long as its corresponding [, opening square bracket, has been escaped. Finally, both parentheses ( and ) need to be escaped to create the group. (END OF EDIT)
Test:
$ cat check.yml
This is a test
Another line
#discovery.seed_hosts: ["host1","host2"]
#discovery7seed_hosts: ["host1","host2"]
OK. Good bye?
$ sed 's/#\(discovery\.seed_hosts: \["\)host1","host2"]/\1${extraNode1}","${extraNode2}","${masterIP}"]/' check.yml
This is a test
Another line
discovery.seed_hosts: ["${extraNode1}","${extraNode2}","${masterIP}"]
#discovery7seed_hosts: ["host1","host2"]
OK. Good bye?
$
You'll need to escape [s and "s with backslashes as:
sudo sed -i "/#discovery.seed_hosts: \[\"host1\",\"host2\"]/s/#discovery.seed_hosts: \[\"host1\",\"host2\"]/discovery.seed_hosts: [\""${extraNode1}\"",\""${extraNode2}\"",\""${masterIP}\""]/" check.yml
I know this question has been asked before, I just can't seem to get the correct syntax for my sed command.
I need to replace OPP/com.user.opp.orchest.po.services.stub-npo/npo-stub with OPP/com.user.opp.orchest.po.services.stub-ica/npo-ica
A snippet of the file I am replacing it is the following:
config.xml
<compareType>PLAIN</compareType>
<pattern>
OPP/com.user.opp.orchest.po.services.stub-npo/npo-stub
</pattern>
<branches>
<com.sonyuser.hudson.plugins.gerrit.trigger.hudsontrigger.data.Branch>
<compareType>ANT</compareType>
<pattern>master</pattern>
</com.sonyuser.hudson.plugins.gerrit.trigger.hudsontrigger.data.Branch>
</branches>
${REPO_MIRROR}/OPP/com.user.opp.orchest.po.services.stub-npo/npo-stub
I have tried the following,
sed -i '/^\/OPP/\com.user.opp.orchest.po.services.stub-npo/\npo-stub\/OPP/\com.user.opp.orchest.po.services.stub-ica/\npo-ica/g' config.xml
In your command, you are missing s for substitution and have wrongly escaped \ character. Also as you replied to my comment, that you want to replace it from anywhere in the file, you don't have to use ^ character in your regex. And dot . in regex means any character so they need to be escaped too.
You can use this command,
sed -i 's/OPP\/com\.user\.opp\.orchest\.po\.services\.stub-npo\/npo-stub/OPP\/com.user.opp.orchest.po.services.stub-ica\/npo-ica/g' yourfilename
You need to specify s command and replace the /\ with \/. There are some other typos here as well (\/ at the start is not necessary). Also, escape dots to match literal dots. A good idea is to use some other delimiter here instead of /, e.g. ,, because you have / chars in the regex and replacement parts.
You may use
sed -i 's,^OPP/com\.user\.opp\.orchest\.po\.services\.stub-npo/npo-stub,OPP/com.user.opp.orchest.po.services.stub-ica/npo-ica,' file
See the online demo
I found a solution for extracting the password from a Mac OS X Keychain item. It uses sed to get the password from the security command:
security 2>&1 >/dev/null find-generic-password -ga $USER | \
sed -En '/^password: / s,^password: "(.*)"$,\1,p'
The code is here in a comment by 'sr105'. The part before the | evaluates to password: "secret". I'm trying to figure out exactly how the sed command works. Here are some thoughts:
I understand the flags -En, but what are the commas doing in this example? In the sed docs it says a comma separates an address range, but there's 3 commas.
The first 'address' /^password: / has a trailing s; in the docs s is only mentioned as the replace command like s/pattern/replacement/. Not the case here.
The ^password: "(.*)"$ part looks like the Regex for isolating secret, but it's not delimited.
I can understand the end part where the back-reference \1 is printed out, but again, what are the commas doing there??
Note that I'm not interested in an easier alternative to this sed example. This will only be part of a larger bash script which will include some more sed parsing in an .htaccess file, so I'd really like to learn the syntax even if it is obscure.
Thanks for your help!
Here is sed command:
sed -En '/^password: / s,^password: "(.*)"$,\1,p'
Commas are used as regex delimiter it can very well be another delimiter like #:
sed -En '/^password: / s#^password: "(.*)"$#\1#p'`
/^password: / finds an input line that starts with password:
s#^password: "(.*)"$#\1#p finds and captures double-quoted string after password: and replaces the entire line with the captured string \1 ( so all that remains is the password )
First, the command extracts passwords from a file (or stream) and prints them to stdout.
While you "normally" might execute a sed command on all lines of a file, sed offers to specify a regex pattern which describes which lines the following command should get applied to.
In your case
/^password: /
is a regex, saying that the command:
s,^password: "(.*)"$,\1,p
should get executed for all lines looking like password: "secret". The command substitutes those lines with the password itself while suppressing the outer lines.
The substitute command might look uncommon but you can choose the delimiter in an sed command, it is not limited to /. In this case , was chosen.
Can anyone explain me how the regular expression works in the sed substitute command.
$ cat path.txt
/usr/kbos/bin:/usr/local/bin:/usr/jbin:/usr/bin:/usr/sas/bin
/usr/local/sbin:/sbin:/bin/:/usr/sbin:/usr/bin:/opt/omni/bin:
/opt/omni/lbin:/opt/omni/sbin:/root/bin
$ sed 's/\(\/[^:]*\).**/\1/g' path.txt
/usr/kbos/bin
/usr/local/sbin
/opt/omni/lbin
From the above sed command they used back reference and save operator concept.
Can anyone explain me how the regular expression especially /[^:]* work in the substitute command to get only the first path in each line.
I think you wrote an extra asterisk * in your sed code, so it should be like this:
$ sed 's/\(\/[^:]*\).*/\1/g' file
/usr/kbos/bin
/usr/local/sbin
/opt/omni/lbin
To change the delimiter will help to understand it a little bit better:
sed 's#\(/[^:]*\).*#\1#g'
The s#something#otherthing#g is a basic sed command that looks for something and changes it for otherthing all over the file.
If you do s#(something)#\1#g then you "save" that something and then you can print it back with \1.
Hence, what it is doing is to get a pattern like /[^:]* and then print is back. /[^:]* means / and then every char except :. So it will get / + all the string until it finds a semicolon :. It will store that piece of the string and then print it back.
Small examples:
# get every char
$ echo "hello123bye" | sed 's#\([a-z]*\).*#\1#g'
hello
# get everything until it finds the number 3
$ echo "hello123bye" | sed 's#\([^3]*\).*#\1#g'
hello12
[^:]*
in regex would match all characters except for :, so it would match until this:
/usr/kbos/bin
also it would match these,
/usr/local/bin
/usr/jbin
/usr/bin
/usr/sas/bin
As, these all contains characters, that are not :
.* match any character, zero or more times.
Thus, this regex [^:]*.*, would match all this expressions:
/usr/kbos/bin:/usr/local/bin:/usr/jbin:/usr/bin:/usr/sas/bin
/usr/local/bin:/usr/jbin:/usr/bin:/usr/sas/bin
/usr/jbin:/usr/bin:/usr/sas/bin
/usr/bin:/usr/sas/bin
However, you get only the first field (ie,/usr/kbos/bin, by using back reference in sed), because, regular expression output the longest possible match found.
I have a file that looks like this:
#"Afghanistan.png",
#"Albania.png",
#"Algeria.png",
#"American_Samoa.png",
I want it to look like this
#"Afghanistan.png",
#"Afghanistan",
#"Albania.png",
#"Albania",
#"Algeria.png",
#"Algeria",
#"American_Samoa.png",
#"American_Samoa",
I thought I could use sed to do this but I can't figure out how to store something in a buffer and then modify it.
Am I even using the right tool?
Thanks
You don't have to get tricky with regular expressions and replacement strings: use sed's p command to print the line intact, then modify the line and let it print implicitly
sed 'p; s/\.png//'
Glenn jackman's response is OK, but it also doubles the rows which do not match the expression.
This one, instead, doubles only the rows which matched the expression:
sed -n 'p; s/\.png//p'
Here, -n stands for "print nothing unless explicitely printed", and the p in s/\.png//p forces the print if substitution was done, but does not force it otherwise
That is pretty easy to do with sed and you not even need to use the hold space (the sed auxiliary buffer). Given the input file below:
$ cat input
#"Afghanistan.png",
#"Albania.png",
#"Algeria.png",
#"American_Samoa.png",
you should use this command:
sed 's/#"\([^.]*\)\.png",/&\
#"\1",/' input
The result:
$ sed 's/#"\([^.]*\)\.png",/&\
#"\1",/' input
#"Afghanistan.png",
#"Afghanistan",
#"Albania.png",
#"Albania",
#"Algeria.png",
#"Algeria",
#"American_Samoa.png",
#"American_Samoa",
This commands is just a replacement command (s///). It matches anything starting with #" followed by non-period chars ([^.]*) and then by .png",. Also, it matches all non-period chars before .png", using the group brackets \( and \), so we can get what was matched by this group. So, this is the to-be-replaced regular expression:
#"\([^.]*\)\.png",
So follows the replacement part of the command. The & command just inserts everything that was matched by #"\([^.]*\)\.png", in the changed content. If it was the only element of the replacement part, nothing would be changed in the output. However, following the & there is a newline character - represented by the backslash \ followed by an actual newline - and in the new line we add the #" string followed by the content of the first group (\1) and then the string ",.
This is just a brief explanation of the command. Hope this helps. Also, note that you can use the \n string to represent newlines in some versions of sed (such as GNU sed). It would render a more concise and readable command:
sed 's/#"\([^.]*\)\.png",/&\n#"\1",/' input
I prefer this over Carles Sala and Glenn Jackman's:
sed '/.png/p;s/.png//'
Could just say it's personal preference.
or one can combine both versions and apply the duplication only on lines matching the required pattern
sed -e '/^#".*\.png",/{p;s/\.png//;}' input