Sed replace text with another text - sed

I want to replace the text disable_functions = with: disable_functions = show_source, system, shell_exec, passthru, exec, phpinfo, popen, proc_open, ini_set in my php.ini file with SED.
This is my SH code:
sed -i /disable_functions =/disable_functions = show_source, system, shell_exec, passthru, exec, phpinfo, popen, proc_open, ini_set/g' /usr/local/lib/php.ini
I don't know why it is not working if I execute the .SH script.
Can someone help me with it?

You miss the search 's tag in your line, use this:
sed -i 's/disable_functions =/disable_functions = show_source, system, shell_exec, passthru, exec, phpinfo, popen, proc_open, ini_set/g' /usr/local/lib/php.ini

Related

sed extra characters at end of l command

I am trying replace value in a config file with sed in cshell.
However it gives me the error:
sed: 1: "/usr/local/etc/raddb/mo ...": extra characters at the end of l command
I am trying the following command:
sed -i "s/private_key_password = .*/private_key_password = test/" /usr/local/etc/raddb/mods-available/eap
I have looked at examples of sed to do this but they all look similar with what I am doing, what is going wrong here?
FreeBSD sed requires an argument after -i to rename the original file to. For example sed -i .orig 's/../../' file will rename he original file to file.orig, and save the modified file to file.
This is different from GNU sed, which doesn't require an argument for the -i flag. See sed(1) for the full documentation. This is one of those useful extensions to the POSIX spec which is unfortunately implemented inconsistently.
Right now, the "s/private_key_password = .*/private_key_password = test/" parts gets interpreted as an argument to -i, and /usr/local/etc/raddb/mods-available/eap gets interpreted as the command. Hence the error.
So you want to use:
sed -i .orig "s/private_key_password = .*/private_key_password = test/" /usr/local/etc/raddb/mods-available/eap
You can then check if the changes are okay with diff and remove /usr/local/etc/raddb/mods-available/eap.orig if they are.

Execute sed command inside TCL script

I am trying to execute sed command inside TCL script . Basically i wanted to remove all empty lines from the input file before reading the file using TCL. so i tried following in my script
exec sed -i '/^\s*$/d' .tmp.PG_Ring
set fid [open ".tmp.PG_Ring" r]
But the script is dumping following Error .
sed: -e expression #1, char 1: unknown command: `''
while executing
"exec sed -i '/^\s*$/d' .tmp.PG_Ring"
(file "pg_ring.tcl" line 1)
could you please provide me work around for this & help me with best way to do this?
That won't work, as single quotes have no special meaning to Tcl at all. Tcl uses braces to mean the same sort of thing (except they nest nicely), so instead you can use this:.
exec sed -i {/^\s*$/d} .tmp.PG_Ring

remove ^M characters from file using sed

I have this line inside a file:
ULNET-PA,client_sgcib,broker_keplersecurities
,KEPLER
I try to get rid of that ^M (carriage return) character so I used:
sed 's/^M//g'
However this does remove everything after ^M:
[root#localhost tmp]# vi test
ULNET-PA,client_sgcib,broker_keplersecurities^M,KEPLER
[root#localhost tmp]# sed 's/^M//g' test
ULNET-PA,client_sgcib,broker_keplersecurities
What I want to obtain is:
[root#localhost tmp]# vi test
ULNET-PA,client_sgcib,broker_keplersecurities,KEPLER
Use tr:
tr -d '^M' < inputfile
(Note that the ^M character can be input using Ctrl+VCtrl+M)
EDIT: As suggested by Glenn Jackman, if you're using bash, you could also say:
tr -d $'\r' < inputfile
still the same line:
sed -i 's/^M//g' file
when you type the command, for ^M you type Ctrl+VCtrl+M
actually if you have already opened the file in vim, you can just in vim do:
:%s/^M//g
same, ^M you type Ctrl-V Ctrl-M
You can simply use dos2unix which is available in most Unix/Linux systems. However I found the following sed command to be better as it removed ^M where dos2unix couldn't:
sed 's/\r//g' < input.txt > output.txt
Hope that helps.
Note: ^M is actually carriage return character which is represented in code as \r
What dos2unix does is most likely equivalent to:
sed 's/\r\n/\n/g' < input.txt > output.txt
It doesn't remove \r when it is not immediately followed by \n and replaces both with just \n. This fails with certain types of files like one I just tested with.
alias dos2unix="sed -i -e 's/'\"\$(printf '\015')\"'//g' "
Usage:
dos2unix file
If Perl is an option:
perl -i -pe 's/\r\n$/\n/g' file
-i makes a .bak version of the input file
\r = carriage return
\n = linefeed
$ = end of line
s/foo/bar/g = globally substitute "foo" with "bar"
In awk:
sub(/\r/,"")
If it is in the end of record, sub(/\r/,"",$NF) should suffice. No need to scan the whole record.
This is the better way to achieve
tr -d '\015' < inputfile_name > outputfile_name
Later rename the file to original file name.
I agree with #twalberg (see accepted answer comments, above), dos2unix on Mac OSX covers this, quoting man dos2unix:
To run in Mac mode use the command-line option "-c mac" or use the
commands "mac2unix" or "unix2mac"
I settled on 'mac2unix', which got rid of my less-cmd-visible '^M' entries, introduced by an Apple 'Messages' transfer of a bash script between 2 Yosemite (OSX 10.10) Macs!
I installed 'dos2unix', trivially, on Mac OSX using the popular Homebrew package installer, I highly recommend it and it's companion command, Cask.
This is clean and simple and it works:
sed -i 's/\r//g' file
where \r of course is the equivalent for ^M.
Simply run the following command:
sed -i -e 's/\r$//' input.file
I verified this as valid in Mac OSX Monterey.
remove any \r :
nawk 'NF+=OFS=_' FS='\r'
gawk 3 ORS= RS='\r'
remove end of line \r :
mawk2 8 RS='\r?\n'
mawk -F'\r$' NF=1

Add text at the end of each line

I'm on Linux command line and I have file with
127.0.0.1
128.0.0.0
121.121.33.111
I want
127.0.0.1:80
128.0.0.0:80
121.121.33.111:80
I remember my colleagues were using sed for that, but after reading sed manual still not clear how to do it on command line?
You could try using something like:
sed -n 's/$/:80/' ips.txt > new-ips.txt
Provided that your file format is just as you have described in your question.
The s/// substitution command matches (finds) the end of each line in your file (using the $ character) and then appends (replaces) the :80 to the end of each line. The ips.txt file is your input file... and new-ips.txt is your newly-created file (the final result of your changes.)
Also, if you have a list of IP numbers that happen to have port numbers attached already, (as noted by Vlad and as given by aragaer,) you could try using something like:
sed '/:[0-9]*$/ ! s/$/:80/' ips.txt > new-ips.txt
So, for example, if your input file looked something like this (note the :80):
127.0.0.1
128.0.0.0:80
121.121.33.111
The final result would look something like this:
127.0.0.1:80
128.0.0.0:80
121.121.33.111:80
Concise version of the sed command:
sed -i s/$/:80/ file.txt
Explanation:
sed stream editor
-i in-place (edit file in place)
s substitution command
/replacement_from_reg_exp/replacement_to_text/ statement
$ matches the end of line (replacement_from_reg_exp)
:80 text you want to add at the end of every line (replacement_to_text)
file.txt the file name
How can this be achieved without modifying the original file?
If you want to leave the original file unchanged and have the results in another file, then give up -i option and add the redirection (>) to another file:
sed s/$/:80/ file.txt > another_file.txt
sed 's/.*/&:80/' abcd.txt >abcde.txt
If you'd like to add text at the end of each line in-place (in the same file), you can use -i parameter, for example:
sed -i'.bak' 's/$/:80/' foo.txt
However -i option is non-standard Unix extension and may not be available on all operating systems.
So you can consider using ex (which is equivalent to vi -e/vim -e):
ex +"%s/$/:80/g" -cwq foo.txt
which will add :80 to each line, but sometimes it can append it to blank lines.
So better method is to check if the line actually contain any number, and then append it, for example:
ex +"g/[0-9]/s/$/:80/g" -cwq foo.txt
If the file has more complex format, consider using proper regex, instead of [0-9].
You can also achieve this using the backreference technique
sed -i.bak 's/\(.*\)/\1:80/' foo.txt
You can also use with awk like this
awk '{print $0":80"}' foo.txt > tmp && mv tmp foo.txt
Using a text editor, check for ^M (control-M, or carriage return) at the end of each line. You will need to remove them first, then append the additional text at the end of the line.
sed -i 's|^M||g' ips.txt
sed -i 's|$|:80|g' ips.txt
sed -i 's/$/,/g' foo.txt
I do this quite often to add a comma to the end of an output so I can just easily copy and paste it into a Python(or your fav lang) array

sed text replace

How can I replace text with other text using GNU sed? I was hacked and am just trying to see if I can remove some of the code that was placed into my php files. The text is of the
eval(base64_decode('blah'));
variety. All of them are identical, I would just like to find and replace all of them in all files. I have tried some commands, but they either needlessly alter and damage text in the files or simply fail to launch at all.
sed -i 's/text/other text/g' filename
(sed -i "s/eval(base64_decode('blah'))/huh/g" filename in your case).
find . -name \*.php -exec sed -i "s/text/other/g" {} \;
You may want to do a dry run and leave off the -i and just direct it to a file as a test first.
On Mac the -i usually doesn't work.