Please help me creating a shell script to search all lines and replace format in a file using sed.
Example - [‘abc,xyz’] to be changed to [‘abc’,’xyz’]
Here you go:
[user#myserver ~]$ echo "['abc,xyz']" | sed "s:,:\',\':g"
['abc','xyz']
Here colon (:) is used as delimiter and backslash (\) has been used as escape character in sed command.
This might work for you (GNU sed):
sed -E ':a;s/(\[('\''[^'\'',]*'\'',)*'\''[^'\'',]*),([^]]*'\''\])/\1'\'','\''\3/;ta' file
Using bash, how would one replace all unquoted characters from a file?
I have a system that I can't modify that spits out CSV files such as:
code;prop1;prop2;prop3;prop4;prop5;prop6
0,1000,89,"a1,a2,a3",33,,
1,,,"a55,a10",1,1 L,87
2,25,1001,a4,,"1,5 L",
I need this to become, for a new system being added
code;prop1;prop2;prop3;prop4;prop5;prop6
0;1000;89;a1,a2,a3;33;;
1;;;a55,a10;1;1 L;87
2;25;1001;a4;1,5 L;
If the quotes can be removed after this substitution happens in one command it would be nice :) But I prefer clarity to complicated one-liners for future maintenance.
Thank you
With sed:
sed -e 's/,/;/g' -e ':loop; s/\("\)\([^;]*\);\([^"]*"\)/\1\2,\3/; t loop'
Test:
$ sed -e 's/,/;/g' -e ':loop; s/\("\)\([^;]*\);\([^"]*"\)/\1\2,\3/; t loop' yourfile
code;prop1;prop2;prop3;prop4;prop5;prop6
0;1000;89;"a1,a2,a3";33;;
1;;;"a55,a10";1;1 L;87
2;25;1001;a4;;"1,5 L";
You want to use a csv parser. Parsing csv with shell tools is hard (you will encounter regular expressions soon, and they rarely get all cases).
There is one in almost every language. I recommend python.
You can also do this using excel/openoffice variants by opening the file and then saving with ; as the separator.
You can used sed:
echo '0,1000,89,"a1,a2,a3",33,,' | sed -e "s|\"||g"
This will replace " with the empty string (deletes it), and you can pipe another sed to replace the , with ;:
sed -e "s|,|;|g"
$ echo '0,1000,89,"a1,a2,a3",33,,' | sed -e "s|\"||g" | sed -e "s|,|;|g"
>> 0;1000;89;a1;a2;a3;33;;
Note that you can use any separator you want instead of | inside the sed command. For example, you can rewrite the first sed as:
sed -e "s-\"--g"
I am in use of bash shell.
I have a file filelist:
cat filelist
../1.txt
../2.txt
...
../100.txt
I want to remove "../" and I tried $ cat filelist | sed s/..\///
, but it gives an error message. How can i remove the slash?
You need quotes:
sed 's/..\///'
. means any character so you have to escape it too:
sed 's/\.\.\///'
and for readability you can use another character for the separator:
sed 's|\.\./||'
You need quotes around the sed argument and also need to include the g global flag. It is also not necessary to cat the file first. You should also escape the periods.
Use:
sed 's/\.\.\///g' filelist
Gives:
1.txt
2.txt
...
100.txt
remove starting ../ (i guess it's the purpose but not specified)
sed 's#^\.\./##' filelist
escaping the dot for avoiding regex meaning
changing default separator /by # to allow a readible / in path
adding ^ for limiting to starting path and not changing something like bad/../folder. If not the wanted behavior, just remove this caret.
direct use of file from sed (no need of cat with sed in this case)
Use an alternate separator.
sed 's|^\.\./||g' filelist
You can also use the -i flag to edit the file in-place.
sed -i 's|^\.\./||g' filelist
I came across the following in our legacy bash code
sed -itmp <something> file.txt
Since I did not understand what it does clearly, I tried the following
Here is the content of file.txt before running sed
dummy={my.java.home}
dummy={my_java_home}
I run now
sed -itmp "s#{my.java.home}#${JAVA_HOME}#g" file.txt
After I run this, I get the following
dummy=/Library/Java/JavaVirtualMachines/jdk1.7.0_71.jdk/Contents/Home
dummy=/Library/Java/JavaVirtualMachines/jdk1.7.0_71.jdk/Contents/Home
I can see how sed replaces. But I fail to understand sed replaces both my_java_home and my.java.home in my original file although I asked it to change only my.java.home when issuing sed command above.
Thanks
Because dot in your regex matches any charcter not only the literal dot. So i suggest you to escape the dots present in your regex , so that it matches my.java.home string only.
sed -itmp "s#{my\.java\.home}#${JAVA_HOME}#g" file.txt
And you won't actually need a tmp parameters.
sed -i "s#{my\.java\.home}#${JAVA_HOME}#g" file.txt
How to insert a newline before a pattern within a line?
For example, this will insert a newline behind the regex pattern.
sed 's/regex/&\n/g'
How can I do the same but in front of the pattern?
Given this sample input file, the pattern to match on is the phone number.
some text (012)345-6789
Should become
some text
(012)345-6789
This works in bash and zsh, tested on Linux and OS X:
sed 's/regexp/\'$'\n/g'
In general, for $ followed by a string literal in single quotes bash performs C-style backslash substitution, e.g. $'\t' is translated to a literal tab. Plus, sed wants your newline literal to be escaped with a backslash, hence the \ before $. And finally, the dollar sign itself shouldn't be quoted so that it's interpreted by the shell, therefore we close the quote before the $ and then open it again.
Edit: As suggested in the comments by #mklement0, this works as well:
sed $'s/regexp/\\\n/g'
What happens here is: the entire sed command is now a C-style string, which means the backslash that sed requires to be placed before the new line literal should now be escaped with another backslash. Though more readable, in this case you won't be able to do shell string substitutions (without making it ugly again.)
Some of the other answers didn't work for my version of sed.
Switching the position of & and \n did work.
sed 's/regexp/\n&/g'
Edit: This doesn't seem to work on OS X, unless you install gnu-sed.
In sed, you can't add newlines in the output stream easily. You need to use a continuation line, which is awkward, but it works:
$ sed 's/regexp/\
&/'
Example:
$ echo foo | sed 's/.*/\
&/'
foo
See here for details. If you want something slightly less awkward you could try using perl -pe with match groups instead of sed:
$ echo foo | perl -pe 's/(.*)/\n$1/'
foo
$1 refers to the first matched group in the regular expression, where groups are in parentheses.
On my mac, the following inserts a single 'n' instead of newline:
sed 's/regexp/\n&/g'
This replaces with newline:
sed "s/regexp/\\`echo -e '\n\r'`/g"
echo one,two,three | sed 's/,/\
/g'
You can use perl one-liners much like you do with sed, with the advantage of full perl regular expression support (which is much more powerful than what you get with sed). There is also very little variation across *nix platforms - perl is generally perl. So you can stop worrying about how to make your particular system's version of sed do what you want.
In this case, you can do
perl -pe 's/(regex)/\n$1/'
-pe puts perl into a "execute and print" loop, much like sed's normal mode of operation.
' quotes everything else so the shell won't interfere
() surrounding the regex is a grouping operator. $1 on the right side of the substitution prints out whatever was matched inside these parens.
Finally, \n is a newline.
Regardless of whether you are using parentheses as a grouping operator, you have to escape any parentheses you are trying to match. So a regex to match the pattern you list above would be something like
\(\d\d\d\)\d\d\d-\d\d\d\d
\( or \) matches a literal paren, and \d matches a digit.
Better:
\(\d{3}\)\d{3}-\d{4}
I imagine you can figure out what the numbers in braces are doing.
Additionally, you can use delimiters other than / for your regex. So if you need to match / you won't need to escape it. Either of the below is equivalent to the regex at the beginning of my answer. In theory you can substitute any character for the standard /'s.
perl -pe 's#(regex)#\n$1#'
perl -pe 's{(regex)}{\n$1}'
A couple final thoughts.
using -ne instead of -pe acts similarly, but doesn't automatically print at the end. It can be handy if you want to print on your own. E.g., here's a grep-alike (m/foobar/ is a regex match):
perl -ne 'if (m/foobar/) {print}'
If you are finding dealing with newlines troublesome, and you want it to be magically handled for you, add -l. Not useful for the OP, who was working with newlines, though.
Bonus tip - if you have the pcre package installed, it comes with pcregrep, which uses full perl-compatible regexes.
In this case, I do not use sed. I use tr.
cat Somefile |tr ',' '\012'
This takes the comma and replaces it with the carriage return.
To insert a newline to output stream on Linux, I used:
sed -i "s/def/abc\\\ndef/" file1
Where file1 was:
def
Before the sed in-place replacement, and:
abc
def
After the sed in-place replacement. Please note the use of \\\n. If the patterns have a " inside it, escape using \".
Hmm, just escaped newlines seem to work in more recent versions of sed (I have GNU sed 4.2.1),
dev:~/pg/services/places> echo 'foobar' | sed -r 's/(bar)/\n\1/;'
foo
bar
echo pattern | sed -E -e $'s/^(pattern)/\\\n\\1/'
worked fine on El Captitan with () support
In my case the below method works.
sed -i 's/playstation/PS4/' input.txt
Can be written as:
sed -i 's/playstation/PS4\nplaystation/' input.txt
PS4
playstation
Consider using \\n while using it in a string literal.
sed : is stream editor
-i : Allows to edit the source file
+: Is delimiter.
I hope the above information works for you 😃.
in sed you can reference groups in your pattern with "\1", "\2", ....
so if the pattern you're looking for is "PATTERN", and you want to insert "BEFORE" in front of it, you can use, sans escaping
sed 's/(PATTERN)/BEFORE\1/g'
i.e.
sed 's/\(PATTERN\)/BEFORE\1/g'
You can also do this with awk, using -v to provide the pattern:
awk -v patt="pattern" '$0 ~ patt {gsub(patt, "\n"patt)}1' file
This checks if a line contains a given pattern. If so, it appends a new line to the beginning of it.
See a basic example:
$ cat file
hello
this is some pattern and we are going ahead
bye!
$ awk -v patt="pattern" '$0 ~ patt {gsub(patt, "\n"patt)}1' file
hello
this is some
pattern and we are going ahead
bye!
Note it will affect to all patterns in a line:
$ cat file
this pattern is some pattern and we are going ahead
$ awk -v patt="pattern" '$0 ~ patt {gsub(patt, "\n"patt)}1' d
this
pattern is some
pattern and we are going ahead
sed -e 's/regexp/\0\n/g'
\0 is the null, so your expression is replaced with null (nothing) and then...
\n is the new line
On some flavors of Unix doesn't work, but I think it's the solution to your problem.
echo "Hello" | sed -e 's/Hello/\0\ntmow/g'
Hello
tmow
This works in MAC for me
sed -i.bak -e 's/regex/xregex/g' input.txt sed -i.bak -e 's/qregex/\'$'\nregex/g' input.txt
Dono whether its perfect one...
After reading all the answers to this question, it still took me many attempts to get the correct syntax to the following example script:
#!/bin/bash
# script: add_domain
# using fixed values instead of command line parameters $1, $2
# to show typical variable values in this example
ipaddr="127.0.0.1"
domain="example.com"
# no need to escape $ipaddr and $domain values if we use separate quotes.
sudo sed -i '$a \\n'"$ipaddr www.$domain $domain" /etc/hosts
The script appends a newline \n followed by another line of text to the end of a file using a single sed command.
In vi on Red Hat, I was able to insert carriage returns using just the \r character. I believe this internally executes 'ex' instead of 'sed', but it's similar, and vi can be another way to do bulk edits such as code patches. For example. I am surrounding a search term with an if statement that insists on carriage returns after the braces:
:.,$s/\(my_function(.*)\)/if(!skip_option){\r\t\1\r\t}/
Note that I also had it insert some tabs to make things align better.
Just to add to the list of many ways to do this, here is a simple python alternative. You could of course use re.sub() if a regex were needed.
python -c 'print(open("./myfile.txt", "r").read().replace("String to match", "String to match\n"))' > myfile_lines.txt
sed 's/regexp/\'$'\n/g'
works as justified and detailed by mojuba in his answer .
However, this did not work:
sed 's/regexp/\\\n/g'
It added a new line, but at the end of the original line, a \n was added.