How do I modify my bash file to achieve the expected result shown below ?
#!/bin/bash
filename=$1
var="$(<$filename)" | tr -d '\n'
sed -i 's/;/,/g' $var
Convert this input file
a,b;c^d"e}
f;g,h!;i8j-
To this output file
a,b,c,d,e,f,g,h,i,j
How to convert separators using regex in bash
You would, well, literally, do exactly that - convert any of the separators using regex. This consists of steps:
most importantly, figure out the exact definition of what consists of a "separator"
writing a regex for it
writing an algorithm for it
running and testing the code
For example, assuming a separator is a sequence of of any of \n,;^"}!8- characters, you could do:
sed -zi 's/[,;^"}!8-]\+/,/g; s/,$/\n/' input_file
Or similar with first tr '\n' , for example when -z is not available with your sed, and then pass the result of tr to sed. The second regex adds a trailing newline on the output instead of a trailing ,.
Additionally, in your code:
var is unset on sed line. Parts of | pipeline are running in a subshell.
var=$(<$filename) contains the contents of the file, whereas sed wants a filename as argument, not file contents.
var=.... | ... is pipeing the result of assignment to tr. The output of assignment is empty, so that line produces nothing, and its output is unused.
Remember to check bash scripts with shellcheck.
For a somewhat portable solution, maybe try
tr -cs A-Za-z , <input_file | sed '$s/,$/\n/' >output_file
The use of \n to force a final newline is still not entirely reliable; there are some sed versions which interpret the sequence as a literal n.
You'd move output_file back on top of input_file after this command if you want to replace the original.
How do I strip multi-line markdown comments, such as the one below, in bash?
some text
<!-- QUESTION:
How do I remove everything
in-between these tags?
-->
some<!-- Including embedded single-line comments such as this --> text
I've tried sed -e 's/<!--((.*?)\n?)+-->//g' $1, which works only with single line, and cat $1 | tr '\n' '\r' | sed -e 's/<!--.*-->//g' | tr '\r' '\n', which removes everything after the first multiline comment.
<!--((.*?)\n?)+--> captures the required area in my text-editor, but
sed -e 's/<!--((.*?)\n?)+-->//g' $1 doesn't work as expected.
Other examples I can find that works with C++ comments are too complicated to decode.
You can accomplish this with a perl one-liner.
Perl Switches:
-0 sets the input record separator to the null character \0
-p prints the result of perl code
-e executes the following code
Inside the regular expression:
g flag means global (perform the replacement as many times as possible)
s flag means treat the input as a multi-line string
Match the characters `<!--` followed by anything up to the characters `-->`
including anything after that till the newline. Replace that with nothing.
In Action:
perl -0pe 's|<!--.+?-->.*?\n||gs;' input
Output:
some text
some text
This is what I have so far, tried multiple ways but can't get it just right.
My goal is to sanitize the input to prevent problems while inputting to mysql from text file
cat 'file.txt' | awk '{gsub(/'"'"'/, '.') ; print $0}' > 'file_sanitized.txt'
What you want is:
awk '{gsub(/\047/,".")}1' file
See http://awk.freeshell.org/PrintASingleQuote.
Since you requested "how to escape single quote using awk" here's a solution using awk:
awk '{ gsub("\x27", ".");print $0}'
where \x27 is the escape sequence representation of the hexadecimal value 27 (a single quote).
For a list of all escape sequences see https://www.gnu.org/software/gawk/manual/html_node/Escape-Sequences.html
I'm not sure if I got the problem correctly. If you want to replace all occurrences of a single quote by a dot, use tr:
tr "'" "." < file.txt > sanitized.txt
If you want to escape the single quote with a backslash use sed like this:
sed "s/'/\\\'/g" file.txt > sanitized.txt
Note: Please take the advice from CharlesDuffy seriously. This is far from a stable and safe solution to escape values for an SQL import.
You have used wrong quoting:
awk '{gsub(/'"'"'/, "."); print}'
I have some data that looks like this
"thumbnailUrl": "http://placehold.it/150/adf4e1"
I want to know how I can get the trailing part of the URL, I want the output to be
adf4e1
I was trying to grep when starting with / and ending with " but I'm only a beginner in shell scripting and need some help.
I came up with a quick and dirty solution, using grep (with perl regex) and cut:
$ cat file
"thumbnailUrl": "http://placehold.it/150/adf4e1"
"anotherUrl": "http://stackoverflow.com/questions/3979680"
"thumbnailUrl": "http://facebook.com/12f"
"randortag": "http://google.com/this/is/how/we/roll/3fk19as1"
$ cat file | grep -o '/\w*"$' | cut -d'/' -f2- | cut -d'"' -f1
adf4e1
3979680
12f
3fk19as1
We could kill this with a thousand little cuts, or just one blow from Awk:
awk -F'[/"]' '{ print $(NF-1); }'
Test:
$ echo '"thumbnailUrl": "http://placehold.it/150/adf4e1"' \
| awk -F'[/"]' '{ print $(NF-1); }'
adf4e1
Filter thorugh Awk using double quotes and slashes as field separators. This means that the trailing part ../adf4e1" is separated as {..}</>{adf4e1}<">{} where curly braces denote fields and angle brackets separators. The Awk variable NF gives the 1-based number of fields and so $NF is the last field. That's not the one we want, because it is blank; we want $(NF-1): the second last field.
"Golfed" version:
awk -F[/\"] '$0=$(NF-1)'
If the original string is coming from a larger JSON object, use something like jq to extract the value you want.
For example:
$ jq -n '{thumbnail: "http://placehold.it/150/adf4e1"}' |
> jq -r '.thumbnail|split("/")[-1]'
adf4e1
(The first command just generates a valid JSON object representing the original source of your data; the second command parses it and extracts the desired value. The split function splits the URL into an array, from which you only care about the last element.)
You can also do this purely in bash using string replacement and substring removal if you wrap your string in single quotes and assign it to a variable.
#!/bin/bash
string='"thumbnailUrl": "http://placehold.it/150/adf4e1"'
string="${string//\"}"
echo "${string##*/}"
adf4e1 #output
You can do that using 'cut' command in linux. Cut it using '/' and keep the last cut. Try it, its fun!
Refer http://www.thegeekstuff.com/2013/06/cut-command-examples
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.