It took me a while to figure out how to do this, so posting in case anyone else is looking for the same.
For adding a newline after a pattern, you can also say:
sed '/pattern/{G;}' filename
Quoting GNU sed manual:
G
Append a newline to the contents of the pattern space, and then append the contents of the hold space to that of the pattern space.
EDIT:
Incidentally, this happens to be covered in sed one liners:
# insert a blank line below every line which matches "regex"
sed '/regex/G'
This sed command:
sed -i '' '/pid = run/ a\
\
' file.txt
Finds the line with: pid = run
file.txt before
; Note: the default prefix is /usr/local/var
; Default Value: none
;pid = run/php-fpm.pid
; Error log file
and adds a linebreak after that line inside file.txt
file.txt after
; Note: the default prefix is /usr/local/var
; Default Value: none
;pid = run/php-fpm.pid
; Error log file
Or if you want to add text and a linebreak:
sed -i '/pid = run/ a\
new line of text\
' file.txt
file.txt after
; Note: the default prefix is /usr/local/var
; Default Value: none
;pid = run/php-fpm.pid
new line of text
; Error log file
A simple substitution works well:
sed 's/pattern.*$/&\n/'
Example :
$ printf "Hi\nBye\n" | sed 's/H.*$/&\nJohn/'
Hi
John
Bye
To be standard compliant, replace \n by backslash newline :
$ printf "Hi\nBye\n" | sed 's/H.*$/&\
> John/'
Hi
John
Bye
sed '/pattern/a\\r' file name
It will add a return after the pattern while g will replace the pattern with a blank line.
If a new line (blank) has to be added at end of the file use this:
sed '$a\\r' file name
Another possibility, e.g. if You don't have an empty hold register, could be:
sed '/pattern/{p;s/.*//}' file
Explanation:
/pattern/{...} = apply sequence of commands, if line with pattern found,
p = print the current line,
; = separator between commands,
s/.*// = replace anything with nothing in the pattern register,
then automatically print the empty pattern register as additional line)
The easiest option -->
sed 'i\
' filename
Related
I got a text like this:
TOKEN = decrypt_aes(
"189272123124aqephkiz3")
And I want to change it into:
TOKEN = "189272123124aqephkiz3"
How can I make this?
I can do this when its in a single line with following command:enter code here`
sed -i "s/decrypt_aes(\(.*\))/\1/g"
But I don`t no how to do when its in multi line
If your file contains only the two lines you showed, then this might help with GNU sed:
sed -i 'N; s/decrypt_aes(\n *//; s/)//' file
From man sed:
N: Append the next line of input into the pattern space.
You can use
sed -i '/.*decrypt_aes($/{N;s/^\([^=]*=\).*\(".*"\))$/\1 \2/}' file
Details:
/.*decrypt_aes($/ - matches a line that ends with decrypt_aes( substring, and if it matches, the block that follows is executed
N - append a newline and the next line to the pattern space
s/^\([^=]*=\).*\(".*"\))$/\1 \2/ - replaces
^\([^=]*=\).*\(".*"\))$ - start of string in the pattern space (^), any zero or more chars other than = and then a = are captured into Group 1 (\1), then any text (.*), then a "..." substring (captured into Group 2 (\2)) and then a ) at the end of string
\1 \2 is the replacement pattern, which is Group 1 value + space + Group 2 value.
See the online demo:
#!/bin/bash
s='TOKEN = decrypt_aes(
"189272123124aqephkiz3")'
sed '/.*decrypt_aes($/{N;s/^\([^=]*=\).*\(".*"\))$/\1 \2/}' <<< "$s"
Output:
TOKEN = "189272123124aqephkiz3"
This might work for you (GNU sed):
sed 'N;s/decrypt_aes.*\(".*"\).*/\1/;P;D' file
Append the following line and if the combined lines matches decrpty_aes followed by "...", replace the match by the string "...".
N.B. If no match is found, each line will be printed as is.
i had generate a list of file, and this had 17417 lines like :
./usr
./usr/share
./usr/share/mime-info
./usr/share/mime-info/libreoffice7.0.mime
./usr/share/mime-info/libreoffice7.0.keys
./usr/share/appdata
./usr/share/appdata/libreoffice7.0-writer.appdata.xml
./usr/share/appdata/org.libreoffice7.0.kde.metainfo.xml
./usr/share/appdata/libreoffice7.0-draw.appdata.xml
./usr/share/appdata/libreoffice7.0-impress.appdata.xml
./usr/share/appdata/libreoffice7.0-base.appdata.xml
./usr/share/appdata/libreoffice7.0-calc.appdata.xml
./usr/share/applications
./usr/share/applications/libreoffice7.0-xsltfilter.desktop
./usr/share/applications/libreoffice7.0-writer.desktop
./usr/share/applications/libreoffice7.0-base.desktop
./usr/share/applications/libreoffice7.0-math.desktop
./usr/share/applications/libreoffice7.0-startcenter.desktop
./usr/share/applications/libreoffice7.0-calc.desktop
./usr/share/applications/libreoffice7.0-draw.desktop
./usr/share/applications/libreoffice7.0-impress.desktop
./usr/share/icons
./usr/share/icons/gnome
./usr/share/icons/gnome/16x16
./usr/share/icons/gnome/16x16/mimetypes
./usr/share/icons/gnome/16x16/mimetypes/libreoffice7.0-oasis-formula.png
The thing is i want to delete the lines like :
./usr
./usr/share
./usr/share/mime-info
./usr/share/appdata
./usr/share/applications
./usr/share/icons
./usr/share/icons/gnome
./usr/share/icons/gnome/16x16
./usr/share/icons/gnome/16x16/mimetypes
and the "." at the start, for the result must be like :
/usr/share/mime-info/libreoffice7.0.mime
/usr/share/mime-info/libreoffice7.0.keys
/usr/share/appdata/libreoffice7.0-writer.appdata.xml
/usr/share/appdata/org.libreoffice7.0.kde.metainfo.xml
/usr/share/appdata/libreoffice7.0-draw.appdata.xml
/usr/share/appdata/libreoffice7.0-impress.appdata.xml
/usr/share/appdata/libreoffice7.0-base.appdata.xml
/usr/share/appdata/libreoffice7.0-calc.appdata.xml
/usr/share/applications/libreoffice7.0-xsltfilter.desktop
/usr/share/applications/libreoffice7.0-writer.desktop
/usr/share/applications/libreoffice7.0-base.desktop
/usr/share/applications/libreoffice7.0-math.desktop
/usr/share/applications/libreoffice7.0-startcenter.desktop
/usr/share/applications/libreoffice7.0-calc.desktop
/usr/share/applications/libreoffice7.0-draw.desktop
/usr/share/applications/libreoffice7.0-impress.desktop
/usr/share/icons/gnome/16x16/mimetypes/libreoffice7.0-oasis-formula.png
This is possible using sed ? or is more practical using another tool
With your list in the filename list, you could do:
sed -n 's/^[.]//;/\/.*[._].*$/p' list
Where:
sed -n suppresses printing of pattern-space; then
s/^[.]// is the substitution form that simply removes the first character '.' from each line; then
/\/.*[._].*$/p matches line that contain a '.' or '_' (optional) after the last '/' with p causing that line to be printed.
Example Use/Output
$ sed -n 's/^[.]//;/\/.*[._].*$/p' list
/usr/share/mime-info/libreoffice7.0.mime
/usr/share/mime-info/libreoffice7.0.keys
/usr/share/appdata/libreoffice7.0-writer.appdata.xml
/usr/share/appdata/org.libreoffice7.0.kde.metainfo.xml
/usr/share/appdata/libreoffice7.0-draw.appdata.xml
/usr/share/appdata/libreoffice7.0-impress.appdata.xml
/usr/share/appdata/libreoffice7.0-base.appdata.xml
/usr/share/appdata/libreoffice7.0-calc.appdata.xml
/usr/share/applications/libreoffice7.0-xsltfilter.desktop
/usr/share/applications/libreoffice7.0-writer.desktop
/usr/share/applications/libreoffice7.0-base.desktop
/usr/share/applications/libreoffice7.0-math.desktop
/usr/share/applications/libreoffice7.0-startcenter.desktop
/usr/share/applications/libreoffice7.0-calc.desktop
/usr/share/applications/libreoffice7.0-draw.desktop
/usr/share/applications/libreoffice7.0-impress.desktop
/usr/share/icons/gnome/16x16/mimetypes/libreoffice7.0-oasis-formula.png
Note, without GNU sed that allows chaining of expressions with ';' you would need:
sed -n -e 's/^[.]//' -e '/\/.*[._].*$/p' list
Assuming you want to delete the line(s) which is included other
pathname(s), would you please try:
sort -r list.txt | awk ' # sort the list in the reverse order
{
sub("^\\.", "") # remove leading dot
s = prev; sub("/[^/]+$", "", s) # remove the rightmost slash and following characters
if (s != $0) print # if s != $0, it means $0 is not a substring of the previous line
prev = $0 # keep $0 for the next line
}'
Result:
/usr/share/mime-info/libreoffice7.0.mime
/usr/share/mime-info/libreoffice7.0.keys
/usr/share/icons/gnome/16x16/mimetypes/libreoffice7.0-oasis-formula.png
/usr/share/applications/libreoffice7.0-xsltfilter.desktop
/usr/share/applications/libreoffice7.0-writer.desktop
/usr/share/applications/libreoffice7.0-startcenter.desktop
/usr/share/applications/libreoffice7.0-math.desktop
/usr/share/applications/libreoffice7.0-impress.desktop
/usr/share/applications/libreoffice7.0-draw.desktop
/usr/share/applications/libreoffice7.0-calc.desktop
/usr/share/applications/libreoffice7.0-base.desktop
/usr/share/appdata/org.libreoffice7.0.kde.metainfo.xml
/usr/share/appdata/libreoffice7.0-writer.appdata.xml
/usr/share/appdata/libreoffice7.0-impress.appdata.xml
/usr/share/appdata/libreoffice7.0-draw.appdata.xml
/usr/share/appdata/libreoffice7.0-calc.appdata.xml
/usr/share/appdata/libreoffice7.0-base.appdata.xml
I have files that need to be removed from comments and white space until keyword . Line number varies . Is it possible to limit multiple continued sed substitutions based on Keyword ?
This removes all comments and white spaces from file :
sed -i -e 's/#.*$//' -e 's/;.*$//' -e '/^$/d' file
For example something like this :
# string1
# string2
some string
; string3
; string4
####
<Keyword_Keep_this_line_and_comments_white_space_after_this>
# More comments that need to be here
; etc.
sed -i '1,/keyword/{/^[#;]/d;/^$/d;}' file
I would suggest using awk and setting a flag when you reach your keyword:
awk '/Keyword/ { stop = 1 } stop || !/^[[:blank:]]*([;#]|$)/' file
Set stop to true when the line contains Keyword. Do the default action (print the line) when stop is true or when the line doesn't match the regex. The regex matches lines whose first non-blank character is a semicolon or hash, or blank lines. It's slightly different to your condition but I think it does what you want.
The command prints to standard output so you should redirect to a new file and then overwrite the original to achieve an "in-place edit":
awk '...' input > tmp && mv tmp input
Use grep -n keyword to get the line number that contains the keyword.
Use sed -i -e '1,N s/#..., when N is the line number that contains the keyword, to only remove comments on the lines 1 to N.
mpu3$ echo * | xargs -n 1 -I {} | tr "|" "/n"
which outputs:
#.txt
ag.txt
bg.txt
bh.txt
bi.txt
bid.txt
dh.txt
dw.txt
er.txt
ha.txt
jo.txt
kc.txt
lfr.txt
lg.txt
ng.txt
pb.txt
r-c.txt
rj.txt
rw.txt
se.txt
sh.txt
vr.txt
wa.txt
is what I have so far. What is missing is the output; I get none. What I really want is to get a list of txt files, use their name up to the extension, process out the "|" and replace it with a LF/CR and put the new file in another directory as [old-name].ics. HALP. THX in advance. - Idiot me.
You can loop over the files and use sed to process the file:
for i in *.txt; do
sed -e 's/|/\n/g' "$i" > other_directory/"${i%.txt}".ics
done
No need to use xargs, especially with echo which would risk the filenames getting word split and having globbing apply to them, so could well do the wrong thing.
Then we use sed and use s to substitute | with \n g makes it a global replace. We redirect that to the other director you want and use bash's parameter expansion to strip off the .txt from the end
Here's an awk solution:
$ awk '
FNR==1 { # for first record of every file
close(f) # close previous file f
f="path_to_dir/" FILENAME # new filename with path
sub(/txt$/,"ics",f) } # replace txt with ics
{
gsub(/\|/,"\n") # replace | with \n
print > f }' *.txt # print to new file
I am trying to get a substring between &DEST= and the next & or a line break.
For example :
MYREQUESTISTO8764GETTHIS&DEST=SFO&ORIG=6546
In this I need to extract "SFO"
MYREQUESTISTO8764GETTHIS&DEST=SANFRANSISCO&ORIG=6546
In this I need to extract "SANFRANSISCO"
MYREQUESTISTO8764GETTHISWITH&DEST=SANJOSE
In this I need to extract "SANJOSE"
I am reading a file line by line, and I need to update the text after &DEST= and put it back in the file. The modification of the text is to mask the dest value with X character.
So, SFO should be replaced with XXX.
SANJOSE should be replaced with XXXXXXX.
Output :
MYREQUESTISTO8764GETTHIS&DEST=XXX&ORIG=6546
MYREQUESTISTO8764GETTHIS&DEST=XXXXXXXXXXXX&ORIG=6546
MYREQUESTISTO8764GETTHISWITH&DEST=XXXXXXX
Please let me know how to achieve this in script (Preferably shell or bash script).
Thanks.
$ cat file
MYREQUESTISTO8764GETTHIS&DEST=SFO&ORIG=6546
MYREQUESTISTO8764GETTHIS&DEST=PORTORICA
MYREQUESTISTO8764GETTHIS&DEST=SANFRANSISCO&ORIG=6546
MYREQUESTISTO8764GETTHISWITH&DEST=SANJOSE
$ sed -E 's/^.*&DEST=([^&]*)[&]*.*$/\1/' file
SFO
PORTORICA
SANFRANSISCO
SANJOSE
should do it
Replacing airports with an equal number of Xs
Let's consider this test file:
$ cat file
MYREQUESTISTO8764GETTHIS&DEST=SFO&ORIG=6546
MYREQUESTISTO8764GETTHIS&DEST=SANFRANSISCO&ORIG=6546
MYREQUESTISTO8764GETTHISWITH&DEST=SANJOSE
To replace the strings after &DEST= with an equal length of X and using GNU sed:
$ sed -E ':a; s/(&DEST=X*)[^X&]/\1X/; ta' file
MYREQUESTISTO8764GETTHIS&DEST=XXX&ORIG=6546
MYREQUESTISTO8764GETTHIS&DEST=XXXXXXXXXXXX&ORIG=6546
MYREQUESTISTO8764GETTHISWITH&DEST=XXXXXXX
To replace the file in-place:
sed -i -E ':a; s/(&DEST=X*)[^X&]/\1X/; ta' file
The above was tested with GNU sed. For BSD (OSX) sed, try:
sed -Ee :a -e 's/(&DEST=X*)[^X&]/\1X/' -e ta file
Or, to change in-place with BSD(OSX) sed, try:
sed -i '' -Ee :a -e 's/(&DEST=X*)[^X&]/\1X/' -e ta file
If there is some reason why it is important to use the shell to read the file line-by-line:
while IFS= read -r line
do
echo "$line" | sed -Ee :a -e 's/(&DEST=X*)[^X&]/\1X/' -e ta
done <file
How it works
Let's consider this code:
search_str="&DEST="
newfile=chart.txt
sed -E ':a; s/('"$search_str"'X*)[^X&]/\1X/; ta' "$newfile"
-E
This tells sed to use Extended Regular Expressions (ERE). This has the advantage of requiring fewer backslashes to escape things.
:a
This creates a label a.
s/('"$search_str"'X*)[^X&]/\1X/
This looks for $search_str followed by any number of X followed by any character that is not X or &. Because of the parens, everything except that last character is saved into group 1. This string is replaced by group 1, denoted \1 and an X.
ta
In sed, t is a test command. If the substitution was made (meaning that some character needed to be replaced by X), then the test evaluates to true and, in that case, ta tells sed to jump to label a.
This test-and-jump causes the substitution to be repeated as many times as necessary.
Replacing multiple tags with one sed command
$ name='DEST|ORIG'; sed -E ':a; s/(&('"$name"')=X*)[^X&]/\1X/; ta' file
MYREQUESTISTO8764GETTHIS&DEST=XXX&ORIG=XXXX
MYREQUESTISTO8764GETTHIS&DEST=XXXXXXXXXXXX&ORIG=XXXX
MYREQUESTISTO8764GETTHISWITH&DEST=XXXXXXX
Answer for original question
Using shell
$ s='MYREQUESTISTO8764GETTHIS&DEST=SFO&ORIG=6546'
$ s=${s#*&DEST=}
$ echo ${s%%&*}
SFO
How it works:
${s#*&DEST=} is prefix removal. This removes all text up to and including the first occurrence of &DEST=.
${s%%&*} is suffix removal_. It removes all text from the first & to the end of the string.
Using awk
$ echo 'MYREQUESTISTO8764GETTHIS&DEST=SFO&ORIG=6546' | awk -F'[=\n]' '$1=="DEST"{print $2}' RS='&'
SFO
How it works:
-F'[=\n]'
This tells awk to treat either an equal sign or a newline as the field separator
$1=="DEST"{print $2}
If the first field is DEST, then print the second field.
RS='&'
This sets the record separator to &.
With GNU bash:
while IFS= read -r line; do
[[ $line =~ (.*&DEST=)(.*)((&.*|$)) ]] && echo "${BASH_REMATCH[1]}fooooo${BASH_REMATCH[3]}"
done < file
Output:
MYREQUESTISTO8764GETTHIS&DEST=fooooo&ORIG=6546
MYREQUESTISTO8764GETTHIS&DEST=fooooo&ORIG=6546
MYREQUESTISTO8764GETTHISWITH&DEST=fooooo
Replace the characters between &DEST and & (or EOL) with x's:
awk -F'&DEST=' '{
printf("%s&DEST=", $1);
xlen=index($2,"&");
if ( xlen == 0) xlen=length($2)+1;
for (i=0;i<xlen;i++) printf("%s", "X");
endstr=substr($2,xlen);
printf("%s\n", endstr);
}' file