SET_VALUE(ab.ms.r.gms_dil_cfg.f().gms_dil_mode, dsad_sd );
How can I use sed to replace only from the SET_VALUE until the , with each letter after _ to be upper case?
result:
SET_VALUE(ab.ms.r.gmsDilCfg.f().gmsDilMode, dsad_sd );
For your input string you may apply the following sed expression + bash variable substitution:
s="SET_VALUE(ab.ms.r.gms_dil_cfg.f().gms_dil_mode, dsad sd )"
res=$(sed '1s/_\([a-z]\)/\U\1/g;' <<< "${s%,*}"),${s#*,}
echo "$res"
The output:
SET_VALUE(ab.ms.r.gmsDilCfg.f().gmsDilMode, dsad_sd );
Got distracted while writing this one up so Roman beat me to the punch, but this has a slight variation so figured I'd post it as another option ...
$ s="SET_VALUE(ab.ms.r.gms_dil_cfg.f().gms_dil_mode, dsad_sd );"
$ sed 's/,/,\n/g' <<< "$s" | sed -n '1{s/_\([a-z]\)/\U\1/g;N;s/\n//;p}'
SET_VALUE(ab.ms.r.gmsDilCfg.f().gmsDilMode, dsad_sd );
s/,/,\n/g : break input into separate lines at the comma (leave comma on first line, push rest of input to a second line)
at this point we've broken our input into 2 lines; the second sed invocation will now be working with a 2-line input
sed -n : refrain from printing input lines as they're processed; we'll explicitly print lines when required
1{...} : for the first line, apply the commands inside the braces ...
s/_\([a-z]\)/\U\1/g : for each pattern we find like '_[a-z]', save the [a-z] in buffer #1, and replace the pattern with the upper case of the contents of buffer #1
at this point we've made the desired edits to line #1 (ie, everything before the comma in the original input), now ...
N : read and append the next line into the pattern space
s/\n// : replace the carriage return with a null character
at this point we've pasted lines #1 and #2 together into a single line
p : print the pattern space
Related
I have been tinkering with this for a while but can't quite figure it out. A sample line within the file looks like this:
"...~236 characters of data...Y YYY. Y...many more characters of data"
How would I use sed or awk to replace spaces with a B character only between positions 236 and 246? In that example string it starts at character 29 and ends at character 39 within the string. I would want to preserve all the text preceding and following the target chunk of data within the line.
For clarification based on the comments, it should be applied to all lines in the file and expected output would be:
"...~236 characters of data...YBBYYY.BBY...many more characters of data"
With GNU awk:
$ awk -v FIELDWIDTHS='29 10 *' -v OFS= '{gsub(/ /, "B", $2)} 1' ip.txt
...~236 characters of data...YBBYYY.BBY...many more characters of data
FIELDWIDTHS='29 10 *' means 29 characters for first field, next 10 characters for second field and the rest for third field. OFS is set to empty, otherwise you'll get space added between the fields.
With perl:
$ perl -pe 's/^.{29}\K.{10}/$&=~tr| |B|r/e' ip.txt
...~236 characters of data...YBBYYY.BBY...many more characters of data
^.{29}\K match and ignore first 29 characters
.{10} match 10 characters
e flag to allow Perl code instead of string in replacement section
$&=~tr| |B|r convert space to B for the matched portion
Use this Perl one-liner with substr and tr. Note that this uses the fact that you can assign to substr, which changes the original string:
perl -lpe 'BEGIN { $from = 29; $to = 39; } (substr $_, ( $from - 1 ), ( $to - $from + 1 ) ) =~ tr/ /B/;' in_file > out_file
To change the file in-place, use:
perl -i.bak -lpe 'BEGIN { $from = 29; $to = 39; } (substr $_, ( $from - 1 ), ( $to - $from + 1 ) ) =~ tr/ /B/;' in_file
The Perl one-liner uses these command line flags:
-e : Tells Perl to look for code in-line, instead of in a file.
-p : Loop over the input one line at a time, assigning it to $_ by default. Add print $_ after each loop iteration.
-l : Strip the input line separator ("\n" on *NIX by default) before executing the code in-line, and append it when printing.
-i.bak : Edit input files in-place (overwrite the input file). Before overwriting, save a backup copy of the original file by appending to its name the extension .bak.
I would use GNU AWK following way, for simplicity sake say we have file.txt content
S o m e s t r i n g
and want to change spaces from 5 (inclusive) to 10 (inclusive) position then
awk 'BEGIN{FPAT=".";OFS=""}{for(i=5;i<=10;i+=1)$i=($i==" "?"B":$i);print}' file.txt
output is
S o mBeBsBt r i n g
Explanation: I set field pattern (FPAT) to any single character and output field seperator (OFS) to empty string, thus every field is populated by single characters and I do not get superfluous space when print-ing. I use for loop to access desired fields and for every one I check if it is space, if it is I assign B here otherwise I assign original value, finally I print whole changed line.
Using GNU awk:
awk -v strt=29 -v end=39 '{ ram=substr($0,strt,(end-strt));gsub(" ","B",ram);print substr($0,1,(strt-1)) ram substr($0,(end)) }' file
Explanation:
awk -v strt=29 -v end=39 '{ # Pass the start and end character positions as strt and end respectively
ram=substr($0,strt,(end-strt)); # Extract the 29th to the 39th characters of the line and read into variable ram
gsub(" ","B",ram); # Replace spaces with B in ram
print substr($0,1,(strt-1)) ram substr($0,(end)) # Rebuild the line incorporating raw and printing the result
}'file
This is certainly a suitable task for perl, and saddens me that my perl has become so rusty that this is the best I can come up with at the moment:
perl -e 'local $/=\1;while(<>) { s/ /B/ if $. >= 236 && $. <= 246; print }' input;
Another awk but using FS="":
$ awk 'BEGIN{FS=OFS=""}{for(i=29;i<=39;i++)sub(/ /,"B",$i)}1' file
Output:
"...~236 characters of data...YBBYYY.BBY...many more characters of data"
Explained:
$ awk ' # yes awk yes
BEGIN {
FS=OFS="" # set empty field delimiters
}
{
for(i=29;i<=39;i++) # between desired indexes
sub(/ /,"B",$i) # replace space with B
# if($i==" ") # couldve taken this route, too
# $i="B"
}1' file # implicit output
With sed :
sed '
H
s/\(.\{236\}\)\(.\{11\}\).*/\2/
s/ /B/g
H
g
s/\n//g
s/\(.\{236\}\)\(.\{11\}\)\(.*\)\(.\{11\}\)/\1\4\3/
x
s/.*//
x' infile
When you have an input string without \r, you can use:
sed -r 's/(.{236})(.{10})(.*)/\1\r\2\r\3/;:a;s/(\r.*) (.*\r)/\1B\2/;ta;s/\r//g' input
Explanation:
First put \r around the area that you want to change.
Next introduce a label to jump back to.
Next replace a space between 2 markers.
Repeat until all spaces are replaced.
Remove the markers.
In your case, where the length doesn't change, you can do without the markers.
Replace a space after 236..245 characters and try again when it succeeds.
sed -r ':a; s/^(.{236})([^ ]{0,9}) /\1\2B/;ta' input
This might work for you (GNU sed):
sed -E 's/./&\n/245;s//\n&/236/;h;y/ /B/;H;g;s/\n.*\n(.*)\n.*\n(.*)\n.*/\2\1/' file
Divide the problem into 2 lines, one with spaces and one with B's where there were spaces.
Then using pattern matching make a composite line from the two lines.
N.B. The newline can be used as a delimiter as it is guaranteed not to be in seds pattern space.
I want to replace only the last string "delay" by "ens_delay" in my file and delete the others one before the last one:
Input file:
alpha_notify_teta=''
alpha_notify_check='YES'
text='CRDS'
delay=''
delay=''
delay=''
textfileooooop=''
alpha_enable='YES'
alpha_hostnames=''
alpha_orange='YES'
alpha_orange_interval='300'
alpha_notification_level='ALL'
expression='YES'
delay='9'
textfileooooop=''
alpha_enable='YES'
alpha_hostnames=''
Output file: (expected value)
alpha_notify_teta=''
alpha_notify_check='YES'
text='CRDS'
textfileooooop=''
alpha_enable='YES'
alpha_hostnames=''
alpha_orange='YES'
alpha_orange_interval='300'
alpha_notification_level='ALL'
expression='YES'
ens_delay='9'
textfileooooop=''
alpha_enable='YES'
alpha_hostnames=''
Here my first command but it doesn't work because it will work only if I have delay as last line.
sed -e '$,/delay/ s/delay/ens_delay/'
My second command will delete all lines contain "delay", even "ens_delay" will be deleted.
sed -i '/delay/d'
Thank you
This might work for you (GNU sed):
sed '/^delay=/,$!b;/^delay=/!H;//{x;s/^[^\n]*\n\?//;/./p;x;h};$!d;x;s/^/ens_/' file
Lines before the first line beginning delay= should be printed as normal. Otherwise, a line beginning delay= is stored in the hold space and subsequent lines that do not begin delay= are appended to it. Should the hold space already contain such lines, the first line is deleted and the remaining lines printed before the hold space is replaced by the current line. At the end of the file, the first line of the hold space is amended to prepend the string ens_ and then the whole of the hold space is printed.
You cannot do this kind of thing with sed. There is no way in sed to "look forward" and tell if there are more matches to the pattern. You can kind of look back, but that won't be sufficient to solve this problem.
This perl script will solve it:
#!/usr/bin/perl
use strict;
use warnings;
my ($seek, $replacement, $last, #new) = (shift, shift, 0);
open(my $fh, shift) or die $!;
my #l = <$fh>;
close($fh) or die $!;
foreach (reverse #l){
if(/$seek/){
if ($last++ == 0){
s/$seek/$replacement/;
} else {
next;
}
}
unshift(#new, $_);
}
print join "", #new;
Call like:
./script delay= ens_delay= inputfile
I chose to entirely eliminate lines which you intended to delete rather than collapse them in to a single blank line. If that is really required then it's a bit more complicated: the first such line in any consecutive set (or rather the last such) must be pushed on to the output list and you have to track whether this has just been done so you know whether to push the next time, too.
You could also solve this problem with awk, python, or any number of other languages. Just not sed.
Have this monster:
sed -e "1,$(expr $(sed -n '/^delay=/=' your_file.txt | tail -1) - 1)"'s/^delay=.*$//' \
-e 's/^delay=/ens_delay=/' your_file.txt
Here:
sed -n '/^delay=/=' your_file.txt | tail -1 return the last line number of the encountered pattern (let's name it X)
expr is used to get the X-1 line
"1,X-1"'[command]' means "perform this command betwen the first and the X-1 line included (I used double quotes to let the expansion getting done)
's/^delay=.*$//' the said [command]
-e 's/^delay=/ens_delay=/' the next expression to perform (will occur only on the last line)
Output:
alpha_notify_teta=''
alpha_notify_check='YES'
text='CRDS'
textfileooooop=''
alpha_enable='YES'
alpha_hostnames=''
alpha_hsm_backup_notification='YES'
alpha_orange='YES'
alpha_orange_interval='300'
alpha_notification_level='ALL'
expression='YES'
ens_delay='9'
textfileooooop=''
alpha_enable='YES'
alpha_hostnames=''
alpha_hsm_backup_notification='YES'
If you want to delete the lines instead of leaving them blank:
sed -e "1,$(expr $(sed -n '/^delay=/=' your_file.txt | tail -1) - 1)"'{/^delay=.*$/d}' \
-e 's/^delay=/ens_delay=/' your_file.txt
As was mentioned elsewhere, sed can't know which occurrence of a substring is the last one. But awk can keep track of things in arrays. For example, the following will delete all duplicate assignments, as well ask making your substitution:
awk 'BEGIN{FS=OFS="="} $1=="delay"{$1="ens_delay"} !($1 in a){o[++i]=$1} {a[$1]=$0} END{for(x=0;x<i;x++) printf "%s\n",a[o[x]]}' inputfile
Or, broken out for easier reading/comments:
BEGIN {
FS=OFS="=" # set the field separator, to help isolate the left hand side
}
$1=="delay" {
$1="ens_delay" # your field substitution
}
!($1 in a) {
o[++i]=$1 # if we haven't seen this variable, record its position
}
{
a[$1]=$0 # record the value of the last-seen occurrence of this variable
}
END {
for (x=0;x<i;x++) # step through the array,
printf "%s\n",a[o[x]] # printing the last-seen values, in the order
} # their variable was first seen in the input file.
You might not care about the order of the variables. If so, the following might be simpler:
awk 'BEGIN{FS=OFS="="} $1=="delay"{$1="ens_delay"} {o[$1]=$0} END{for(i in o) printf "%s\n", o[i]}' inputfile
This simply stores the last-seen line in an array whose key is the variable name, then prints out the content of the array in an unknown order.
Assuming I understand your specifications properly, this should do what you need. Given infile x,
$: last=$( grep -n delay x|tail -1|sed 's/:.*//' )
This grep's the file for all lines with delay and returns them with the line number prepended with a colon. The tail -1 grabs the last of those lines, ignoring all the others. sed 's/:.*//' strips the colon and the actual line content, leaving only the number (here it was 14.)
That all evaluates out to assign 14 as $last.
$: sed '/delay/ { '$last'!d; '$last' s/delay/ens_delay/; }' x
alpha_notify_teta=''
alpha_notify_check='YES'
text='CRDS'
textfileooooop=''
alpha_enable='YES'
alpha_hostnames=''
alpha_orange='YES'
alpha_orange_interval='300'
alpha_notification_level='ALL'
expression='YES'
ens_delay='9'
textfileooooop=''
alpha_enable='YES'
alpha_hostnames=''
Apologies for the ugly catenation. What this does is writes the script using the value of $last so that the result looks like this to sed:
$: sed '/delay/ { 14!d; 14 s/delay/ens_delay/; }' x
sed reads leading numbers as line selectors, so what this script of commands do -
First, sed automatically prints lines unless told not to, so by default it would just print every line. The script modifies that.
/delay/ {...} is a pattern-based record selector. It will apply the commands between the {} to all lines that match /delay/, which is why it doesn't need another grep - it handles that itself. Inside the curlies, the script does two things.
First, 14!d says (only if this line has delay, which it will) that if the line number is 14, do not (the !) delete the record. Since all the other lines with delay won't be line 14 (or whatever value of the last one the earlier command created), those will get deleted, which automatically restarts the cycle and reads the next record.
Second, if the line number is 14, then it won't delete, and so will progress to the s/delay/ens_delay/ which updates your value.
For all lines that don't match /delay/, sed just prints them as-is.
I want to extract the first column of the last line of a text file. Instead of output the content of interest in another file and read it in again, can I just use some command to read it into a variable directly?
For exampole, if my file is like this:
...
123 456 789(this is the last line)
What I want is to read 123 into a variable in my shell script. How can I do that?
One approach is to extract the line you want, read its columns into an array, and emit the array element you want.
For the last line:
#!/bin/bash
# ^^^^- not /bin/sh, to enable arrays and process substitution
read -r -a columns < <(tail -n 1 "$filename") # put last line's columns into an array
echo "${columns[0]}" # emit the first column
Alternately, awk is an appropriate tool for the job:
line=2
column=1
var=$(awk -v line="$line" -v col="$column" 'NR == line { print $col }' <"$filename")
echo "Extracted the value: $var"
That said, if you're looking for a line close to the start of a file, it's often faster (in a runtime-performance sense) and easier to stick to shell builtins. For instance, to take the third column of the second line of a file:
{
read -r _ # throw away first line
read -r _ _ value _ # extract third value of second line
} <"$filename"
This works by using _s as placeholders for values you don't want to read.
I guess with "first column", you mean "first word", do you?
If it is guaranteed, that the last line doesn't start with a space, you can do
tail -n 1 YOUR_FILE | cut -d ' ' -f 1
You could also use sed:
$> var=$(sed -nr '$s/(^[^ ]*).*/\1/p' "file.txt")
The -nr tells sed to not output data by default (-n) and use extended regular expressions (-r to avoid needing to escape the paranthesis otherwise you have to write \( \))). The $ is an address that specifies the last line. The regular expression anchors the beginning of the line with the first ^, then matches everything that is not a space [^ ]* and puts that the result into a capture group ( ) and then gets rid of the rest of the line .* by replacing the line with the capture group \1, then print p to print the line.
I have a text file in which I want to find all of ID:= "abc123" when it finds that I want it to take that value of abc123 and create a new line and have a set string, newId:= "abc123 How can I do this within terminal?
I'd like to use bash, below are some examples, find the string '"ID": ", copy the value (abc123) and make a new line with this data.
"ID": "abc123"
"newID": "abc123"
You can do this:
sed -e 's/^"ID": "\(.*\)"/&\
"newID": "\1"/' myfile.txt
First, I'll try to explain the regular expression that searches for matches:
^ Matches the start of the line
"ID": " Matches that exact string
\(.*\) Matches a sequence of zero or more (*) of any character (.). Placing this expression between backslashed parenthesis creates a "capture", which allows us to store the resulting part of the match into an auxiliary variable \1.
" Matches the double-quote character
When it finds a match, it replaces it with:
& the match itself. This operator is an auxiliary variable that represents what was matched.
\<new-line> the backslash followed by an actual new line character escapes a new line, ie. it allows us to print a new line character into the replacement
"newId": " prints that exact string
\1 prints the contents of our capture, so it prints the ID we found
" prints a double quote character.
Hope this helps =)
Try doing this :
sed -r 's#^"ID": "([a-Z0-9]+)"#"newID": "\1"#' file.txt
sed : the executable
-r : extented mode (no need to backslash parenthesis)
s : we perform a substitution, skeleton is s#origin#replacement# (the separator can be anything)
^ : means start of line in regex
( ) : parenthesis is a capture
"newID": is the start of the new string
\1 : is the end of the substituted string (the captured string)
Considering your question is very vague I made some assumptions which will become apparent in my implementation.
INPUT FILE -- call it t
ID="one"
dkkd
ID="two"
ffkjf
ID="three"
ldl
Command ran on input file
for line in `cat t`; do newID=`echo $line | grep ID | cut -d= -f2`; if [[ "$newID" != "" ]]; then echo $line >> t2; echo newID=$newID >> t2; else echo $line >> t2; fi; done
OUTPUT FILE -- Name is t2 (apparent from the command)
ID="one"
newID="one"
dkkd
ID="two"
newID="two"
ffkjf
ID="three"
newID="three"
ldl
Basically this command goes line by line in the file (in this case called t) looks for an ID line. If it finds one it gets its value, prints the original line with the ID and then prints another one with a newID following right after. If the line in question does not have and ID then it just prints the line it self.
Things to note:
If you have any other line in the file that contains "ID" in it but is not the normal ID that you requested, this will not work.
I have a string which i need to insert at a specific position in a file :
The file contains multiple semicolons(;) i need to insert the string just before the last ";"
Is this possible with SED ?
Please do post the explanation with the command as I am new to shell scripting
before :
adad;sfs;sdfsf;fsdfs
string = jjjjj
after
adad;sfs;sdfsf jjjjj;fsdfs
Thanks in advance
This might work for you:
echo 'adad;sfs;sdfsf;fsdfs'| sed 's/\(.*\);/\1 jjjjj;/'
adad;sfs;sdfsf jjjjj;fsdfs
The \(.*\) is greedy and swallows the whole line, the ; makes the regexp backtrack to the last ;. The \(.*\) make s a back reference \1. Put all together in the RHS of the s command means insert jjjjj before the last ;.
sed 's/\([^;]*\)\(;[^;]*;$\)/\1jjjjj\2/' filename
(substitute jjjjj with what you need to insert).
Example:
$ echo 'adad;sfs;sdfsf;fsdfs;' | sed 's/\([^;]*\)\(;[^;]*;$\)/\1jjjjj\2/'
adad;sfs;sdfsfjjjjj;fsdfs;
Explanation:
sed finds the following pattern: \([^;]*\)\(;[^;]*;$\). Escaped round brackets (\(, \)) form numbered groups so we can refer to them later as \1 and \2.
[^;]* is "everything but ;, repeated any number of times.
$ means end of the line.
Then it changes it to \1jjjjj\2.
\1 and \2 are groups matched in first and second round brackets.
For now, the shorter solution using sed : =)
sed -r 's#;([^;]+);$#; jjjjj;\1#' <<< 'adad;sfs;sdfsf;fsdfs;'
-r option stands for extented Regexp
# is the delimiter, the known / separator can be substituted to any other character
we match what's finishing by anything that's not a ; with the ; final one, $ mean end of the line
the last part from my explanation is captured with ()
finally, we substitute the matching part by adding "; jjjj" ans concatenate it with the captured part
Edit: POSIX version (more portable) :
echo 'adad;sfs;sdfsf;fsdfs;' | sed 's#;\([^;]\+\);$#; jjjjj;\1#'
echo 'adad;sfs;sdfsf;fsdfs;' | sed -r 's/(.*);(.*);/\1 jjjj;\2;/'
You don't need the negation of ; because sed is by default greedy, and will pick as much characters as it can.
sed -e 's/\(;[^;]*\)$/ jjjj\1/'
Inserts jjjj before the part where a semicolon is followed by any number of non-semicolons ([^;]*) at the end of the line $. \1 is called a backreference and contains the characters matched between \( and \).
UPDATE: Since the sample input has no longer a ";" at the end.
Something like this may work for you:
echo "adad;sfs;sdfsf;fsdfs"| awk 'BEGIN{FS=OFS=";"} {$(NF-1)=$(NF-1) " jjjjj"; print}'
OUTPUT:
adad;sfs;sdfsf jjjjj;fsdfs
Explanation: awk starts with setting FS (field separator) and OFS (output field separator) as semi colon ;. NF in awk stands for number of fields. $(NF-1) thus means last-1 field. In this awk command {$(NF-1)=$(NF-1) " jjjjj" I am just appending jjjjj to last-1 field.