How to insert string to the beginning of the last line?
I want to add a time stamp to a text file which contains multiple lines
var1 = `date`
LINE1
LINE2
LINE3
...
(INSERT var1 here) LASTLINE
sed 's/^/test line /g' textfile inserts characters to the beginning of every line but how can I specifically modify the last line only?
Thanks
Going forward:
sed '$s/^/sample text /' textfile works, but only when inserting regular strings. If I try
var1 = "sample text"
and use substition, here are the problems I encounter
using single quotes in sed does not expand variables, so sed '$s/^/$var1/' textfile will insert the string $var1 into the beginning of the last line.
To enable variable substitution I tried using double quotes. It works when I specify the exact line number. something like:
sed "5s/^/$var1/" textfile
But when I try sed "$s/^/$var1" text file, it returns an error:
sed: -e expression #1, char 5: extra characters after command
Can someone help me please?
Like this:
sed '$s/^/test line /' textfile
$ indicates last line. Similarly, you can insert into a any specific line by putting the line number in place of $
But when I try sed "$s/^/$var1" text file, it returns an error:
It returns an error because the shell attempts to expand $s since you've used double quotes. You need to escape the $ in $s.
sed "\$s/^/$var1/" filename
sedshould be the best tool, but awk can do this too:
awk '{a[++t]=$0} END {for (i=1;i<t;i++) print a[i];print v$0}' v="$var1" file
It will insert value of var1 in front of last line
Another variation
awk 'NR==f {$0=v$0}1' v="$var1" f=$(wc -l file)
PS you do not need to specify file after awk, not sure why. If you do so, it reads it double.
This command would work for you:
sed -i "5s/^/$var1 /" text file
Related
Newbie to unix/shell/bash. I have a file name CellSite whose 6th line is as below:
btsName = "RV74XC038",
I want to extract the string from 6th line that is between double quotes (i.e.RV74XC038) and save it to a variable. Please note that the 6th line starts with 4 blank spaces. And this string would vary from file. So I am looking for a solution that would extract a string from 6th line between the double quotes.
I tried below. But does not work.
str2 = sed '6{ s/^btsName = \([^ ]*\) *$/\1/;q } ;d' CellSite;
Any help is much appreciated. TIA.
sed is a stream editor.
For just parsing files, you want to look into awk. Something like this:
awk -F \" '/btsName/ { print $2 }' CellSite
Where:
-F defines a "field separator", in your case the quotation marks "
the entire script consists of:
/btsName/ act only on lines that contain the regex "btsName"
from that line print out the second field; the first field will be everything before the first quotation marks, second field will be everything from the first quotes to the second quotes, third field will be everything after the second quotes
parse through the file named "CellSite"
There are possibly better alternatives, but you would have to show the rest of your file.
Using sed
$ str2=$(sed '6s/[^"]*"\([^"]*\).*/\1/' CellSite)
$ echo "$str2"
RV74XC038
You can use the following awk solution:
btsName=$(awk -F\" 'NR==6{print $2; exit}' CellSite)
Basically, get to the sixth line (NR==6), print the second field value (" is used to split records (lines) into fields) and then exit.
See the online demo:
#!/bin/bash
CellSite='Line 1
Line 2
Line 3
btsName = "NO74NO038",
Line 5
btsName = "RV74XC038","
Line 7
btsName = "no11no000",
'
btsName=$(awk -F\" 'NR==6{print $2; exit}' <<< "$CellSite")
echo "$btsName" # => RV74XC038
This might work for you (GNU sed):
var=$(sed -En '6s/.*"(.*)".*/\1/p;6q' file)
Simplify regexs and turn off implicit printing.
Focus on the 6th line only and print the value between double quotes, then quit.
Bash interpolates the sed invocation by means of the $(...) and the value extracted defines the variable var.
Have to write a script which updates the file in this way.
raw file:
<?blah blah blah?>
<pen>
<?pineapple?>
<apple>
<pen>
Final file:
<?blah blah blah?><pen>
<?pineapple?><apple><pen>
Where ever in the file if the new line charter is not followed by
<?
We have to remove the newline in order to append it at the end of previous line.
Also it will be really helpful if you explain how your sed works.
Perl solution:
perl -pe 'chomp; substr $_, 0, 0, "\n" if $. > 1 && /^<\?/'
-p reads the input line by line, printing each line after changes
chomp removes the final newline
substr with 4 arguments modifies the input string, here it prepends newline if it's not the first line ($. is the input line number) and the line starts with <?.
Sed solution:
sed ':a;N;$!ba;s/\n\(<[^?]\)/\1/g' file > newfile
The basic idea is to replace every
\n followed by < not followed by ?
with what you matched except the \n.
When you are happy with a solution that puts every <? at the start of a line, you can combine tr with sed.
tr -d '\n' < inputfile| sed 's/<?/\n&/g;$s/$/\n/'
Explanation:
I use tr ... < inputfile and not cat inputfile | tr ... avoiding an additional catcall.
The sed command has 2 parts.
In s/<?/\n&/g it will insert a newline and with & it will insert the matched string (in this case always <?, so it will only save one character).
With $s/$/\n/ a newline is appended at the end of the last line.
EDIT: When you only want newlines before <? when you had them already,
you can use awk:
awk '$1 ~ /^<\?/ {print} {printf("%s",$0)} END {print}'
Explanation:
Consider the newline as the start of the line, not the end. Then your question transposes into "write a newline when the line starts with <?. You must escape the ? and use ^ for the start of the line.
awk '$1 ~ /^<\?/ {print}'
Next print the line you read without a newline character.
And you want a newline at the end.
I use the following sed command in order to replace string in CSV line
( the condition to replace the string is to match the number in the beginning of the CSV file )
SERIAL_NUM=1
sed "/$SERIAL_NUM/ s//OK/g" file.csv
the problem is that I want to match only the number that start in the beginning of the line ,
but sed match other lines that have this number
example:
in this example I want to replace the word - STATUS to OK but only in line that start with 1 ( before the "," separator )
so I do this
SERIAL_NUM=1
more file.csv
1,14556,43634,266,242,def,45,STATUS
2,4345,1,43,57,24,657,SD,STATUS
3,1,WQ,435,676,90,3,44f,STATUS
sed -i "/$SERIAL_NUM/ s/STATUS/OK/g" file.csv
more file.csv
1,14556,43634,266,242,def,45,OK
2,4345,1,43,57,24,657,SD,OK
3,1,WQ,435,676,90,3,44f,OK
but sed also replace the STATUS to OK also in line 2 and line 3 ( because those lines have the number 1 )
please advice how to change the sed syntax in order to match only the number that start the line before the "," separator
remark - solution can be also with perl line liner or awk ,
You can use anchor ^ to make sure $SERIAL_NUM only matches at start and use , after that to make sure there is a comma followed by this number:
sed "/^$SERIAL_NUM,/s/STATUS/OK/g" file.csv
Since this answer was ranked fifth in the Stackoverflow perl report but had no perl content, I thought it would be useful to add the following - instead of removing the perl tag :-)
#!/usr/bin/env perl
use strict;
use warnings;
while(<DATA>){
s/STATUS/OK/g if /^1\,/;
print ;
}
__DATA__
1,14556,43634,266,242,def,45,STATUS
2,4345,1,43,57,24,657,SD,STATUS
3,1,WQ,435,676,90,3,44f,STATUS
or as one line:
perl -ne 's/STATUS/OK/g if /^1\,/;' file.csv
My Stress.k file is as follows
180.4430
*INCLUDE
$# filename
*STRESS_INITIALIZATION
*END
I want it to be like
180.4430
*INCLUDE
$# filename
*STRESS_INITIALIZATION
*/home/hassan/534.k
*END
for that I used sed as follows
a="$(cat flow.k)"
sed -i -e '/*END/i \*/home/hassan/$a.k ' Stress.k
where flow.k has only a single number like 534.k or something . Here sed put the line before END but it doesn't take the value of a , instead it puts the same alphabet and it doesn't understand $a.k.
Please also tell me how to delete the second last line or the line with a string hassan for example so that I can delete it first and the for the next step I use it to enter my required line.
if possible please also suggest the alternatives.
best regards
bash variables are only replaced when in double quotes, e.g.
sed -i -e "/*END/i \*/home/hassan/$a.k " Stress.k
Use double quotes to allow the variable to be expanded.
sed -i -e "/*END/i \*/home/hassan/$a.k " Stress.k
To replace the string, do it as you read in the file:
a=$(sed 's/534/100/' flow.k)
To delete a line:
sed '/hassan/d' inputfile
To read a file into the stream after the current line:
sed '/foo/r filename' inputfile
UPDATED:
Using sed, how can I insert (NOT SUBSTITUTE) a new line on only the first match of keyword for each file.
Currently I have the following but this inserts for every line containing Matched Keyword and I want it to only insert the New Inserted Line for only the first match found in the file:
sed -ie '/Matched Keyword/ i\New Inserted Line' *.*
For example:
Myfile.txt:
Line 1
Line 2
Line 3
This line contains the Matched Keyword and other stuff
Line 4
This line contains the Matched Keyword and other stuff
Line 6
changed to:
Line 1
Line 2
Line 3
New Inserted Line
This line contains the Matched Keyword and other stuff
Line 4
This line contains the Matched Keyword and other stuff
Line 6
You can sort of do this in GNU sed:
sed '0,/Matched Keyword/s//New Inserted Line\n&/'
But it's not portable. Since portability is good, here it is in awk:
awk '/Matched Keyword/ && !x {print "Text line to insert"; x=1} 1' inputFile
Or, if you want to pass a variable to print:
awk -v "var=$var" '/Matched Keyword/ && !x {print var; x=1} 1' inputFile
These both insert the text line before the first occurrence of the keyword, on a line by itself, per your example.
Remember that with both sed and awk, the matched keyword is a regular expression, not just a keyword.
UPDATE:
Since this question is also tagged bash, here's a simple solution that is pure bash and doesn't required sed:
#!/bin/bash
n=0
while read line; do
if [[ "$line" =~ 'Matched Keyword' && $n = 0 ]]; then
echo "New Inserted Line"
n=1
fi
echo "$line"
done
As it stands, this as a pipe. You can easily wrap it in something that acts on files instead.
If you want one with sed*:
sed '0,/Matched Keyword/s//Matched Keyword\nNew Inserted Line/' myfile.txt
*only works with GNU sed
This might work for you:
sed -i -e '/Matched Keyword/{i\New Inserted Line' -e ':a;n;ba}' file
You're nearly there! Just create a loop to read from the Matched Keyword to the end of the file.
After inserting a line, the remainder of the file can be printed out by:
Introducing a loop place holder :a (here a is an arbitrary name).
Print the current line and fetch the next into the pattern space with the ncommand.
Redirect control back using the ba command which is essentially a goto to the a place holder. The end-of-file condition is naturally taken care of by the n command which terminates any further sed commands if it tries to read passed the end-of-file.
With a little help from bash, a true one liner can be achieved:
sed $'/Matched Keyword/{iNew Inserted Line\n:a;n;ba}' file
Alternative:
sed 'x;/./{x;b};x;/Matched Keyword/h;//iNew Inserted Line' file
This uses the Matched Keyword as a flag in the hold space and once it has been set any processing is curtailed by bailing out immediately.
If you want to append a line after first match only, use AWK instead of SED as below
awk '{print} /Matched Keyword/ && !n {print "New Inserted Line"; n++}' myfile.txt
Output:
Line 1
Line 2
Line 3
This line contains the Matched Keyword and other stuff
New Inserted Line
Line 4
This line contains the Matched Keyword and other stuff
Line 6