I need to extract the string between quotation marks in a file.
For example: my file is called test.txt and it has the following content:
"Hello_World"
I am reading it as follows from bash:
string="$(head -1 test.txt)"
echo $string
This prints "Hello_World", but I need Hello_World.
Any help will be appreciated. Thanks.
You can do this in pure bash without having to spawn any external programs:
read -r line < test.txt ; line=${line#\"} ; line=${line%\"} ; echo $line
The read actually reads in the entire line, and the two assignments actually strip off any single quote at the start or end of the line.
I assumed you didn't want to strip out any quotes within the string itself so I've limited it to one at either end.
It also allows you to successfully read lines without a leading quote, trailing quote, or both.
You can use tr:
echo "$string " | tr -d '"'
From man tr:
DESCRIPTION
The tr utility copies the standard input to the standard output with substitution or deletion of selected characters.
The following options are available:
-C Complement the set of characters in string1, that is ``-C ab'' includes every character except for `a' and `b'.
-c Same as -C but complement the set of values in string1.
-d Delete characters in string1 from the input.
You can simply use sed to read the first line and also filter out ", try following command,
sed -n '1 s/"//gp' test.txt
Brief explanation,
-n: suppress automatic print
1: Match only the first line
s/"//gp: filter out ", and then print the line
Related
Would like to insert line number at specific location in file
e.g.
apple
ball
should be
(1) apple
(2) ball
Using command
sed '/./=' <FileName>| sed '/./N; s/\n/ /'
It generates
1 Apple
2 Ball
1st solution: This should be an easy task for awk.
awk '{print "("FNR") "$0}' Input_file
2nd solution: With pure sed as per OP's attempt try:
sed '=' Input_file | sed 'N; s/^/(/;s/\n/) /'
Easy to do with perl instead:
perl -ne 'print "($.) $_"' foo.txt
If you want to modify the file in-place instead of just printing out the numbered lines on standard output:
perl -ni -e 'print "($.) $_"' foo.txt
Many ways are there to insert line numbers in a file
some of them are :-
1.Using cat command
cat -n file.txt > newfile.txt
2.Using nl command
nl -b a file.txt
Awk and perl both are very usefull and powerfull. But if, like me, you are reluctant to learn yet another programming language, you can complete this task with the bash commands you probably know already.
With bash you can
increment a sequence number n: $((++n))
read all lines from a file foo into a variable l: while read -r l;do ...;done <foo, where the option -r serves to treat backslashes as just characters.
print formatted output to a line: printf "plain text %i %s\n" number string
Now suppose you want to enclose your sequence number in parentheses, and format them to 8 digits with leading zeroes, then you combine all this to get:
n=0;while read -r l;do printf "(%08i) %s\n" $((++n)) "$l";done <foo >numberedfoo
Note that you do not need to initialize the variable n to use it as a sequence number further on. But if you experiment with this command a few times without reinitializing n, your lines will be numbered from where your previous try stopped incrementing.
Finally, if you don't like the C-like formatting syntax of printf, just use plain echo, and leave the formatting to bash variable expansion. Here is how to format a number like in the command above (do type a space before the -, and a ; before the echo) :
nformat="0000000$n"; echo "(${nformat: -8}) ...";
I am new to shell scripting. I am using ksh.
I have this particular line in my script which I use to append text in a variable q to the end of a particular line given by the variable a
containing the line number .
sed -i ''$a's#$#'"$q"'#' test.txt
Now the variable q can contain a large amount of text, with all sorts of special characters, such as !##$%^&*()_+:"<>.,/;'[]= etc etc, no exceptions. For now, I use a couple of sed commands in my script to remove any ' and " in this text (sed "s/'/ /g" | sed 's/"/ /g'), but still when I execute the above command I get the following error
sed: -e expression #1, char 168: unterminated `s' command
Any sed, awk, perl, suggestions are very much appreciated
The difficulty here is to quote (escape) the substitution separator characters # in the sed command:
sed -i ''$a's#$#'"$q"'#' test.txt
For example, if q contains # it will not work. The # will terminate the replacement pattern prematurely. Example: q='a#b', a=2, and the command expands to
sed -i 2s#$#a#b# test.txt
which will not append a#b to the end of line 2, but rather a#.
This can be solved by escaping the # characters in q:
sed -i 2s#$#a\#b# test.txt
However, this escaping could be cumbersome to do in shell.
Another approach is to use another level of indirection. Here is an example of using a Perl one-liner. First q is passed to the script in quoted form. Then, within the script the variable assigned to a new internal variable $q. Using this approach there is no need to escape the substitution separator characters:
perl -pi -E 'BEGIN {$q = shift; $a = shift} s/$/$q/ if $. == $a' "$q" "$a" test.txt
Do not bother trying to sanitize the string. Just put it in a file, and use sed's r command to read it in:
echo "$q" > tmpfile
sed -i -e ${a}rtmpfile test.txt
Ah, but that creates an extra newline that you don't want. You can remove it with:
sed -e ${a}rtmpfile test.txt | awk 'NR=='$a'{printf $0; next}1' > output
Another approach is to use the patch utility if present in your system.
patch test.txt <<-EOF
${a}c
$(sed "${a}q;d" test.txt)$q
.
EOF
${a}c will be replaced with the line number followed by c which means the operation is a change in line ${a}.
The second line is the replacement of the change. This is the concatenated value of the original text and the added text.
The sole . means execute the commands.
This question already has answers here:
Using different delimiters in sed commands and range addresses
(3 answers)
Closed 6 years ago.
I'm trying to replace text in a file with the output of another command. Unfortunately, the outputted text contains characters bash expands. For example, I'm running the following script to change the file (somestring references output that would break the sed command):
#!/bin/bash
somestring='$6$sPnfj/lnXwZVrec7$fCnL9uy1oWIMZduInKTHBAxhsQxGCsBpm2XfVFFqDPHKidrd93yfjbYvKgYexXHVcvkKdu9lbfy16Ek5GvKy/1'
sed '0,/^title/s/^title*/'"$somestring"'\n&/' $HOME/example.txt
sed fails with this error:
sed: -e expression #1, char 30: unknown option to `s'
I think bash is substuting the contents of $somestring when building the sed command, but is then trying to expand the resulting text. I can't put the entire sed script in single quotes, I need bash to expand it the first time, just not the second. Any suggestions? Thanks
here the forward slash / is the problem. If it's the only issue you can set sed to use a different delimiter.
for example
$ somestring="abc/def"; echo xxx | sed 's/xxx/'"$somestring"'/'
sed: -e expression #1, char 11: unknown option to `s'
$ somestring="abc/def"; echo xxx | sed 's_xxx_'"$somestring"'_'
abc/def
you also need to worry about & and \ chars and escape them if can appear in the replacement text.
If you can't control the the replacement string, either you have to sanitize with another sed script or, alternatively use r command to read it from a file. For example,
$ seq 5 | sed -e '/3/{r replace' -e 'd}'
1
2
3slashes///1ampersand&and2backslashes\\end
4
5
where
$ cat replace
3slashes///1ampersand&and2backslashes\\end
You have several errors here:
the string somestring has characters that are significative for sed command (the most important being '/' that you are using as a delimiter) You can escape it, by substituting it with a previous
somestring=$(echo "$somestring" | sed -e 's/\//\\\//g')
that will convert your / chars to \/ sequences.
you are using sed '0,/^title/s/^title*/'"$somestring"'\n&/' $HOME/example.txt which is looking to substitute the string titl followed by any number of e characters by that $somestring value, followed by a new line and the original one. Unfortunately, sed(1) doesn't allow you to use newline characters in the pattern substitution side of the s command, but you can afford the result by using the i command with a text consisting of you pattern (preceding any new line by a \ to interpret it as literal):
Finally the script leads to:
#!/bin/bash
somestring='$6$sPnfj/lnXwZVrec7$fCnL9uy1oWIMZduInKTHBAxhsQxGCsBpm2XfVFFqDPHKidrd93yfjbYvKgYexXHVcvkKdu9lbfy16Ek5GvKy/1'
somestring=$(echo "$somestring" | sed -e 's/\//\\\//g')
sed '/^title/i\
'"$somestring\\
" $HOME/example.txt
If your shell is Bash, you can use parameter substitution to replace the problematic /:
somestring="{somestring//\//\\/}"
That looks scary, but is easier to understand if you look at the version that replaces x with __:
somestring="${somestring//x/__}"
It might be easier to use (say) underscore as the delimiter for your sed s command, and then the substitution above would be
somestring="${somestring//_/\\_}"
If you already have backslashes, you'll need to first replace those:
somestring="${somestring//\\/\\\\}"
somestring="{somestring//\//\\/}"
If there were other characters that needed escaping (e.g. on the search side of s///), then you could extend the above appropriately.
This URL provides the cleanest answer:
Command to escape a string in bash
printf "%q" "$someVariable"
will escape any characters you need escaped for you.
I was trying to do these few operations/commands on a single line and assign it to a variable. I have it working about 90% of the way except for one part of it.
I was unaware you could do this, but I read that you can nest $(..) inside other $(..).... So I was trying to do that to get this working, but can't seem to get it the rest of the way.
So basically, what I want to do is:
1. Take the output of a file and assign it to a variable
2. Then pre-pend some text to the start of that output
3. Then append some text to the end of the output
4. And finally remove newlines and replace them with "\n" character...
I can do this just fine in multiple steps but I would like to try and get this working this way.
So far I have tried the following:
My 1st attempt, before reading about nested $(..):
MY_VAR=$(echo -n "<pre style=\"display:inline;\">"; cat input.txt | sed ':a;N;$!ba;s/\n/\\n/g'; echo -n "</pre>")
This one worked 99% of the way except there was a newline being added between the cat command's output and the last echo command. I'm guessing this is from the cat command since sed removed all newlines except for that one, maybe...?
Other tries:
MY_VAR=$( $(echo -n "<pre style=\"display:inline;\">"; cat input.txt; echo -n "</pre>") | sed ':a;N;$!ba;s/\n/\\n/g')
MY_VAR="$( echo $(echo -n "<pre style=\"display:inline;\">"; cat input.txt; echo "</pre>") | sed ':a;N;$!ba;s/\n/\\n/g' )"
MY_VAR="$( echo "$(echo -n "<pre style=\"display:inline;\">"; cat input.txt; echo "</pre>")" | sed ':a;N;$!ba;s/\n/\\n/g' )"
*Most these others were tried with and without the extra double-quotes surrounding the different $(..) parts...
I had a few other attempts, but they didn't have any luck either... On a few of the other attempts above, it seemed to work except sed was NOT inserting the replacement part of it. The output was correct for the most part, except instead of seeing "\n" between lines it just showed each of the lines smashed together into one line without anything to separate them...
I'm thinking there is something small I am missing here if anyone has any idea..?
*P.S. Does Bash have a name for the $(..) structure? It's hard trying to Google for that since it doesn't really search symbols...
You have no need to nest command substitutions here.
your_var='<pre style="display:inline;">'"$(<input.txt)"'</pre>'
your_var=${your_var//$'\n'/'\n'}
"$(<input.txt)" expands to the contents of input.txt, but without any trailing newline. (Command substitution always strips trailing newlines; printf '%s' "$(cat ...)" has the same effect, albeit less efficiently as it requires a subshell, whereas cat ... alone does not).
${foo//bar/baz} expands to the contents of the shell variable named foo, with all instances of bar replaced with baz.
$'\n' is bash syntax for a literal newline.
'\n' is bash syntax for a two-character string, beginning with a backslash.
Thus, tying all this together, it first generates a single string with the prefix, the contents of the file, and the suffix; then replaces literal newlines inside that combined string with '\n' two-character sequences.
Granted, this is multiple lines as implemented above -- but it's also much faster and more efficient than anything involving a command substitution.
However, if you really want a single, nested command substitution, you can do that:
your_var=$(printf '%s' '<pre style="display:inline;">' \
"$(sed '$ ! s/$/\\n/g' <input.txt | tr -d '\n')" \
'</pre>')
The printf %s combines its arguments without any delimiter between them
The sed operation adds a literal \n to the end of each line except the last
The tr -d '\n' operation removes literal newlines from the file
However, even this approach could be done more efficiently without the nesting:
printf -v your_var '%s' '<pre style="display:inline;">' \
"$(sed '$ ! s/$/\\n/g' <input.txt | tr -d '\n')" \
'</pre>')
...which has the printf assign its results directly to your_var, without any outer command substitution required (and thus saving the expense of the outer subshell).
Say I have some arbitrary multi-line text file:
sometext
moretext
lastline
How can I remove only the last character (the e, not the newline or null) of the file without making the text file invalid?
A simpler approach (outputs to stdout, doesn't update the input file):
sed '$ s/.$//' somefile
$ is a Sed address that matches the last input line only, thus causing the following function call (s/.$//) to be executed on the last line only.
s/.$// replaces the last character on the (in this case last) line with an empty string; i.e., effectively removes the last char. (before the newline) on the line.
. matches any character on the line, and following it with $ anchors the match to the end of the line; note how the use of $ in this regular expression is conceptually related, but technically distinct from the previous use of $ as a Sed address.
Example with stdin input (assumes Bash, Ksh, or Zsh):
$ sed '$ s/.$//' <<< $'line one\nline two'
line one
line tw
To update the input file too (do not use if the input file is a symlink):
sed -i '$ s/.$//' somefile
Note:
On macOS, you'd have to use -i '' instead of just -i; for an overview of the pitfalls associated with -i, see the bottom half of this answer.
If you need to process very large input files and/or performance / disk usage are a concern and you're using GNU utilities (Linux), see ImHere's helpful answer.
truncate
truncate -s-1 file
Removes one (-1) character from the end of the same file. Exactly as a >> will append to the same file.
The problem with this approach is that it doesn't retain a trailing newline if it existed.
The solution is:
if [ -n "$(tail -c1 file)" ] # if the file has not a trailing new line.
then
truncate -s-1 file # remove one char as the question request.
else
truncate -s-2 file # remove the last two characters
echo "" >> file # add the trailing new line back
fi
This works because tail takes the last byte (not char).
It takes almost no time even with big files.
Why not sed
The problem with a sed solution like sed '$ s/.$//' file is that it reads the whole file first (taking a long time with large files), then you need a temporary file (of the same size as the original):
sed '$ s/.$//' file > tempfile
rm file; mv tempfile file
And then move the tempfile to replace the file.
Here's another using ex, which I find not as cryptic as the sed solution:
printf '%s\n' '$' 's/.$//' wq | ex somefile
The $ goes to the last line, the s deletes the last character, and wq is the well known (to vi users) write+quit.
After a whole bunch of playing around with different strategies (and avoiding sed -i or perl), the best way i found to do this was with:
sed '$! { P; D; }; s/.$//' somefile
If the goal is to remove the last character in the last line, this awk should do:
awk '{a[NR]=$0} END {for (i=1;i<NR;i++) print a[i];sub(/.$/,"",a[NR]);print a[NR]}' file
sometext
moretext
lastlin
It store all data into an array, then print it out and change last line.
Just a remark: sed will temporarily remove the file.
So if you are tailing the file, you'll get a "No such file or directory" warning until you reissue the tail command.
EDITED ANSWER
I created a script and put your text inside on my Desktop. this test file is saved as "old_file.txt"
sometext
moretext
lastline
Afterwards I wrote a small script to take the old file and eliminate the last character in the last line
#!/bin/bash
no_of_new_line_characters=`wc '/root/Desktop/old_file.txt'|cut -d ' ' -f2`
let "no_of_lines=no_of_new_line_characters+1"
sed -n 1,"$no_of_new_line_characters"p '/root/Desktop/old_file.txt' > '/root/Desktop/my_new_file'
sed -n "$no_of_lines","$no_of_lines"p '/root/Desktop/old_file.txt'|sed 's/.$//g' >> '/root/Desktop/my_new_file'
opening the new_file I created, showed the output as follows:
sometext
moretext
lastlin
I apologize for my previous answer (wasn't reading carefully)
sed 's/.$//' filename | tee newFilename
This should do your job.
A couple perl solutions, for comparison/reference:
(echo 1a; echo 2b) | perl -e '$_=join("",<>); s/.$//; print'
(echo 1a; echo 2b) | perl -e 'while(<>){ if(eof) {s/.$//}; print }'
I find the first read-whole-file-into-memory approach can be generally quite useful (less so for this particular problem). You can now do regex's which span multiple lines, for example to combine every 3 lines of a certain format into 1 summary line.
For this problem, truncate would be faster and the sed version is shorter to type. Note that truncate requires a file to operate on, not a stream. Normally I find sed to lack the power of perl and I much prefer the extended-regex / perl-regex syntax. But this problem has a nice sed solution.