my goal is to sed the 100th line and convert it to a string, then separate the data of the sentence to word
#!/bin/bash
fid=log.txt;
sentence=`expr sed -n '100p' ${fid}`;
for word in $sentence
do
echo $word
done
but apparently this has failed.
expr: syntax error
would somebody please let me know what have I done wrong? previously for number it worked.
The expr does not seem to serve a useful purpose here, and if it did, a sed command would certainly not be a valid or useful thing to pass to it, under most circumstances. You should probably just take it out.
However, the following loop is also problematic. Unquoted variables in shell script are very frequently an error. In this case, you can't quote the thing you pass to the for loop (that would cause the loop to only run once, with the loop variable set to the quoted string) but you also cannot prevent the shell from performing wildcard expansion on the unquoted string. So if the string happened to contain *, the shell will expand that to a list of files in the current directory, for example.
Fortunately, this can all be done in an only slightly more complicated sed script.
sed '100!d;s/[ \t]\+/\n/g;q' "$fid"
That is, if the line number is not 100, delete this line and start over with the next line. Otherwise, we are at line 100; replace runs of whitespace with newlines, (print) and quit.
(The backslash escape codes \t and \n are not universally portable; and \+ for repetition is also an optional extension. I believe there are also sed variants which dislike semicolon as a command separator. Consult your sed manual page, experiment, and if everything else fails, maybe switch to Awk or Perl. Just in case, here is a version which works even on Mac OSX:
sed '100!d
s/[ ][ ]*/\
/g;q' log.txt
The stuff inside the square brackets are a space and a literal tab; in Bash, with default keybindings, type ctrl-V, tab to produce a literal tab.)
Incidentally, this also gets rid of the variable capture antipattern. There are good reasons to capture output to a variable, but if it can be avoided, you often end up with a simpler, more robust and efficient, as well as more idiomatic and elegant script. (I see no reason to put the log file name in a variable, either, in this isolated case; but in a larger script, it might make sense.)
I don't think you need expr command in this case.
expr is used to do calculations. Something like:
expr 1 + 1
Just this one is fine:
sentence=`sed -n '100p' ${fid}`;
#!/bin/bash
fid=log.txt;
sentence=$(sed -n '100p' ${fid});
for word in $sentence
do
echo $word
done
put a dollar sign and parenthesis solve the problem
Related
I need to generate filename from three parts, two strings, and one variable.
for f in `cat files.csv`; do echo fastq/$f\_1.fastq.gze; done
files.csv has the following lines:
Sample_11
Sample_12
I need to generate the following:
fastq/Sample_11_1.fastq.gze
fastq/Sample_12_1.fastq.gze
My problem is that I got the below files:
_1.fastq.gze_11
_1.fastq.gze_12
the string after the variable deletes the string before it.
I appreciate any help
Regards
By the way your idiom: for f in cat files.csv should be avoid. Refer: Dangerous Backticks
while read f
do
echo "fastq/${f}/_1.fastq.gze"
done < files.csv
You can make it a one-liner with xargs and printf.
xargs printf 'fastq/%s_1.fastq.gze\n' <files.csv
The function of printf is to apply the first argument (the format string) to each argument in turn.
xargs says to run this command on as many files as it can fit onto the command line (splitting it up into multiple invocations if the input file is too large to fit all the arguments onto a single command line, subject to the ARG_MAX constant in your kernel).
Your best bet, generally, is to wrap the variable name in braces. So, in this case:
echo fastq/${f}_1.fastq.gz
See this answer for some details about the general concept, as well.
Edit: An additional thought looking at the now-provided output makes me think that this isn't a coding problem at all, but rather a conflict between line-endings and the terminal/console program.
Specifically, if the CSV file ends its lines with just a carriage return (ASCII/Unicode 13), the end of Sample_11 might "rewind" the line to the start and overwrite.
In that case, based loosely on this article, I'd recommend replacing cat (if you understandably don't want to re-architect the actual script with something like while) with something that will strip the carriage returns, such as:
for f in $(tr -cd '\011\012\040-\176' < temp.csv)
do
echo fastq/${f}_1.fastq.gze
done
As the cited article explains, Octal 11 is a tab, 12 a line feed, and 40-176 are typeable characters (Unicode will require more thinking). If there aren't any line feeds in the file, for some reason, you probably want to replace that with tr '\015' '\012', which will convert the carriage returns to line feeds.
Of course, at that point, better is to find whatever produces the file and ask them to put reasonable line-endings into their file...
The problem
I have multiple property lines in a single string separated by \n like this:
LINES2="Abc1.def=$SOME_VAR\nAbc2.def=SOMETHING_ELSE\n"$LINES
The LINES variable
might contain an undefined set of characters
may be empty. If it is empty, I want to avoid the trailing \n.
I am open for any command line utility (sed, tr, awk, ... you name it).
Tryings
I tried this to no avail
sed -z 's/\\n$//g' <<< $LINES2
I also had no luck with tr, since it does not accept regex.
Idea
There might be an approach to convert the \n to something else. But since $LINES can contain arbitrary characters, this might be dangerous.
Sources
I skim read through the following questions
How can I replace a newline (\n) using sed?
sed with literal string--not input file
Here's one solution:
LINES2="Abc1.def=$SOME_VAR"$'\n'"Abc2.def=SOMETHING_ELSE${LINES:+$'\n'$LINES}"
The syntax ${name:+value} means "insert value if the variable name exists and is not empty." So in this case, it inserts a newline followed by $LINES if $LINES is not empty, which seems to be precisely what you want.
I use $'\n' because "\n" is not a newline character. A more readable solution would be to define a shell variable whose value is a single newline.
It is not necessary to quote strings in shell assignment statements, since the right-hand side of an assignment does not undergo word-splitting nor glob expansion. Not quoting would make it easier to interpolate a $'\n'.
It is not usually advisable to use UPPER-CASE for shell variables because the shell and the OS use upper-case names for their own purposes. Your local variables should normally be lower case names.
So if I were not basing the answer on the command in the question, I would have written:
lines2=Abc1.def=$someVar$'\n'Abc2.def=SOMETHING_ELSE${lines:+$'\n'$lines}
I have a block of text I'm trying to edit like this, in a script:
First, I tried
VAR2=`echo $VAR | sed 's/\n/\n\t/g'
It removes the newlines, but doesn't add the newline or tab back in.
Is this some stupid mistake? Not escaping something I should?
Two things:
You have to prevent shell expansion of $VAR, or the newlines will be lost before you have a chance to handle them
sed works in a line-based manner. It treats every line individually, and it doesn't see the newlines between them (unless you do special things).
The first can be handled by quoting $VAR, the second problem I would circumvent by reformulating the problem as "insert a tab to the beginning of every line but the first." That leaves us with:
VAR2=$(echo "$VAR" | sed '1!s/^/\t/')
Where the sed code means: Under the condition 1! (which is the case when we're not handling the first line), do s/^/\t/ -- i.e., replace the empty string at the beginning of the line with a tab.
Note that to look at the result of the substitution, you'll have to quote it as well, or it'll be shell-expanded, and the inserted whitespaces will be lost. That is to say,
echo "$VAR2"
will show the result you want, while
echo $VAR2
will lose all formatting (and potentially do silly things, if there are special characters such as $ in the paragraph).
I'm having an extremely hard time figuring out how to echo this:
I keep getting this error:
bash: ![alt: event not found
Using double quotes around it does not work. The using single quotes around it does work, however, I also need to echo strings that have single quotes within them. I wouldn't be able to wrap the string with single quotes then.
Is there a way to echo a string of ANY content?
Thanks.
EDIT: Here is some context. I am making a Markdown renderer that grabs the content of a code editor, then appends every line of the code individually into a text file. I am doing this by doing this:
echo TheLineOfMarkdown > textfile.txt
Unlike in many programing languages, '...' and "..." in Bash do not represent "strings" per se; they quote/escape whatever they contain, but they do not create boundaries that separate arguments. So, for example, these two commands are equivalent:
echo foobar
echo "fo"ob'ar'
So if you need to quote some of an argument with single-quotes, and a different part of the argument has to contain single-quotes — no problem.
For example:
echo '![alt text](https://... "What'"'"'s up, Doc?")'
Another option is to use \, which is similar to '...' except that it only quotes a single character. It can even be used inside double-quotes:
echo "\![alt text](https://... \"What's up, Doc?\")"
For more information, see §3.1.2 "Quoting" in the Bash Reference Manual.
! is annoying. My advice: Use \!.
! invokes history completion, which is also performed inside double-quotes. So you need to single-quote the exclamation mark, but as you say that conflicts with the need to not single-quote other single-quotes.
Remember that you can mix quotes:
$ echo '!'"'"'"'
!'"
(That's just one argument.) But in this case, the backslash is easier to type and quite possibly more readable.
Wow, this one has really got me. Gonna need some tricky sed skill here I think. Here is the output value of command text I'm trying to replace:
...
fast
n : abstaining from food
The value I'd like to replace it with, is:
...
Noun
: abstaining from food
This turns out to be tricker that I thought. Because 'fast' is listed a number of times and because it is listed in other places at the beginning of the line. So I came up with this to define the range:
sed '/fast/,/^ n : / s/fast/Noun/'
Which I thought would do, but... Unfortunately, this doesn't end the replacement and the rest of the output following this match are replaced with Noun. How to get sed to stop replacement after the match? Even better, can I find a two line pattern match and replace it?
Try this:
sed "h; :b; \$b ; N; /^${1}\n n/ {h;x;s//Noun\n/; bb}; \$b ; P; D"
Unfortunately, Paul's answer reads the whole file in which makes any additional processing you might want to do difficult. This version reads the lines in pairs.
By enclosing the sed script in double quotes instead of single quotes, you can include shell variables such as positional parameters. I would recommend surrounding them with curly braces so they are set apart from the adjacent characters. When using double quotes, you'll have to be careful of the shell wanting to do its various expansions. In this example, I've escaped the dollar signs that signify the last line of the input file for the branch commands. Otherwise the shell will try to substitute the value of a variable $b which is likely to be null thus making sed unhappy.
Another technique would be to use single quotes and close and open them each time you have a shell variable:
sed 'h; :b; $b ; N; /^'${1}'\n n/ {h;x;s//Noun\n/; bb}; $b ; P; D'
# ↑open close↑ ↑open close↑
I'm assuming that the "[/code]" in your expected result is a typo. Let me know if it's not.
This seems to do what you want:
sed -e ':a;N;$!ba;s/fast\n n/Noun\n/'
I essentially stole the answer from here.
This might work for you:
sed '$!N;s/^fast\n\s*n :/Noun\n :/;P;D' file
...
Noun
: abstaining from food