For example
bwa sampe ref.fa r1.sai r2.sai r1.fq r2.fq | samtools view -bSho out.bam -;
What is the purpose of the "-;" characters at the end? What do they do? Why are they necessary?
The semicolon ends the command (to end the pipeline is, I believe, technically the right way to say it). You could follow it with another command if you wanted to, as in
bwa sampe ref.fa r1.sai r2.sai r1.fq r2.fq | samtools view -bSho out.bam -; echo Here is another command.
Otherwise, the semicolon is harmless but probably unnecessary.
Regarding the - hyphen that precedes the semicolon, for samtools and many other commands it means to use standard input in place of an input file (or, in some cases, standard output in place of an output file). This is typical Linux/Unix usage.
(Thanks to #phatfingers for verifying the usage of the samtools command.)
Related
I am quite new to bash and trying to type some text into multiple files with a single command using brace expansion.
I tried: cat > file_{1..100} to write into 100 files some text that I will type in the terminal. I get the following error:
bash: file_{1..100}: ambiguous redirect
I also tried: cat > "file_{1..100}" but that creates a singe file named: file_{1..100}.
I tried: cat > `file_{1..100}` but that gives the error:
file_1: command not found
How can I achieve this using brace expansion? Maybe there are other ways using other utilities and/or pipelines. But I want to know if that is possible using only simple brace expansion or not.
You can't do this with cat alone. It only writes its output to its standard output, and that single file descriptor can only be associated with a single file.
You can however do it with tee file_{1..100}.
You may wish to consider using tee file_{01..100} instead, so that the filenames are zero-padded to all have the same width: file_001, file_002, ... This has the advantage that lexicographic order will agree with numerical order, and so ls, *, etc, will process them in numerical order. Without this, you have the situation that file_2 comes after file_10 in lexicographic order.
target could be only a pipe, not a multiple files.
If you want redirect output to multiple files, use tee
cat | tee file_{1..100}
Don't forget to check man tee, for example if you want to append to the files, you should add -a option (tee -a file_{1..100})
This types the string or text into file{1..4}
echo "hello you just knew me by kruz" > file{1..4}
Use to remove them
rm file*
I'm trying to display the output of an AWS lambda that is being captured in a temporary text file, and I want to remove that file as I display its contents. Right now I'm doing:
... && cat output.json && rm output.json
Is there a clever way to combine those last two commands into one command? My goal is to make the full combined command string as short as possible.
For cases where
it is possible to control the name of the temporary text file.
If file is not used by other code
Possible to pass "/dev/stdout" as the.name of the output
Regarding portability: see stack exchange how portable ... /dev/stdout
POSIX 7 says they are extensions.
Base Definitions,
Section 2.1.1 Requirements:
The system may provide non-standard extensions. These are features not required by POSIX.1-2008 and may include, but are not limited to:
[...]
• Additional character special files with special properties (for example, /dev/stdin, /dev/stdout, and /dev/stderr)
Using the mandatory supported /dev/tty will force output into “current” terminal, making it impossible to pipe the output of the whole command into different program (or log file), or to use the program when there is no connected terminals (cron job, or other automation tools)
No, you cannot easily remove the lines of a file while displaying them. It would be highly inefficient as it would require removing characters from the beginning of a file each time you read a line. Current filesystems are pretty good at truncating lines at the end of a file, but not at the beginning.
A simple but extremely slow method would look like this:
while [ -s output.json ]
do
head -1 output.json
sed -i 1d output.json
done
While this algorithm is plain and simple, you should know that each time you remove the first line with sed -i 1d it will copy the whole content of the file but the first line into a temporary file, resulting in approximately 0.5*n² lines written in total (where n is the number of lines in your file).
In theory you could avoid this by do something like that:
while [ -s output.json ]
do
line=$(head -1 output.json)
printf -- '%s\n' "$line"
fallocate -c -o 0 -l $((${#len}+1)) output.json
done
But this does not account for variable newline characters (namely DOS-formatted newlines) and fallocate does not always work on xfs, among other issues.
Since you are trying to consume a file alongside its creation without leaving a trace of its existence on disk, you are essentially asking for a pipe functionality. In my opinion you should look into how your output.json file is produced and hopefully you can pipe it to a script of your own.
I'am new in Linux and I want to write a bash script that can read in a file name of a directory that starts with LED + some numbers.(Ex.: LED5.5.002)
In that directory there is only one file that will starts with LED. The problem is that this file will every time be updated, so the next time it will be for example LED6.5.012 and counting.
I searched and tried a little bit and came to this solution:
export fspec=/home/led/LED*
LedV=`basename $fspec`
echo $LedV
If I give in those commands one by one in my terminal it works fine, LedV= LED5.5.002 but if i run it in a bash scripts it gives the result: LedV = LED*
I search after another solution:
a=/home/led/LED*
LedV=$(basename $a)
echo $LedV
but here again the same, if i give it in one by one it's ok but in a script: LedV = LED*.
It's probably something small but because of my lack of knowledge over Linux I cannot find it. So can someone tell what is wrong?
Thanks! Jan
Shell expansions don't happen on scalar assignments, so in
varname=foo*
the expansion of "$varname" will literally be "foo*". It's more confusing when you consider that echo $varname (or in your case basename $varname; either way without the double quotes) will cause the expansion itself to be treated as a glob, so you may well think the variable contains all those filenames.
Array expansions are another story. You might just want
fspec=( /path/LED* )
echo "${fspec[0]##*/}" # A parameter expansion to strip off the dirname
That will work fine for bash. Since POSIX sh doesn't have arrays like this, I like to give an alternative approach:
for fspec in /path/LED*; do
break
done
echo "${fspec##*/}"
pwd
/usr/local/src
ls -1 /usr/local/src/mysql*
/usr/local/src/mysql-cluster-gpl-7.3.4-linux-glibc2.5-x86_64.tar.gz
/usr/local/src/mysql-dump_test_all_dbs.sql
if you only have 1 file, you will only get 1 result
MyFile=`ls -1 /home/led/LED*`
I am trying to understand how
sed 's/\^\[/\o33/g;s/\[1G\[/\[27G\[/' /var/log/boot
worked and what the pieces mean. The man page I read just confused me more and I tried the info sai Id but had no idea how to work it! I'm pretty new to Linux. Debian is my first distro but seemed like a rather logical place to start as it is a root of many others and has been around a while so probably is doing stuff well and fairly standardized. I am running Wheezy 64 bit as fyi if needed.
The sed command is a stream editor, reading its file (or STDIN) for input, applying commands to the input, and presenting the results (if any) to the output (STDOUT).
The general syntax for sed is
sed [OPTIONS] COMMAND FILE
In the shell command you gave:
sed 's/\^\[/\o33/g;s/\[1G\[/\[27G\[/' /var/log/boot
the sed command is s/\^\[/\o33/g;s/\[1G\[/\[27G\[/' and /var/log/boot is the file.
The given sed command is actually two separate commands:
s/\^\[/\o33/g
s/\[1G\[/\[27G\[/
The intent of #1, the s (substitute) command, is to replace all occurrences of '^[' with an octal value of 033 (the ESC character). However, there is a mistake in this sed command. The proper bash syntax for an escaped octal code is \nnn, so the proper way for this sed command to have been written is:
s/\^\[/\033/g
Notice the trailing g after the replacement string? It means to perform a global replacement; without it, only the first occurrence would be changed.
The purpose of #2 is to replace all occurrences of the string \[1G\[ with \[27G\[. However, this command also has a mistake: a trailing g is needed to cause a global replacement. So, this second command needs to be written like this:
s/\[1G\[/\[27G\[/g
Finally, putting all this together, the two sed commands are applied across the contents of the /var/log/boot file, where the output has had all occurrences of ^[ converted into \033, and the strings \[1G\[ have been converted to \[27G\[.
I have a trivial error that I cant seem to get around. Im trying to return the various section numbers of lets say "man" since it resides in all the sections. I am using the -s command but am having problems. Every time I use it I keep getting "what manual page do you want". Any help?
In the case of getting the section number of a command, you want something like man -k "page_name" | awk -F'-' "/^page_name \(/ {print $1}", replacing any occurrence of page_name with whatever command you're needing.
This won't work for all systems necessarily as the format for the "man" output is "implementation-defined". In other words, the format on FreeBSD, OS X, various flavours of Linux, etc. may not be the same. For example, mine is:
page_name (1) - description
If you want the section number only, I'm sure there is something you can do such as saving the result of that line in a shell variable and use parameter expansion to remove the parentheses around the section number:
man -k "page_name" | awk -F'-' "/^page_name \(/ {print $1}" | while IFS= read sect ; do
sect="${sect##*[(]}"
sect="${sect%[)]*}"
printf '%s\n' "$sect"
done
To get the number of sections a command appears in, add | wc -l at the end on the same line as the done keyword. For the mount command, I have 3:
2
2freebsd
8
You've misinterpreted the nature of -s. From man man:
-S list, -s list, --sections=list
List is a colon- or comma-separated list of `order specific' manual sections to search. This option overrides the
$MANSECT environment variable. (The -s
spelling is for compatibility with System V.)
So when man sees man -s man it thinks you want to look for a page in section "man" (which most likely doesn't exist, since it is not a normal section), but you didn't say what page, so it asks:
What manual page do you want?
BTW, wrt "man is just the test case cuz i believe its in all the sections" -- nope, it is probably only in one, and AFAIK there isn't any word with a page in all sections. More than 2 or 3 would be very unusual.
The various standard sections are described in man man too.
The correct syntax requires an argument. Typically you're looking for either
man -s 1 man
to read the documentation for the man(1) command, or
man -s 7 man
to read about the man(7) macro package.
If you want a list of standard sections, the former contains that. You may have additional sections installed locally, though. A directory listing of /usr/local/share/man might reveal additional sections, for example.
(Incidentally, -s is not a "command" in this context, it's an option.)