Is it possible to execute node.js app from .sh script, return some variable and continue the .sh script?
Something like:
#!/bin/sh
SOME_VARIABLE = node app.js
echo ${SOME_VARIABLE}
Firstly, ensure you're using bash, not sh, as there are significant differences in functionality between the two.
One simple solution is command substitution, although be aware that trailing newlines in the command output will be stripped. Also, when echoing, to protect the contents of the variable (such as spaces and glob characters) from metaprocessing by the shell, you have to double-quote it:
#!/bin/bash
output=$(node app.js);
echo "$output";
Another solution is process substitution in more recent versions of bash. You could even collect the output as an array of lines in this case:
#!/bin/bash
exec 3< <(node app.js);
lines=();
while read -r; do lines+=("$REPLY"); done <&3;
exec 3<&-;
echo "${lines[#]}";
Related
I have following images.
10.jpg
11.jpg
12.jpg
I want to remove above images. I used following shell script file.
for file in /home/scrapping/imgs/*
do
COUNT=$(expr $COUNT + 1)
STRING="/home/scrapping/imgs/""Img_"$COUNT".jpg"
echo $STRING
mv "$file" "$STRING"
done
So, replaced file name
Img_1.jpg
Img_2.jpg
Img_3.jpg
But, I want to replace the file name like this:
Img_10.jpg
Img_11.jpg
Img_12.jpg
So, How to set COUNT value 10 to get my own output?
The expr syntax is pretty outdated, POSIX shell allows you to do arithmetic evaluation with $(()) syntax. You can just do
#!/usr/bin/env bash
count=10
for file in /home/scrapping/imgs/*; do
[ -f "$file" ] || continue
mv "$file" "/home/scrapping/imgs/Img_$((count++)).jpg"
done
Also from the errors reported in the comments, you seem to be running it from the dash shell. It does not seem to have all the features complying to the standard POSIX shell. Run it with the sh or the bash shell.
And always use lowercase letters for user defined variables in your shell script. Upper case letters are primarily for the environment variables managed by the shell itself.
With rename command you can suffix your files with Img_:
rename 's/^/Img_/' *
The ^ means replace the start of the filename with Img_, i.e: adds a suffix.
I am having a problem where cmd1 works, but not cmd2 in my Bash script ending in .sh. I have made the Bash script executable.
Additionally, I can execute cmd2 just fine from my Bash terminal. I have tried to make a minimally reproducible example, but my larger goal is to run a complicated executable with command line arguments and pass output to a file that may or may not exist (rather than displaying the output in the terminal).
Replacing > with >> also gives the same error in the script, but not the terminal.
My Bash script:
#!/bin/bash
cmd1="cat test.txt"
cmd2="cat test.txt > a"
echo $cmd1
$cmd1
echo $cmd2
$cmd2
test.txt has the words "dog" and "cat" on two separate lines without quotes.
Short answer: see BashFAQ #50: I'm trying to put a command in a variable, but the complex cases always fail!.
Long answer: the shell expands variable references (like $cmd1) toward the end of the process of parsing a command line, after it's done parsing redirects (like > a is supposed to be) and quotes and escapes and... In fact, the only thing it does with the expanded value is word splitting (e.g. treating cat test.txt > a as "cat" followed by "test.txt", ">", and finally "a", rather than a single string) and wildcard expansion (e.g. if $cmd expanded to cat *.txt, it'd replace the *.txt part with a list of matching files). (And it skips word splitting and wildcard expansion if the variable is in double-quotes.)
Partly as a result of this, the best way to store commands in variables is: don't. That's not what they're for; variables are for data, not commands. What you should do instead, though, depends on why you were storing the command in a variable.
If there's no real reason to store the command in a variable, then just use the command directly. For conditional redirects, just use a standard if statement:
if [ -f a ]; then
cat test.txt > a
else
cat test.txt
fi
If you need to define the command at one point, and use it later; or want to use the same command over and over without having to write it out in full each time, use a function:
cmd2() {
cat test.txt > a
}
cmd2
It sounds like you may need to be able to define the command differently depending on some condition, you can actually do that with a function as well:
if [ -f a ]; then
cmd() {
cat test.txt > a
}
else
cmd() {
cat test.txt
}
fi
cmd
Alternately, you can wrap the command (without redirect) in a function, then use a conditional to control whether it redirects:
cmd() {
cat test.txt
}
if [ -f a ]; then
cmd > a
else
cmd
fi
It's also possible to wrap a conditional redirect into a function itself, then pipe output to it:
maybe_redirect_to() {
if [ -f "$1" ]; then
cat > "$1"
else
cat
fi
}
cat test.txt | maybe_redirect_to a
(This creates an extra cat process that isn't really doing anything useful, but if it makes the script cleaner, I'd consider that worth it. In this particular case, you could minimize the stray cats by using maybe_redirect_to a < test.txt.)
As a last resort, you can store the command string in a variable, and use eval to parse it. eval basically re-runs the shell parsing process from the beginning, meaning that it'll recognize things like redirects in the string. But eval has a well-deserved reputation as a bug magnet, because it's easy for it to treat parts of the string you thought were just data as command syntax, which can cause some really weird (& dangerous) bugs.
If you must use eval, at least double-quote the variable reference, so it runs through the parsing process just once, rather than sort-of-once-and-a-half as it would unquoted. Here's an example of what I mean:
cmd3="echo '5 * 3 = 15'"
eval "$cmd3"
# prints: 5 * 3 = 15
eval $cmd3
# prints: 5 [list of files in the current directory] 3 = 15
# ...unless there are any files with shell metacharacters in their names, in
# which case something more complicated might happen.
BashFAQ #50 discusses some other possible reasons and solutions. Note that the array approach will not work here, since arrays also get expanded after redirects are parsed.
If you pop an 'eval' in front of $cmd2 it should work as expected:
#!/bin/bash
cmd2="cat test.txt > a"
eval $cmd2
If you're not sure about the operation of a script you could always use the debug mode to see if you can determine the error.
bash -x scriptname
This will run the command and display the output of variable evaluations. Hopefully this will reveal any issues with syntax.
I want to run a program (when executed it produces logdata) out of a shell script and write the output into a text file. I failed to do so :/
$prog is the executed prog -> socat /dev/ttyUSB0,b9600 STDOUT
$log/$FILE is just path to a .txt file
I had a Perl script to do this:
open (S,$prog) ||die "Cannot open $prog ($!)\n";
open (R,">>","$log") ||die "Cannot open logfile $log!\n";
while (<S>) {
my $date = localtime->strftime('%d.%m.%Y;%H:%M:%S;');
print "$date$_";
}
I tried to do this in a shell script like this
#!/bin/sh
FILE=/var/log/mylogfile.log
SOCAT=/usr/bin/socat
DEV=/dev/ttyUSB0
BAUD=,b9600
PROG=$SOCAT $DEV$BAUD STDOUT
exec 3<&0
exec 0<$PROG
while read -r line
do
DATE=`date +%d.%m.%Y;%H:%M:%S;`
echo $DATE$line >> $FILE
done
exec 0<&3
Doesn't work at all...
How do I read the output of that prog and pipe it into my text file using a shell script? What did I do wrong (if I didn't do everything wrong)?
Final code:
#!/bin/sh
FILE=/var/log/mylogfile.log
SOCAT=/usr/bin/socat
DEV=/dev/ttyUSB0
BAUD=,b9600
CMD="$SOCAT $DEV$BAUD STDOUT"
$CMD |
while read -r line
do
echo "$(date +'%d.%m.%Y;%H:%M:%S;')$line" >> $FILE
done
To read from a process, use process substitution
exec 0< <( $PROG )
/bin/sh doesn't support it, so use /bin/bash instead.
To assign several words to a variable, quote or backslash whitespace:
PROG="$SOCAT $DEV$BAUD STDOUT"
Semicolon is special in shell, quote it or backslash it:
DATE=$(date '+%d.%m.%Y;%H:%M:%S;')
Moreover, no exec's are needed:
while ...
...
done < <( $PROG )
You might even add > $FILE after done instead of adding each line separately to the file.
Original answer
You haven't shown the error messages — which would have been helpful.
Your problem, though, is probably this line:
DATE=`date +%d.%m.%Y;%H:%M:%S;`
where the semicolons mark the end of a command, and there likely isn't a command %H that does anything useful, etc.
You need quotes around the format argument to date, and I'd use single quotes for this job:
DATE=$(date +'%d.%m.%Y;%H:%M:%S;')
or even replace the two lines in the body of the loop with:
echo "$(date +'%d.%m.%Y;%H:%M:%S;')$line" >> $FILE
The double quotes prevent a variety of problems.
That assumes you fix a bunch of other problems, such as the setting of the variables FILE and prog. Also, I'd probably use:
exec > $FILE
to initially zap the output file and then all subsequent standard output would go to that file, so the echo line becomes:
echo "$(date +'%d.%m.%Y;%H:%M:%S;')$line"
Amended answer
The question was originally missing lots of key information. It eventually got updated to include the complete code.
The problem I identified originally remains an issue, but you weren't running into it because the input redirection was not working. If you want the input to come from a process, use a pipe, or possibly process substitution. However, note that you have #!/bin/sh as your shebang line, and /bin/sh won't recognized process substitution; either change the shebang or use the pipe notation. Note that process substitution has advantages if the loop is setting variables that need to be accessed after the loop is complete.
$SOCAT $DEV$BAUD STDOUT |
while read -r line
do
…
done
or
while read -r line
do
…
done < <($SOCAT $DEV$BAUD STDOUT)
Note that your code contains the line:
PROG=$SOCAT $DEV$BAUD STDOUT
This runs the command identified by $DEV$BAUD with the argument STDOUT and the environment variable PROG set to the value of $SOCAT. That is not what you wanted.
You could use an array:
PROG=($SOCAT $DEV$BAUD STDOUT)
and then run:
"${PROG[#]}"
either in the pipe line:
"${PROG[#]}" |
while read -r line
do
…
done
or with process substitution:
while read -r line
do
…
done < <("${PROG[#]}")
Note that unless there is code after the final exec 0<&3, there was no particular virtue in the redirections involving file descriptor 3. You should also close 3 when you're done with it:
exec 0<&3 3>&-
The 'final' code includes the lines:
CMD="$SOCAT $DEV$BAUD STDOUT"
$CMD |
while read -r line
This works OK because there are no spaces in the arguments to the command. That's a common case, but beware of spaces in arguments and file paths.
I'm new to shell programming. I intend to get directory name after zip file was extracted. The print statement of it is
$test.sh helloworld.zip
helloworld
Let's take a look at test.sh:
#! /bin/sh
length=echo `expr index "$1" .zip`
a=$1
echo $(a:0:length}
However I got the Bad substitution error from the compiler.
And when I mention about 'shell'.I just talking about shell for I don't know the difference between bash or the others.I just using Ubuntu 10.04 and using the terminal. (I am using bash.)
If your shell is a sufficiently recent version of bash, that parameter expansion notation should work.
In many other shells, it will not work, and a bad substitution error is the way the shell says 'You asked for a parameter substitution but it does not make sense to me'.
Also, given the script:
#! /bin/sh
length=echo `expr index "$1" .zip`
a=$1
echo $(a:0:length}
The second line exports variable length with value echo for the command that is generated by running expr index "$1" .zip. It does not assign to length. That should be just:
length=$(expr index "${1:?}" .zip)
where the ${1:?} notation generates an error if $1 is not set (if the script is invoked with no arguments).
The last line should be:
echo ${a:0:$length}
Note that if $1 holds filename.zip, the output of expr index $1 .zip is 2, because the letter i appears at index 2 in filename.zip. If the intention is to get the base name of the file without the .zip extension, then the classic way to do it is:
base=$(basename $1 .zip)
and the more modern way is:
base=${1%.zip}
There is a difference; if the name is /path/to/filename.zip, the classic output is filename and the modern one is /path/to/filename. You can get the classic output with:
base=${1%.zip}
base=${base##*/}
Or, in the classic version, you can get the path with:
base=$(dirname $1)/$(basename $1 .zip)`.)
If the file names can contain spaces, you need to think about using double quotes, especially in the invocations of basename and dirname.
Try running it with bash.
bash test.sh helloworld.zip
-likewise-
"try changing the first line to #!/bin/bash" as comment-answered by – #shellter
Try that in bash :
echo $1
len=$(wc -c <<< "$1")
a="${1}.zip"
echo ${a:0:$len}
Adapt it to fit your needs.
I just started learning PHP. I'm following phpacademy's tutorials which I would recommend to anyone. Anyways, I'm using XAMPP to test out my scripts. I'm trying to write a bash script that will start XAMPP and then open firefox to the localhost page if it finds a specific string, "XAMPP for Linux started.", that has been redirected from the terminal to the file xampp.log. I'm having a problem searching the file. I keep getting a:
grep: for: No such file or directory
I know the file exists, I think my syntax is wrong. This is what I've got so far:
loaded=$false
string="XAMPP for Linux started."
echo "Starting Xampp..."
sudo /opt/lampp/lampp start 2>&1 > ~/Documents/xampp.log
sleep 15
if grep -q $string ~/Documents/xampp.log; then
$loaded=$true
echo -e "\nXampp successfully started!"
fi
if [$loaded -eq $true]; then
echo -e "Opening localhost..."
firefox "http://localhost/"
else
echo -e "\nXampp failed to start."
echo -e "\nHere's what went wrong:\n"
cat ~/Documents/xampp.log
fi
In shell scripts you shouldn't write $variable, since that will do word expansion on the variable's value. In your case, it results in four words.
Always use quotes around the variables, like this:
grep -e "$string" file...
The -e is necessary when the string might start with a dash, and the quotes around the string keep it as one word.
By the way: when you write shell programs, the first line should be set -eu. This enables *e*rror checking and checks for *u*ndefined variables, which will be useful in your case. For more details, read the Bash manual.
You are searching for a string you should put wihtin quotes.
Try "$string" instead of $string
There are a couple of problems:
quote variables if you want to pass them as a simple argument "$string"
there is no $true and $false
bash does variable expansion, it substitutes variable names with their values before executing the command. $loaded=$true should be loaded=true.
you need spaces and usually quotes in the if: if [$loaded -eq $true] if [ "$loaded" -eq true ]. in this case the variable is set so it won't cause problems but in general don't rely on that.