I'm writing a script which needs to use the shell's brace expansion, but nothing I've tried works. For (a contrived) instance, say I have a variable containing the string
thing{01..02}
and I (obviously) want to expand it to
thing01 thing02
from inside the script, how can I do that?
(For anyone who assumes this is a duplicate of this other question, please read them more carefully. That question is regarding working from the shell, not a shell script, and doesn't require the ability to expand arbitrary expressions.)
$ echo thing{01,02}
thing01 thing02
Make sure that braceexpand is turned on with set -o braceexpand.
Related
I have a shell script that I'm trying to write to a file using multiple variables, but one of them is being ignored.
#!/bin/bash
dir=/folder
name=bob
date=`date +%Y`
command > $dir/$name_$date.ext
The $name is being ignored. How can I fix this?
Have you noticed that the _ was "ignored" as well? That's a big hint.
If you use set -u, you'll see the following:
-bash: name_: unbound variable
The way bash parses it, the underscore is part of the variable name.
There are several ways to fix the problem.
The cleanest is the ${var} construct which separate the variable name from its surroundings.
You can also use quotation in various ways to force the right parsing, e.g.: "$dir/$name""_$date.ext"
And in case your variables might contain spaces (now, or in the future) use quotation for words.
command >"$dir/${name}_$date.ext"
command >"${dir}/${name}_${date}.ext"
Both these are fine, just pick one style and stick to it.
I have a shell script in my home directory called "echo". I added my home directory to my path, so that this echo would replace the other one.
To do this, I used: export PATH=/home/me:$PATH
When I do which echo, it shows the one I want. /home/me/echo
But when I actually do something like echo asdf it uses the system echo.
Am I doing something wrong?
which is an external command, so it doesn't have access to your current shell's built-in commands, functions, or aliases. In fact, at least on my system, /usr/bin/which is a shell script, so you can examine it and see how it works.
If you want to know how your shell will interpret a command, use type rather than which. If you're using bash, type -a will print all possible meanings in order of precedence. Consult your shell's documentation for details.
For most shells, built-in commands take precedence over commands in your $PATH. The whole point of having a built-in echo, for example, is that it's faster than loading /bin/echo into memory.
If you want your own echo command to override the shell's built-in echo, you can define it as a shell function.
On the other hand, overriding the built-in echo command doesn't strike me as a good idea in the first place. If it behaves the same as the built-in echo, there's not much point. If it doesn't, then it could break scripts that use echo expecting it to work a certain way. If possible, I suggest giving your command a different way. If it's an enhanced version of echo, you could even call it Echo.
It is likely using the shell's builtin.
If you want the one in your path you can do
`which echo` asdf
From this little article that explains the rules, here's a list in descending order of precedence:
Aliases
Shell functions
Shell builtin commands
Hash tables
PATH variable
echo is a shell builtin command (al least in bash) and PATH has the lowest priority. I guess you'll need to create a function or an alias.
I'm working on what should be a very simple BASH script. What I want to do is a pull an image from a webcamera using curl and write it to a file whose name is datestamped.
#! /bin/bash
DATE=$(date +%Y-%m-%d_%H-%M)
DIRECTORY1=home/manager/security_images/Studio_1/
TARGET1=${DIRECTORY1}${DATE}.jpg
curl http://web#192.168.180.211/snapshot.cgi > $TARGET1
When I try to run this I am told that there is no such file or directory. I believe this is due to an error in my escaping but I have tried seemingly every combination of quotation marks around the variables at each stage and still can't get it to work. I just don't understand what is going wrong and could really use some pointers towards what I'm doing wrong.
Many thanks
No, it’s just a typo.
DIRECTORY1=home/manager/security_images/Studio_1/
^^
Should be
DIRECTORY1=/home/manager/security_images/Studio_1/
of course.
As for escaping, even though only safe characters are used now, so quotes are technically superfluous, quoting out all $variables in every line by default is a good habit in shell scripting, there are very few cases when you do not want to use them.
Double quoting the redirection target should be enough:
curl http://web#192.168.180.211/snapshot.cgi > "$TARGET1"
Just make sure the path to it exists. You can run your script with
set -xv
to see how variables are interpolated.
Here is what it finally took to get my code in my makefile to work
Line 5 is the question area
BASE=50
INCREMENT=1
FORMATTED_NUMBER=${BASE}+${INCREMENT}
all:
echo $$((${FORMATTED_NUMBER}))
why do i have to add two $ and two (( )) ?
Formatted_Number if i echo it looks like "50+1" . What is the logic that make is following to know that seeing $$(("50+1")) is actually 51?
sorry if this is a basic question i'm new to make and dont fully understand it.
First, whenever asking questions please provide a complete example. You're missing the target and prerequisite here so this is not a valid makefile, and depending on where they are it could mean very different things. I'm assuming that your makefile is something like this:
BASE=50
INCREMENT=1
FORMATTED_NUMBER=${BASE}+${INCREMENT}
all:
echo $$((${FORMATTED_NUMBER}))
Makefiles are interesting in that they're a combination of two different formats. The main format is makefile format (the first five lines above), but inside a make recipe line (that's the last line above, which is indented with a TAB character) is shell script format.
Make doesn't know anything about math. It doesn't interpret the + in the FORMATTED_NUMBER value. Make variables are all strings. If you want to do math, you have to do it in the shell, in a shell script, using the shell's math facilities.
In bash and other modern shells, the syntax $(( ...expression... )) will perform math. So in the shell if you type echo $((50+1)) (go ahead and try it yourself) it will print 51.
That's why you need the double parentheses ((...)): because that's what the shell wants and you're writing a shell script.
So why the double $? Because before make starts the shell to run your recipe, it first replaces all make variable references with their values. That's why the shell sees 50+1 here: before make started the shell it expanded ${FORMATTED_NUMBER} into its value, which is ${BASE}+${INCREMENT}, then it expanded those variables so it ends up with 50+1.
But what if you actually want to use a $ in your shell script (as you do here)? Then you have to tell make to not treat the $ as introducing a make variable. You do this by doubling it, so if make sees $$ then it does not think that's a make variable, and sends a single $ to the shell.
So for the recipe line echo $$((${FORMATTED_NUMBER})) make actually invokes a shell script echo $((50+1)).
You can use this in BASH:
FORMATTED_NUMBER=$((BASE+INCREMENT))
Is using non BASH use:
FORMATTED_NUMBER=`echo "$BASE + $INCREMENT" | bc`
I have a simple script
...
dir=`pwd`
echo $dir
cd ./selenium-grid-1.0.8/
CMD="ant -Dport=$1 -Dhost=$2 -DhubURL=http://172.16.1.137:4444 -Denvironment="$3"-DseleniumArgs="-firefoxProfileTemplate C:/software/rc_user_ffprofile -multiWindow" launch-remote-control"
echo $CMD
$CMD 2>&1
#End
Whenever i run this command, i get: ./register_rc.sh: line 16: C:/software/rc_user_ffprofile: is a directory
this directory has to be an argument to the -firefoxProfileTemplate option. How do i include that in this string without it baffing??
help
thnx
I believe your command should read:
CMD="ant -Dport=$1 -Dhost=$2 -DhubURL=http://172.16.1.137:4444 -Denvironment=\"$3\"-DseleniumArgs=\"-firefoxProfileTemplate C:/software/rc_user_ffprofile -multiWindow\" launch-remote-control"
The backslashes are used to "escape" the quotation marks.
The answers here telling to escape your quotes are wrong. That will pass those quotes directly to ant, I doubt that's what you want.
What's the reason to store the command in a variable? It's a very bad idea. Why can't you just write that command as is? If you want to achieve modularity or code reuse, then define a function.
If you want to display executed commands, use set -x.
Looks like you're mixing your quotes up. Take a look at the syntax highlighting that StackOverflow did for you.
I recommend generating the CMD variable in multiple steps, and make sure you \-escape your quotes.