I am using following shebang in my script:
#!/bin/sh
To check whether I haven't used any bash syntax, I checked my script with checkbashisms script and it says
possible bashism in my_script.sh line 3 (echo -e):
echo -e "hello world"
And I am running this script on embedded board with busybox shell.
How to resolve this?
POSIX allows echo -e as the default behavior. You may do
var="string with escape\nsequence"
printf "%s\n" "$var" # not interpreting backslashes
printf "%b\n" "$var" # interpreting backslashes
This should pass the checkbashisms test
help echo
-e enable interpretation of the following backslash escapes
...
If your code doesn't need to interpret backslash escapes then you can safely remove the -e flag.
Sometimes, people tend to use -e just for interpreting \n as newline. If that's the case, know that you can have multiline strings just by quoting them and writing as it is. An example:
var="Hi,
I am
a multiline
string"
But if you do need to interpret backslash escapes, then using printf will be your best bet.
Though I would recommend using printf whether you need backslash interpretation or not.
With printf:
printf "%s\n" "$var" # no backslash interpretation
printf "$var" # backslash interpretation occurs
Related
There are a lot of examples here how to source a dotenv file in bash but has anyone one a method that achieves the same with dash (which is the default shell for minimal Debian installations)?
The solution should look like this:
$ some foo my-command-using-env-vars
e.g.
$ env $(cat .env) my-command-using-env-vars
And it is important that the solution supports multiline values with spaces like:
SSH_PRIVATE_KEY="-----BEGIN RSA PRIVATE KEY-----\nfoo\nbar\baz"
and special characters like hash within quotes:
SPECIAL="foo#bar"
It seems that your problem is not so much that you're using dash but that you want to support \n escapes. The following works in dash and supports \n escapes:
eval "$(echo $(cat .env))" my-command-using-env-vars
That's because unlike in bash the built-in echo in dash supports backslash escapes even without the -e option. The following works in both bash and dash, provided that the non-built-in, system echo supports the -e option:
eval "$(env echo -e $(cat .env))" my-command-using-env-vars
Note that both approaches will also handle other backslash escapes (either POSIX or GNU), possibly in a different way than you expect.
Some technical notes:
$(cat .env)
performs Field Splitting, converting any newline in file .env into spaces.
"$(env echo -e ...)"
expands backslash escapes regardless of the current shell by invoking echo -e via env. The double quotes disable field splitting, so that newlines are preserved.
I would like to use the bash here doc syntax to build a long string. I would like the heredoc to ignore newlines/spaces/tabs even when I use newlines for code clarity.
I thought this would work:
#!/bin/bash
#http://unix.stackexchange.com/questions/20035/how-to-add-newlines-into-variables-in-bash-script
IFS= read -r -d '' NS_LOG<<-EOF
*=error|warn|prefix_node|prefix_func
:PointToPointNetDevice
:ClockTest
:ClockPerfect
:TcpTestSuite
:TcpRxBuffer
:TcpTxBuffer
:TcpHeader=*
:TcpL4Protocol
:TraceHelper:PointToPointHelper
EOF
echo $NS_LOG
export NS_LOG
but somewhere bash appends spaces between lines and instead of having the desired
*=error|warn|prefix_node|prefix_func:PointToPointNetDevice:ClockTest:ClockPerfect:Clock
I have when running $ ./launch_myscript.sh:
*=error|warn|prefix_node|prefix_func :PointToPointNetDevice :ClockTest :ClockPerfect :Clock etc...
My bash --version:
GNU bash, version 4.3.30(1)-release (x86_64-pc-linux-gnu)
I just saw in the recommended posts this one Bash: Why is echo adding extra space?. How can I prevent NS_LOG from being considered as several arguments? Ultimately the goal is to export that variable.
Your read command is very explicitly treating newlines as data: By clearing IFS and passing -d '', you tell read not to treat whitespace characters as special; since they're not special, they go into the output variable like everything else. However, you can take them out later:
IFS= read -r -d '' NS_LOG <<'EOF'
...content...
EOF
NS_LOG=${NS_LOG//[[:space:]]/} ## replace all whitespace with the empty string
printf '%s\n' "$NS_LOG" ## the quotes are important!
See this snippet run, and its output, at http://ideone.com/fWhzBB.
Notes:
<<'EOF' prevents expansions from occurring within the heredoc itself; with <<EOF, $foo, $(foo), etc. would be special.
<<- only trims leading tab characters, not any other form of whitespace; it's typically safer to do without.
echo $foo string-splits and glob-expands the contents of $foo, passing each word created by this process as a separate argument; echo then places spaces between each argument. echo "$foo" ensures that the entire expansion is treated as a single word. See BashPitfalls #14.
Using echo with nontrivial or unknown data is advised against in the relevant portion of the POSIX specification; printf is the preferred substitute. POSIX echo is explicitly allowed to behave in undefined ways when content contains backslash literals, and the BSD- and AT&T-derived forms of the command are explicitly incompatible, both with each other and with the common GNU implementation (providing an -e flag, which the POSIX spec requires to simply print -e on its output).
things aren't that difficult. here is another solution:
var="`tr -d '[:space:] <<EOF'
your
text
with
lot of spaces
EOF`"
result:
$ echo "$var"
yourtextwithlotofspaces
When trying to use ANSI color escapes from a shell script, I was completely stuck, as the escape sequences (\e) were printed verbatim to the output.
#!/bin/sh
GREEN="\e[32m"
RED="\e[31m"
CLEAR="\e[0m"
printf "${GREEN}test passed${CLEAR}\n"
printf "${RED}test failed${CLEAR}\n"
Produces
\e[32mtest passed\e[0m
\e[31mtest failed\e[0m
\e is not recognized by POSIX sh (as mentioned by honzasp), but \033 is.
GREEN='\033[32m'
CLEAR='\033[0m'
printf "${GREEN}testpassed${CLEAR}\n"
Generally, it's safer to not expand parameters inside the first argument to printf (consider, for example FOO="hello %s"; printf "$FOO bar \n" baz;). However, this requires you to embed an actual escape character in your parameters, rather than a string that printf interprets as an escape character.
GREEN=$(printf '\033[32m')
CLEAR=$(printf '\033[0m')
printf '%stest passed%s' "$GREEN" "$CLEAR"
The solution is to use #!/bin/bash instead of #!/bin/sh in the first line, because raw sh's printf doesn't understand the escapes.
I would like to use echo in bash to print out a string of characters followed by only a carriage return. I've looked through the man page and have found that echo -e will make echo interpret backslash escape characters. Using that I can say echo -e 'hello\r' and it will print like this
$>echo -e 'hello\r'
hello
$>
So it looks like it handled the carriage return properly. I also found echo -n in the man page will stop echo from inserting a newline character and it looks like it works when I do this
$>echo -n 'hello\r'
hello\r$>
The problem I'm having is in combining both -e and -n. I've tried each of echo -e -n 'hello\r', echo -n -e 'hello\r', echo -en 'hello\r', and echo -ne 'hello\r' and nothing gets printed like so:
$>echo -ne 'hello\r'
$>
Is there something I'm missing here or can the -e and -n options not be used together?
I think it's working, you're just not seeing it. This is your command:
$> echo -ne 'hello\r'
Because of the carriage return (\r), that will leave the cursor at the start of the same line on the terminal where it wrote the hello - which means that's where the next thing output to the terminal will be written. So if your actual prompt is longer than the $> you show here, it will overwrite the hello completely.
This sequence will let you see what's actually happening:
echo -ne 'hello\r'; sleep 5; echo 'good-bye'
But for better portability to other shells, I would avoid using options on echo like that. Those are purely bashisms, not supported by the POSIX standard. The printf builtin, however, is specified by POSIX. So if you want to display strings with no newline and parsing of backslash sequences, you can just use it:
printf '%s\r' 'hello'
There are numerous different implementations of the echo command. There's one built into most shells (with different behavior for each), and the behavior of /bin/echo varies considerably from one system to another.
Rather than echo, use printf. It's built into bash as well as being available as an external command, and its behavior is much more consistent across implementations. (The major variation is that the GNU coreutils printf recognizes --help and --version options.)
Just use:
printf 'hello\r'
I'd like you to introduce you to printf.
OP, meet printf. printf. This is the OP...
Whenever you are trying to do anything unusual with output in BASH, you should switch to printf. In fact, I use printf all the time, so my scripts will run under both BASH and Kornshell.
Although BASH and Kornshell are 99% the same, the echo command is different. In Kornshell, it's deprecated, and you're supposed to use the builtin print. However, there's no print in BASH. Using printf solves the problem because it works (sort of) the same in both shells.
Since printf doesn't automatically end each line with a \n, you don't have to worry about how to prevent the \n from being appended. Heck, if you want a \n, you have to put it yourself. In your case:
printf `hello\r`
However, for safety reasons, you should really do this:
printf '%s\r' 'hello'
This is a habit I've gotten into when I'm using printf. Imagine if I want to print out $FOO, and I do this:
printf "$FOO\n"
That will normally work, but what if $FOO has a dash at the beginning, or has a % sign in it. The above won't do what I want to do. However, if I do this:
printf '%s\n' "$FOO"
This will print out $FOO no matter what $FOO has in it.
POSIX 7 says it is not possible:
If the first operand is -n, or if any of the operands contain a backslash character, the results are implementation-defined.
On Ubuntu 12.04 for example:
Bash's built-in echo echo -e '\n' interprets the newline, echo '\n' does not.
/bin/sh's built-in echo does not have the -e option, and interprets the newline even without it. This can byte when writing Makefiles, which use /bin/sh by default.
POSIX 7 solution: printf as the others mentioned.
Quoting. It's the wave of the future. Always quote.
echo -ne 'hello\r'
Use double quotes if you put a variable inside that you want expanded.
I've read the man pages on echo, and it tells me that the -e parameter will allow an escaped character, such as an escaped n for newline, to have its special meaning. When I type the command
$ echo -e 'foo\nbar'
into an interactive bash shell, I get the expected output:
foo
bar
But when I use this same command (i've tried this command character for character as a test case) I get the following output:
-e foo
bar
It's as if echo is interpretting the -e as a parameter (because the newline still shows up) yet also it interprets the -e as a string to echo. What's going on here? How can I prevent the -e showing up?
You need to use #!/bin/bash as the first line in your script. If you don't, or if you use #!/bin/sh, the script will be run by the Bourne shell and its echo doesn't recognize the -e option. In general, it is recommended that all new scripts use printf instead of echo if portability is important.
In Ubuntu, sh is provided by a symlink to /bin/dash.
Different implementations of echo behave in annoyingly different ways. Some don't take options (i.e. will simply echo -e as you describe) and automatically interpret escape sequences in their parameters. Some take flags, and don't interpret escapes unless given the -e flag. Some take flags, and interpret different escape sequences depending on whether the -e flag was passed. Some will cause you to tear your hair out if you try to get them to behave in a predictable manner... oh, wait, that's all of them.
What you're probably seeing here is a difference between the version of echo built into bash vs /bin/echo or maybe vs. some other shell's builtin. This bit me when Mac OS X v10.5 shipped with a bash builtin echo that echoed flags, unlike what all my scripts expected...
In any case, there's a solution: use printf instead. It always interprets escape sequences in its first argument (the format string). The problems are that it doesn't automatically add a newline (so you have to remember do that explicitly), and it also interprets % sequences in its first argument (it is, after all, a format string). Generally, you want to put all the formatting stuff in the format string, then put variable strings in the rest of the arguments so you can control how they're interpreted by which % format you use to interpolate them into the output. Some examples:
printf "foo\nbar\n" # this does what you're trying to do in the example
printf "%s\n" "$var" # behaves like 'echo "$var"', except escapes will never be interpreted
printf "%b\n" "$var" # behaves like 'echo "$var"', except escapes will always be interpreted
printf "%b\n" "foo\nbar" # also does your example
Use
alias echo /usr/bin/echo
to force 'echo' invoking coreutils' echo which interpret '-e' parameter.
Try this:
import subprocess
def bash_command(cmd):
subprocess.Popen(['/bin/bash', '-c', cmd])
code="abcde"
// you can use echo options such as -e
bash_command('echo -e "'+code+'"')
Source: http://www.saltycrane.com/blog/2011/04/how-use-bash-shell-python-subprocess-instead-binsh/