In Bash (or other shells) how can I print an environment variable which has a multi-line value?
text='line1
line2'
I know a simple usual echo $text won't work out of the box.
Would some $IFS tweak help?
My current workaround is something like ruby -e 'print ENV["text"]'.
Can this be done in pure shell? I was wondering if env command would take an unresolved var name but it does not seem to.
Same solution as always.
echo "$text"
export TEST="A\nB\nC"
echo $TEST
gives output:
A\nB\nC
but:
echo -e $TEST
A
B
C
So, the answer seems to be the '-e' parameter to echo, assuming that I understand your question correctly.
I'm writing a bash script which has to pass a variable to another program:
./program $variable
The problem is, it is absolutely necessary for $variable to be passed as a single parameter, which isn't the case if it contains whitespace.
variable=Hello World
./program $variable
-> program receives two arguments: 'Hello' 'World'
Quoting it doesn't do anything at all (well done, bash devs):
variable=Hello World
./program "$variable"
-> program receives: 'Hello' 'World'
Double quoting it does crazy stuff (well done, bash devs):
variable=Hello World
./program "'$variable'"
-> program receives: "'Hello" "World'"
Is there an easy way to do this? Heck, is there a way to do this at all?
Update: Okay, since the problem doesn't seem to be bash, here's some additional info.
The program I'm passing arguments to is a python script. Without modifying the arguments in any way, I get
print sys.argv
-> ['/usr/bin/program', "'Hello", "World'"]
How can I fix that?
Edit: No, I haven't tried
variable="Hello World"
because I never declare $variable. It's not being declared inside my bash function and I'm not allowed to modify it.
Edit: Okay, I got it to work that way.
local temp="$variable"
./program "$temp"
I'd like to know why it works that way and not any other way, though.
did you try with var="hello world"?
i tried this in my solaris box.
> setenv var "hello world"
> cat temp.sh
#!/bin/sh
echo $1
echo $2
> ./temp.sh "$var"
hello world
>
as you can see the $2 is not printed.$var is considered as only one argument.
When you call your script pass the arguments within quotes.
Example script:
#!/bin/bash
for arg in "$#"; do
echo "arg: $1";
shift;
done
When you call it with:
./program "parameter with multiple words" parameter2 parameter3 "another parameter"
The output should be:
arg: parameter with multiple words
arg: parameter2
arg: parameter3
arg: another parameter
Have a look on http://tldp.org/LDP/Bash-Beginners-Guide/html/sect_03_04.html .
The problem is that the expansion of variables is done before of the command line parameters hence your behavior.
You might work it arround with setting IFS to something weird as
IFS='###' V='foo bar baz'; ./program $V
The problem seems to be inside the "program"
variable="Hello World" # quotes are needed because of the space
./program "$variable" # here quotes again
and inside the program
echo "program received $# arguments:"
i=1
for arg in "$#" # again quotes needed
do echo "arg $((i=i+1)): '$arg'" # and again
done
This is almost certainly a problem in the way you are reading the variable in your program.
For instance suppose this is your script (just one line for testing):
echo "$1"
Let's call it echo.sh. If you run echo.sh "test best", you will get test best.
But if your program says
echo $1
you might get behaviour like what you're seeing.
I want the following:
$ DOMAIN=chron echo snippet-www.$DOMAIN.com-head.html
to output:
snippet.www.chron.com-head.html
but for the life of me, I can't figure out how do this except via the two commands:
$ export DOMAIN=chron
$ echo snippet-www.$DOMAIN.com-head.html
Isn't there a way to get this to work as one command?
I have stumbled upon the answer!!
Just add a semicolon (;) after the variable assignment.
$ DOMAIN=chron; echo snippet-www.$DOMAIN.com-head.html
snippet-www.chron.com-head.html
$ ( export DOMAIN=chron ; echo snippet-www.$DOMAIN.com-head.html )
This makes $DOMAIN an environment variable (which doesn't matter for this example, but might for other similar commands), and it limits its lifetime to the parentheses.
Your answer:
$ DOMAIN=chron; echo snippet-www.$DOMAIN.com-head.html
causes $DOMAIN to be a (non-exported) shell variable, and retains the setting for later commands.
this should work
DOMAIN=chron eval 'echo snippet-www.$DOMAIN.com-head.html'
I have two variables in bash that complete the name of another one, and I want to expand it but don't know how to do it
I have:
echo $sala
a
echo $i
10
and I want to expand ${a10} in this form ${$sala$i} but apparently the {} scape the $ signs.
There are a few ways, with different advantages and disadvantages. The safest way is to save the complete parameter name in a single parameter, and then use indirection to expand it:
tmp="$sala$i" # sets $tmp to 'a10'
echo "${!tmp}" # prints the parameter named by $tmp, namely $a10
A slightly simpler way is a command like this:
eval echo \${$sala$i}
which will run eval with the arguments echo and ${a10}, and therefore run echo ${a10}. This way is less safe in general — its behavior depends a bit more chaotically on the values of the parameters — but it doesn't require a temporary variable.
Use the eval.
eval "echo \${$sala$i}"
Put the value in another variable.
result=$(eval "echo \${$sala$i}")
The usual answer is eval:
sala=a
i=10
a10=37
eval echo "\$$sala$i"
This echoes 37. You can use "\${$sala$i}" if you prefer.
Beware of eval, especially if you need to preserve spaces in argument lists. It is vastly powerful, and vastly confusing. It will work with old shells as well as Bash, which may or may not be a merit in your eyes.
You can do it via indirection:
$ a10=blah
$ sala=a
$ i=10
$ ref="${sala}${i}"
$ echo $ref
a10
$ echo ${!ref}
blah
However, if you have indexes like that... an array might be more appropriate:
$ declare -a a
$ i=10
$ a[$i]="test"
$ echo ${a[$i]}
test
How do I print a newline? This merely prints \n:
$ echo -e "Hello,\nWorld!"
Hello,\nWorld!
Use printf instead:
printf "hello\nworld\n"
printf behaves more consistently across different environments than echo.
Make sure you are in Bash.
$ echo $0
bash
All these four ways work for me:
echo -e "Hello\nworld"
echo -e 'Hello\nworld'
echo Hello$'\n'world
echo Hello ; echo world
echo $'hello\nworld'
prints
hello
world
$'' strings use ANSI C Quoting:
Words of the form $'string' are treated specially. The word expands to string, with backslash-escaped characters replaced as specified by the ANSI C standard.
You could always do echo "".
For example,
echo "Hello,"
echo ""
echo "World!"
On the off chance that someone finds themselves beating their head against the wall trying to figure out why a coworker's script won't print newlines, look out for this:
#!/bin/bash
function GET_RECORDS()
{
echo -e "starting\n the process";
}
echo $(GET_RECORDS);
As in the above, the actual running of the method may itself be wrapped in an echo which supersedes any echos that may be in the method itself. Obviously, I watered this down for brevity. It was not so easy to spot!
You can then inform your comrades that a better way to execute functions would be like so:
#!/bin/bash
function GET_RECORDS()
{
echo -e "starting\n the process";
}
GET_RECORDS;
Simply type
echo
to get a new line
POSIX 7 on echo
http://pubs.opengroup.org/onlinepubs/9699919799/utilities/echo.html
-e is not defined and backslashes are implementation defined:
If the first operand is -n, or if any of the operands contain a <backslash> character, the results are implementation-defined.
unless you have an optional XSI extension.
So I recommend that you should use printf instead, which is well specified:
format operand shall be used as the format string described in XBD File Format Notation [...]
the File Format Notation:
\n <newline> Move the printing position to the start of the next line.
Also keep in mind that Ubuntu 15.10 and most distros implement echo both as:
a Bash built-in: help echo
a standalone executable: which echo
which can lead to some confusion.
str='hello\nworld'
$ echo | sed "i$str"
hello
world
You can also do:
echo "hello
world"
This works both inside a script and from the command line.
On the command line, press Shift+Enter to do the line break inside the string.
This works for me on my macOS and my Ubuntu 18.04 (Bionic Beaver) system.
For only the question asked (not special characters etc) changing only double quotes to single quotes.
echo -e 'Hello,\nWorld!'
Results in:
Hello,
World!
There is a new parameter expansion added in Bash 4.4 that interprets escape sequences:
${parameter#operator} - E operator
The expansion is a string that is the value of parameter with
backslash escape sequences expanded as with the $'…' quoting
mechanism.
$ foo='hello\nworld'
$ echo "${foo#E}"
hello
world
I just use echo without any arguments:
echo "Hello"
echo
echo "World"
To print a new line with echo, use:
echo
or
echo -e '\n'
This could better be done as
x="\n"
echo -ne $x
-e option will interpret backslahes for the escape sequence
-n option will remove the trailing newline in the output
PS: the command echo has an effect of always including a trailing newline in the output so -n is required to turn that thing off (and make it less confusing)
My script:
echo "WARNINGS: $warningsFound WARNINGS FOUND:\n$warningStrings
Output:
WARNING : 2 WARNINGS FOUND:\nWarning, found the following local orphaned signature file:
On my Bash script I was getting mad as you until I've just tried:
echo "WARNING : $warningsFound WARNINGS FOUND:
$warningStrings"
Just hit Enter where you want to insert that jump. The output now is:
WARNING : 2 WARNINGS FOUND:
Warning, found the following local orphaned signature file:
If you're writing scripts and will be echoing newlines as part of other messages several times, a nice cross-platform solution is to put a literal newline in a variable like so:
newline='
'
echo "first line${newline}second line"
echo "Error: example error message n${newline}${usage}" >&2 #requires usage to be defined
If the previous answers don't work, and there is a need to get a return value from their function:
function foo()
{
local v="Dimi";
local s="";
.....
s+="Some message here $v $1\n"
.....
echo $s
}
r=$(foo "my message");
echo -e $r;
Only this trick worked on a Linux system I was working on with this Bash version:
GNU bash, version 2.2.25(1)-release (x86_64-redhat-linux-gnu)
You could also use echo with braces,
$ (echo hello; echo world)
hello
world
This got me there....
outstuff=RESOURCE_GROUP=[$RESOURCE_GROUP]\\nAKS_CLUSTER_NAME=[$AKS_CLUSTER_NAME]\\nREGION_NAME=[$REGION_NAME]\\nVERSION=[$VERSION]\\nSUBNET-ID=[$SUBNET_ID]
printf $outstuff
Yields:
RESOURCE_GROUP=[akswork-rg]
AKS_CLUSTER_NAME=[aksworkshop-804]
REGION_NAME=[eastus]
VERSION=[1.16.7]
SUBNET-ID=[/subscriptions/{subidhere}/resourceGroups/makeakswork-rg/providers/Microsoft.Network/virtualNetworks/aks-vnet/subnets/aks-subnet]
Sometimes you can pass multiple strings separated by a space and it will be interpreted as \n.
For example when using a shell script for multi-line notifcations:
#!/bin/bash
notify-send 'notification success' 'another line' 'time now '`date +"%s"`
With jq:
$ jq -nr '"Hello,\nWorld"'
Hello,
World
Additional solution:
In cases, you have to echo a multiline of the long contents (such as code/ configurations)
For example:
A Bash script to generate codes/ configurations
echo -e,
printf might have some limitation
You can use some special char as a placeholder as a line break (such as ~) and replace it after the file was created using tr:
echo ${content} | tr '~' '\n' > $targetFile
It needs to invoke another program (tr) which should be fine, IMO.