Consider this dummy Windows batch script:
echo %1
Supposed just to echo to the terminal its first argument.
Assume its path in resp. Windows, Cygwin style is:
c:\test\win.bat
/cygdrive/c/test/win.bat
From Cygwin bash:
$ c:\test\win.bat "hello world"
"hello world"
So quotes correctly identify a single argument.
But now let us introduce spaces in path:
"c:\te st\win.bat"
/cygdrive/c/te\ st/win.bat
Then:
$ /cygdrive/c/te\ st/win.bat "hello world"
Gives:
"C:\te" is not recognized as an internal or external command, operable program or batch file.
The same happens with:
$ "/cygdrive/c/te st/win.bat" "hello world"
It should be noted this:
$ /cygdrive/c/te\ st/win.bat "hello"
hello
That is hello is now passed to win.bat unquoted (and with "/cygdrive/c/te st/win.bat" "hello" either).
How can I have spaces both in the path and the argument?
$ echo "echo %~1" > /cygdrive/c/te\ st/win.bat
$ cat /cygdrive/c/te\ st/win.bat
echo %~1
$ cmd /c $(echo "c:\te st\win.bat"| sed 's/ /^ /g') "aaa bbb"
C:\Users\Me>echo aaa bbb
aaa bbb
When we use shell expansion, it gives all the expanded word in one line. For example:
#!/bin/bash
data="Hello\ {World,Rafi}"
eval echo $data
This produces the following output:
Hello World Hello Rafi
Is it possible to output each line on a separate line like this?
Hello World
Hello Rafi
If I understand you right, you want to generate multiple words using brace expansion ({...}), then print each word on a separate line.
If you don't absolutely have to store "Hello\ {World,Rafi}" in a variable, you can do this with printf shell-builtin
printf "%s\n" "Hello "{Rafi,World}
Some explanation:
The format string (here: %s\n) is reused until all the arguments to printf is used up (Reference).
%s\n consumes 1 argument
"Hello "{Rafi,World} returns 2 words/arguments i.e. "Hello Rafi" and "Hello World"
So, this printf command is equivalent to
printf "%s\n%s\n" "Hello Rafi" "Hello World"
except you don't have to type all that up.
#!/bin/bash
data="Hello\ {World'\n',Rafi'\n',Kamal'\n'}"
eval echo -e "$data"
echo -e will evaluate newline characters.
Same as Antarus'a answer, except that echo has "-n". From http://unixhelp.ed.ac.uk/CGI/man-cgi?echo
-n do not output the trailing newline
#!/bin/bash
data="Hello\ {World,Rafi}'\n'"
eval echo -n -e "$data"
Actually, your problem is not the expansion but the echo command. Depending on your system, you might get what you want by
#!/bin/bash
data="Hello\ {World\\\\n,Rafi}"
eval echo -e "$data"
It is different solution but a very clean one.
#!/bin/bash
names="World Rafi"
for name in $names
do
echo Hello $name
done
Instead of using eval (which is dangerous and really not a good practice — see my comment in your post), another strategy would be to use an array. The following will do exactly what you want, in a clean and safe way:
data=( "Hello "{World,Rafi} )
printf "%s\n" "${data[#]}"
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 need to write a bash script that prints out its command line arguments in sorted order, one per line.
I wrote this script and it works fine, but is there any other way? Especially without outputting it to a file and sorting.
#!/bin/bash
for var in $*
do
echo $var >> file
done
sort file
rm file
Test run of the program:
$ script hello goodbye zzz aa
aa
goodbye
hello
zzz
You can pipe the for-loop to sort, or just do
printf '%s\n' "$#" | sort
#!/bin/bash
for var in "$#"; do
echo "$var"
done | sort
You want to use $# in quotes (instead of $*) to accommodate arguments with spaces, such as
script hello "goodbye, cruel world"
The pipe gets rid of the need for a temporary file.
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.