I have a variable in a shell script in which I'd like to format the data. The variable stores new data during every iteration of a loop. Each time the new data is stored, I'd like to insert a new line character. Here is how I'm trying to store the data into the variable.
VARIABLE="$VARIABLE '\n' SomeData"
Unfortunately, the output includes the literal '\n' Any help would be appreciative.
Try $'\n':
VAR=a
VAR="$VAR"$'\n'b
echo "$VAR"
gives me
a
b
VAR=$VAR$'\n'b must be surrounded by double quotes. Which results in a literal '\n'.
VAR="$VAR"$'\n'"some data" has $VAR in double-quotes, \n in $'' so it'll get interpreted as a newline, and then some data back in double-quotes. As long as there's no space between them, they'll all be concatenated together and the result assigned to VAR.
VAR="$VAR"$'\n'"Audit Source: $B Audit Path: $C" The spaces for formatting was messing things up. New Code:VAR="$VAR"$'\n'"Audit Source:$B"$'\t'"Audit Path:$C" No spaces here! Anyone know why the spacing causes issues? Even though the Original Code has everything surrounded by "?A common technique is:
nl='
'
VARIABLE="PreviousData"
VARIABLE="$VARIABLE${nl}SomeData"
echo "$VARIABLE"
PreviousData
SomeData
Also common, to prevent inadvertently having your string start with a newline:
VARIABLE="$VARIABLE${VARIABLE:+$nl}SomeData"
(The expression ${VARIABLE:+$nl} will expand to a newline if and only if VARIABLE is set and non-empty.)
bash may be OK, most shells will not be OK.SomeData to VARIABLES Such as VAR="$VARIABLE${nl}SomeData" In this answer, there are two separate variables (VAR and VARIABLE I'm trying to append VARIABLE to VARIABLEVAR="one"
VAR="$VAR.\n.two"
echo -e $VAR
one.
.two
-e makes echo treat escaped characters as special characters, without -e you would see entire value on one line with "\n" in the middleOther than $'\n' you can use printf also like this:
VARIABLE="Foo Bar"
VARIABLE=$(printf "${VARIABLE}\nSomeData")
echo "$VARIABLE"
OUTPUT:
Foo Bar
SomeData
I had a problem with all the other solutions: when using a # followed by SPACE (quite common when writing in Markdown) both would get split onto a new line.
So, another way of doing it would involve using single quotes so that the "\n" get rendered.
FOO=$'# Markdown Title #\n'
BAR=$'Be *brave* and **bold**.'
FOOBAR="$FOO$BAR"
echo "$FOOBAR"
Output:
# Markdown Title #
Be *brave* and **bold**.
Single quote All special characters between these quotes lose their special meaning.
https://www.tutorialspoint.com/unix/unix-quoting-mechanisms.htm
So the syntax you use does something different that you want to achieve.
This is what you need:
The $'\X' construct makes the -e option in echo unnecessary.
https://linux.die.net/abs-guide/escapingsection.html
echo -e "something\nsomething"
or
echo "something"$'\n'"something"
It's a lot simpler than you think:
VARIABLE="$VARIABLE
SomeData"
Building upon the first two solutions, I'd do like shown below. Concatenating strings with the '+=' operator, somehow looks clearer to me.
Also rememeber to use printf as opposed to echo, you will save yourself so much trouble
sometext="This is the first line"
sometext+=$'\n\n'
sometext+="This is the second line AFTER the inserted new lines"
printf '%s' "${sometext}"
Outputs:
This is the first line
This is the third line AFTER the inserted new line
Your problem is in the echo command, in ash you have to use the option -e to expand special characters. This should work for you:
VAR="First line"
VAR="$VAR\nSecond line"
echo -e $VAR
This outputs
First line
Second line
