How can I escape double quotes inside a double string in Bash?
For example, in my shell script
#!/bin/bash
dbload="load data local infile \"'gfpoint.csv'\" into table $dbtable FIELDS TERMINATED BY ',' ENCLOSED BY '\"' LINES TERMINATED BY \"'\n'\" IGNORE 1 LINES"
I can't get the ENCLOSED BY '\"'
with double quote to escape correctly. I can't use single quotes for my variable, because I want to use variable $dbtable
.
$dbtable
, there's a risk. This would be very uncommon, though, as end users don't typically SSH into a machine to load their data.
Use a backslash:
echo "\"" # Prints one " character.
A simple example of escaping quotes in the shell:
$ echo 'abc'\''abc'
abc'abc
$ echo "abc"\""abc"
abc"abc
It's done by finishing an already-opened one ('
), placing the escaped one (\'
), and then opening another one ('
).
Alternatively:
$ echo 'abc'"'"'abc'
abc'abc
$ echo "abc"'"'"abc"
abc"abc
It's done by finishing already opened one ('
), placing a quote in another quote ("'"
), and then opening another one ('
).
More examples: Escaping single-quotes within single-quoted strings
echo "abc\"abc"
is sufficient to produce abc"abc
as in Peter answer.
Keep in mind that you can avoid escaping by using ASCII codes of the characters you need to echo.
Example:
echo -e "This is \x22\x27\x22\x27\x22text\x22\x27\x22\x27\x22"
This is "'"'"text"'"'"
\x22
is the ASCII code (in hex) for double quotes and \x27
for single quotes. Similarly you can echo any character.
I suppose if we try to echo the above string with backslashes, we will need a messy two rows backslashed echo... :)
For variable assignment this is the equivalent:
a=$'This is \x22text\x22'
echo "$a"
# Output:
This is "text"
If the variable is already set by another program, you can still apply double/single quotes with sed or similar tools.
Example:
b="Just another text here"
echo "$b"
Just another text here
sed 's/text/"'\0'"/' <<<"$b" #\0 is a special sed operator
Just another "0" here #this is not what i wanted to be
sed 's/text/\x22\x27\0\x27\x22/' <<<"$b"
Just another "'text'" here #now we are talking. You would normally need a dozen of backslashes to achieve the same result in the normal way.
echo 'export PS1='\[\033[00;31m\]${?##0}$([ $? -ne 0 ] && echo \x22 \x22)\[\033[00;32m\]\u\[\033[00m\]@\[\033[00;36m\]\h\[\033[00m\][\[\033[01;33m\]\d \t\[\033[00m\]] \[\033[01;34m\]\w\n\[\033[00m\]$( [ ${EUID} -ne 0 ] && echo \x22$\x22 || echo \x22#\x22 ) '' >> ~/.profile
Bash allows you to place strings adjacently, and they'll just end up being glued together.
So this:
echo "Hello"', world!'
produces
Hello, world!
The trick is to alternate between single and double-quoted strings as required. Unfortunately, it quickly gets very messy. For example:
echo "I like to use" '"double quotes"' "sometimes"
produces
I like to use "double quotes" sometimes
In your example, I would do it something like this:
dbtable=example
dbload='load data local infile "'"'gfpoint.csv'"'" into '"table $dbtable FIELDS TERMINATED BY ',' ENCLOSED BY '"'"'"' LINES "'TERMINATED BY "'"'\n'"'" IGNORE 1 LINES'
echo $dbload
which produces the following output:
load data local infile "'gfpoint.csv'" into table example FIELDS TERMINATED BY ',' ENCLOSED BY '"' LINES TERMINATED BY "'\n'" IGNORE 1 LINES
It's difficult to see what's going on here, but I can annotate it using Unicode quotes. The following won't work in Bash – it's just for illustration:
dbload=
‘load data local infile "
’“'gfpoint.csv'
”‘" into
’“table $dbtable FIELDS TERMINATED BY ',' ENCLOSED BY '
”‘"
’“' LINES
”‘TERMINATED BY "
’“'\n'
”‘" IGNORE 1 LINES
’
The quotes like “ ‘ ’ ” in the above will be interpreted by bash. The quotes like " '
will end up in the resulting variable.
If I give the same treatment to the earlier example, it looks like this:
echo
“I like to use
”
‘"double quotes"
’
“sometimes
”
echo
example, that I've not bothered to explain, is that here I'm actually giving echo three separate arguments, because the strings aren't touching. but the echo
command prints all three, separated by spaces, so it sort of does what you'd expect. The difference is, if you type echo a b c
it will output a b c
, whereas if you type echo 'a b c'
it will print a b c
.
Store the double quote character in a variable:
dqt='"'
echo "Double quotes ${dqt}X${dqt} inside a double quoted string"
Output:
Double quotes "X" inside a double quoted string
Check out printf...
#!/bin/bash
mystr="say \"hi\""
Without using printf
echo -e $mystr
Output: say "hi"
Using printf
echo -e $(printf '%q' $mystr)
Output: say \"hi\"
printf
escapes more characters as well, such as '
, (
and )
printf %q
generates strings ready for eval
, not formatted for echo -e
.
printf
with a useless use of echo
. Both your examples have broken quoting. The proper fix is to double-quote the variable.
Make use of $"string".
In this example, it would be,
dbload=$"load data local infile \"'gfpoint.csv'\" into table $dbtable FIELDS TERMINATED BY ',' ENCLOSED BY '\"' LINES TERMINATED BY \"'\n'\" IGNORE 1 LINES"
Note (from the man page):
A double-quoted string preceded by a dollar sign ($"string") will cause the string to be translated according to the current locale. If the current locale is C or POSIX, the dollar sign is ignored. If the string is translated and replaced, the replacement is double-quoted.
For use with variables that might contain spaces in you Bash script, use triple quotes inside the main quote, e.g.:
[ "$(date -r """$touchfile""" +%Y%m%d)" -eq "$(date +%Y%m%d)" ]
Add "\"
before double quote to escape it, instead of \
#! /bin/csh -f
set dbtable = balabala
set dbload = "load data local infile "\""'gfpoint.csv'"\"" into table $dbtable FIELDS TERMINATED BY ',' ENCLOSED BY '"\""' LINES TERMINATED BY "\""'\n'"\"" IGNORE 1 LINES"
echo $dbload
# load data local infile "'gfpoint.csv'" into table balabala FIELDS TERMINATED BY ',' ENCLOSED BY '"' LINES TERMINATED BY "''" IGNORE 1 LINES
csh
answer to a bash
question? The two are completely distinct and incompatible.
Success story sharing
x=ls; if [ -f "$(which "\""$x"\"")" ]; then echo exists; else echo broken; fi;
gives broken whereas... [ -f "$(which $x)" ]; ...
or... [ -f $(which "$x") ]; ...
work just fine. Issues would arise when either$x
or the result of$(which "$x")
gives anything with a space or other special character. A workaround is using a variable to hold the result ofwhich
, but is bash really incapable of escaping a quote or am I doing something wrong?grep -oh "\"\""$counter"\""\w*"
as part of a bash syntax where in$counter
is a variable. it doesn't like it any thoughts