How can I escape a double quote inside double quotes?
BashQuotesBash Problem Overview
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.
Bash Solutions
Solution 1 - Bash
Use a backslash:
echo "\"" # Prints one " character.
Solution 2 - Bash
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
Solution 3 - Bash
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.
Solution 4 - Bash
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”
Solution 5 - Bash
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
Solution 6 - Bash
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"
Solution 7 - Bash
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.
Solution 8 - Bash
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)" ]
Solution 9 - Bash
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