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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.
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Within double quotes, backslashes that are followed by one of these characters are removed. Backslashes preceding characters without a special meaning are left unmodified. A double quote may be quoted within double quotes by preceding it with a backslash.
If you need to use the double quote inside the string, you can use the backslash character. Notice how the backslash in the second line is used to escape the double quote characters. And the single quote can be used without a backslash.
You can put a backslash character followed by a quote ( \" or \' ). This is called an escape sequence and Python will remove the backslash, and put just the quote in the string. Here is an example. The backslashes protect the quotes, but are not printed.
Simply type echo 'text "hey"' - wrap the entire string in single quotes (').
Use a backslash:
echo "\"" # Prints one " character. If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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