How to escape single quote using awk in a bash script

Question

This is what I have so far, tried multiple ways but can\'t get it just right. My goal is to sanitize the input to prevent problems while inputting to mysql from text file

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Answer 1

What you want is:

awk '{gsub(/\047/,".")}1' file

See http://awk.freeshell.org/PrintASingleQuote.

Answer 2

You have used wrong quoting:

awk '{gsub(/'"'"'/, "."); print}'

Answer 3

Since you requested "how to escape single quote using awk" here's a solution using awk:

awk '{ gsub("\x27", ".");print $0}'

where \x27 is the escape sequence representation of the hexadecimal value 27 (a single quote).

For a list of all escape sequences see https://www.gnu.org/software/gawk/manual/html_node/Escape-Sequences.html

Answer 4

I'm not sure if I got the problem correctly. If you want to replace all occurrences of a single quote by a dot, use tr:

tr "'" "." < file.txt > sanitized.txt

If you want to escape the single quote with a backslash use sed like this:

sed "s/'/\\\'/g" file.txt > sanitized.txt

Note: Please take the advice from CharlesDuffy seriously. This is far from a stable and safe solution to escape values for an SQL import.